How to Prepare for Quantitative Aptitude for CAT (2014)

Block I: Numbers

...BACK TO SCHOOL

• Chapters in this Block: Number Systems and Progressions
• Block Importance – 20–25%

The importance of this block can be gauged from the table below:

As you can see from the table above, doing well in this block alone could give you a definite edge and take you a long way to qualifying the QA section. Although the online CAT has significantly varied the relative importance to this block, the importance of this block remains high. Besides, there is a good chance that once the IIMs get their act together in the context of the online CAT and its question databases—the pre eminence of this block of chapters might return.

Hence, understanding the concepts involved in these chapters properly and strengthening your problem solving experience could go a long way towards a good score.

Before we move into the individual chapters of this block, let us first organise our thinking by looking at the core concepts that we had learnt in school with respect to these chapters.

Pre-assessment Test

This test consists of 25 questions based on the chapters of BLOCK I (Number Systems and Progressions). Do your best in trying to solve each question.

The time limit to be followed for this test is 30 minutes. However, after the 30 minutes is over continue solving till you have spent enough time and paid sufficient attention to each question. After you finish thinking about each and every question of the test, check your scores. Then go through the SCORE INTERPRETATION ALGORITHM given at the end of the test to understand the way in which you need to approach the chapters inside this block.

Solutions

1.The only value that will satisfy will be 2.

2.¼ + 1/6 will give you 5/12.

3.The possible values are 1, 2, 3, 4, 6 and 12. (i.e. the factors of 12)

4.k will be a number of the form 7n + 3. Hence, if you take the value of n as 9, k + 2n will become 7n + 3 + 18 = 7n + 21. This number will be divisible by 7. The numbers 3, 5 and 7 do not provide us with this solution.

5.2⁷³ – 2⁷² – 2⁷¹ = 2⁷¹ (2² – 2 – 1) = 2⁷¹(1). Hence option (c) is correct.

6.Solve through options.

7.The closest value will be option (b), since the percentage change will be lowest when the largest number is reduced by one.

8.This is a property of prime numbers greater than 5.

9.He could have dealt a total of 40 cards, in which case Mushtaq would get 9 cards. On getting one card from Mushtaq, the ratio would become 4:1, while on giving away one card to Mushtaq, the ratio would become 3:1.

10.Looking at the options you realise that the correct answer should be a multiple of 25 and 50 both. The option that satisfies the condition of increasing the number of boxes by 3 is 150. (This is found through trial and error.)

11.Trial and error gives you option 3 as the correct answer.

12.Given that the number must have a 57 in it and should be even at the same time, the only numbers possible are 570, 572, 574, 576 and 578. Also, if there is no 5 in the number, you will get 360 more numbers.

13.203 becomes → 2.3² + 0 + 3.3⁰ = 21 → 2.3¹ + 1.3⁰ = 7. Hence, clearly two steps are required.

14.Trial and error will give option (b) as the correct answer, since 13 × 62 = 26 × 31

15.The solutions are defined as the sum of digits of the answer. Hence, 10 is correct.

16.There are no common factors between 55 and 124. Hence the answer should be 124.

17.At k = –1, the two equations become inconsistent with respect to each other and there will then be no solution to the system of equations.

18.Try with 5, 7, 11. In each case the remainder is 6.

19.Trial and error gives you the answer as 3 Option (b) is correct.

20.Picking up 4 coins will ensure that he wins the game.

21.Option (a) is correct (since the average of any two numbers lies between the numbers.

22.8/2 + 5/2 = 6.5.

23.The answer will be 56 × 12 = 672 → 675. Hence, ` 6.75.

24.The given condition is satisfied only for 24.

25.The answer will be given by 8 + 2 = 10.

(This logic is explained in the Number Systems chapter)

SCORE INTERPRETATION ALGORITHM FOR PRE-ASSESSMENT TEST OF BLOCK I

(Use a similar process for Blocks II to VI on the basis of your performance).

If You Scored: < 7: (In Unlimited Time)

Step One

Go through the block one Back to School Section carefully. Grasp each of the concepts explained in that part carefully. In fact I would recommend that you go back to your Mathematics school books (ICSE/ CBSE) Class 8, 9 and 10 if you feel you need it.

Step Two

Move into the first chapter of the block, viz, Number Systems. When you do so, concentrate on clearly understanding each of the concepts explained in the chapter theory.

Then move onto the LOD 1 exercises. These exercises will provide you with the first level of challenge. Try to solve each and every question provided under LOD 1 of Number systems. While doing so do not think about the time requirement. Once you finish solving LOD 1, revise the questions and their solution processes.

Step Three

After finishing LOD 1 of number systems, move into Chapter 2 of this block—Progressions and repeat the process, viz: chapter theory comprehensively followed by solving LOD 1 questions.

Step Four

Go to the first and second review tests given at the end of the block and solve them. While doing so, first look at the score you get within the mentioned time limit. Then continue to solve the test further without a time limit and try to evaluate the improvement in your unlimited time score.

In case the growth in your score is not significant, go back to the theory of each chapter and review each of the LOD 1 questions for both the chapters.

Step Five

Move to LOD 2 and repeat the process that you followed in LOD 1—first in the chapter of Number Systems, then in the chapter on Progressions. Concentrate on understanding each and every question and its underlying concept.

Step Six

Go to the third to fifth review tests given at the end of the block and solve them. Again, while doing so measure your score within the provided time limit first and then continue to solve the test further without a time limit and try to evaluate the improvement that you have had in your score.

Step Seven

Move to LOD 3 only after you have solved and understood each of the questions in LOD 1 and LOD 2. Repeat the process that you followed in LOD 1—first in the chapter of Number Systems, then with the Chapter on Progressions.

If You Scored: 7–15 (In Unlimited Time)

Although you are better than the person following the instructions above, obviously there is a lot of scope for the development of your score. You will need to work both on your concepts as well as speed. Initially emphasize more on the concept development aspect of your preparations, then move your emphasis onto speed development. The following process is recommended for you:

Step One

Go through the block one Back to School Section carefully. Revise each of the concepts explained in that part. Going through your VIIIth, IXth and Xth standard books will be an optional exercise for you. It will be recommended in case you scored in single digits, while if your score is in two digits, I leave the choice to you.

Step Two

Move into the first chapter of the block. Viz Number Systems. When you do so, concentrate on clearly understanding each of the concepts explained in the chapter theory.

Then move onto the LOD 1 exercises. These exercises will provide you with the first level of challenge. Try to solve each and every question provided under LOD 1 of Number Systems. Once you finish solving LOD 1, revise the questions and their solution processes.

Step Three

After finishing LOD 1 of number systems, move into Chapter 2 of this block—Progressions and repeat the process, viz: Chapter theory comprehensively followed by solving LOD 1 questions.

Step Four

Go to the first and second review tests given at the end of the block and solve them. While doing so, first look at the score you get within the time limit mentioned. Then continue to solve the test further without a time limit and try to evaluate the improvement in your score.

Step Five

Move to LOD 2 and repeat the process that you followed in LOD 1—first with the chapter on Number Systems, then with the chapter on Progressions.

Step Six

Go to the third to fifth review tests given at the end of the block and solve it. Again while doing so measure your score within the provided time limit first and then continue to solve the test further without a time limit and try to evaluate the improvement that you have had in your score.

In case the growth in your score is not significant, go back to the theory of both the chapters and re-solve LOD 1 and LOD 2 of both the chapters. While doing so concentrate more on the LOD 2 questions.

Step Seven

Move to LOD 3 and repeat the process that you followed in LOD 1—first with the chapter on Number Systems, then with the Chapter on Progressions.

If You Scored 15+ (In Unlimited Time)

Obviously you are much better than the first two categories of students. Hence unlike them, your focus should be on developing your speed by picking up the shorter processes explained in this book. Besides, you might also need to pick up concepts that might be hazy in your mind. The following process of development is recommended for you:

Step One

Quickly review the concepts given in the block one Back to School Section. Only go deeper into a concept in case you find it new. Going back to school level books is not required for you.

Step Two

Move into the first chapter of the block: Number Systems. Go through the theory explained there carefully. Concentrate specifically on clearly understanding the concepts which are new to you. Work out the short cuts and in fact try to expand your thinking by trying to think of alternative (and expanded) lines of questioning with respect to the concept you are studying.

Then move onto the LOD 1 exercises. Solve each and every question provided under LOD 1 of Number Systems. While doing so, try to think of variations that you can visualize in the same questions and how you would handle them.

Step Three

After finishing LOD 1 of number systems, move into Chapter 2 of this block, (Progressions) and repeat the process, viz: Chapter theory with emphasis on picking up things that you are unaware of, followed by solving LOD 1 questions and thinking about their possible variations.

Step Four

Move to LOD 2 and repeat the process that you followed in LOD 1—first in the chapter of Number Systems, then with the Chapter on Progressions.

Step Five

Go to the first to fifth review tests—given at the end of the block and solve it. While doing so, first look at the score you get within the time limit mentioned. Then continue to solve the test further without a time limit and try to evaluate the improvement in your score.

Step Six

Move to LOD 3 and repeat the process that you followed in LOD 1—first in the chapter of Number Systems, then with the Chapter on Progressions.

THE NUMBER LINE

Core Concepts

1. The concept of the number line is one of the most crucial concepts in Basic Numeracy.

The number line is a line that starts from zero and goes towards positive infinity when it moves to the right and towards negative infinity when it moves to the left.

The difference between the values of any two points on the number line also gives the distance between the points.

Thus, for example if we look at the distance between the points + 3 and – 2, it will be given by their difference. 3 – (– 2) = 3 + 2 = 5.

1. Types of numbers—

We will be looking at the types of numbers in detail again when we go into the chapter of number systems. Let us first work out in our minds the various types of numbers. While doing so do not fail to notice that most of these number types occur in pairs (i.e. the definition of one of them, defines the other automatically).

Note here that the number line is one of the most critical concepts in understanding and grasping numeracy and indeed mathematics.

Multiples on the Number Line

All tables and multiples of every number can be visualised on the number line. Thus, multiples of 7 on the number line would be seen as follows:

In order to visualize this you can imagine a frog jumping consistently 7 units every time. If it lands on –14, the next landing would be on –7, then on 0, then 7 and finally at 14. This is how you can visualise the table of 7 (in mathematical terms we can also refer to this as 7n – meaning the set of numbers which are multiples of 7 or in other words the set of numbers which are divisible by 7).

You can similarly visualise 5n, 4n, 8n and so on—practically all tables on the number line as above.

What do We Understand by 7n + 1 and Other such Notations & the Equivalence of 7n + 1 and 7n – 6

Look at the following figure closely:

Our 7n frog is made to first land on –13 and then asked to keep doing it’s stuff (jumping 7 to the right everytime). What is the result? It lands on –6, + 1, 8, 15 and it’s next landing would be on 22, 29 and so forth. These numbers cannot be described as 7n, but rather they all have a single property which is constant for all numbers.

They can be described as: “One more than a multiple of 7” and in mathematical terms such numbers are also called as 7n + 1. Alternately these numbers also have the property that they are “6 less than multiples of 7” and in mathematical terms such numbers can also be called as 7n – 6.

That is why in mathematics, we say that the set of numbers represented by 7n + 1 is the same as the set of numbers represented by 7n – 6.

The Implication in Terms of Remainders

This concept can also be talked about in the context of remainders.

When a number which can be described as 7n + 1 or 7n – 6 (like the numbers 8, 15, –6, –13, –20, –27, –34, –41…) is divided by 7, the remainder in every case is seen to be 1. For some people reconciling the fact that the remainder when –27 is divided by 7 the remainder is 1, seems difficult on the surface. Note that this needs to be done because about remainders we should know that remainders are always non-negative.

However, the following thinking would give you the remainder in every case:

27/7, remainder is 6.

–27/7, remainder is –6. In the context of dividing by 7, a remainder of –6 means a remainder of 7 – 6 = 1.

Let us look at another example:

What is the remainder when –29 is divided by 8.

First reaction 29/8 Æ remainder 5, –29/8 Æ remainder –5, hence actual remainder is 8 – 5 = 3.

The student is advised to practice more such situations and get comfortable in converting positive remainders to negative remainders and vice versa.

Even and Odd Numbers

The meaning of 2n and 2n + 1: 2n means a number which is a multiple of the number 2. Since, this can be visualised as a frog starting from the origin and jumping 2 units to the right in every jump, you can also say that this frog represents 2n.

(Note: Multiples of 2, are even numbers. Hence, 2n is also used to denote even numbers.)

So, what does 2n + 1 mean?

Well, simply put, if you place the above frog on the point represented by the number 1 on the number line then the frog will reach points such as 3, 5, 7, 9, 11 …..and so on. This essentially means that the points the frog now reaches are displaced by 1 unit to the right of the 2n frog. In mathematical terms, this is represented as 2n + 1.

In other words, 2n + 1 also represents numbers which leave a remainder of 1, when divided by 2. (Note: This is also the definition of an odd number. Hence, in Mathematics (2n + 1) is used to denote an odd number. Also note that taken together 2n and 2n + 1 denote the entire set of integers. i.e. all integers from – • to + • on the number line can be denoted by either 2n or 2n + 1. This happens because when we divide any integer by 2, there are only two results possible with respect to the remainder obtained, viz: A remainder of zero (2n) or a remainder of one (2n + 1).

This concept can be expanded to represent integers with respect to any number. Thus, in terms of 3, we can only have three types of integers 3n, 3n + 1 or 3n + 2 (depending on whether the integer leaves a remainder 0, 1 or 2 respectively when divided by 3.) Similarly, with respect to 4, we have 4 possibilities—4n, 4n + 1, 4n + 2 or 4n + 3.

Needless to say, from these representations above, the representations 2n and 2n + 1 (which can also be represented as 2n – 1) have great significance in Mathematics as they represent even and odd numbers respectively. Similarly, we use the concept of 4n to check whether a year is a leap year or not.

These representations can be seen on the number line as follows:

Representation of 2n and 2n + 1:

Representation of 3n, 3n + 1 and 3n + 2:

Representation of 4n, 4n + 1, 4n + 2 and 4n + 3:

One Particular Number can be a Multiple of more than 1 Number—The Concept of Common Multiples

Of course one of the things you should notice as you go through the above discussion is that individual numbers can indeed be multiples of more than 1 number—and actually often are.

Thus, for instance the number 14 is a multiple of both 7 and 2—hence 14 can be called as a common multiple of 2 and 7. Obviously, I think you can visualise more such numbers which can be classified as common multiples of 2 and 7?? 28,42,56 and in fact the list is infinite—i.e. the numbers never end. Thus, the common multiples of 2 and 7 can be represented by the infinite set:

{14,28,42,56,70…………1400……..14000……140000……and so on}

The LCM and It’s Significance

From the above list, the number 14 (which is the lowest number in the set of Common multiples of 2 and 7) has a lot of significance in Mathematics. It represents what is commonly known as the Least Common Multiple (LCM)of 2 and 7. It is the first number which is a multiple of both 2 and 7 and there are a variety of questions in numeracy which you would come across—not only when you solve questions based on number systems (where the LCM has it’s dedicated set of questions), but applications of the LCM are seen even in chapters like Time, Speed Distance, Time and Work etc.

So what did school teach us about the process of finding LCM?

Before you start to review/ relearn that process you first need to know about prime factors of a number.

I hope at this point you recognize the difference between finding factors of a number and prime factors of a number.

Simply put, finding factors of a number means finding the divisors of the number. Thus, for instance the factors of the number 80 would be the numbers 1, 2, 4, 5, 8, 10, 16, 20, 40 and 80. On the other hand finding the prime factors of the same number would mean writing the number 80 as 2⁴ × 5¹. This form of writing the number is also called the STANDARD FORM or the CANONICAL FORM of the number.

The school process of writing down the standard form of a number:

Now this is something I would think most of you would remember and recognize:

Finding the prime factors of the number 80

The prime factors of the number 80 are: 2 × 2 × 2 × 2 × 5 = 2⁴ × 5¹.

Exercise for Self-practice

Write down the standard form of the following numbers.

Finding the LCM of two or more numbers: The school process

Step 1: Write down the prime factor form of all the numbers;

Let us say that you have 3 numbers whose standard forms are:

2⁴ × 3² × 5³ × 7¹

2³ × 3¹ × 5² × 11²

2⁴ × 3² × 5¹ × 13¹

To write down the LCM of these numbers write down all the prime numbers and multiply them with their highest available powers. The resultant number would be the LCM of these numbers.

Thus, in the above case the LCM would be:

2⁴ × 3² × 5³ × 7¹ × 11² × 13¹

Divisors (Factors) of a number

As we already mentioned, finding the factors or divisors of a number are one and the same thing. In order to find factors of a number, the key is to spot the factors below the square root of the number. Once you have found them, the factors above the square root would be automatically seen. Consider this for factors of the number 80:

Once you can visualise the list on the left, the factors on the right would be seen automatically.

Common Divisors Between 2 Numbers

List of Common Divisors

When we write down the factors of two numbers, we can look for the common elements within the two lists.

For instance, the factors of 80 are (1, 2, 4, 5, 8, 10, 16, 20, 40 and 80) while the factors of 144 are (1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72 and 144)

If you observe the two lists closely you will get the following list of common factors or common divisors between the two:

List of common factors of 80 and 144: (1, 2, 4, 8, 16)

The number 16 being the highest in this list of common factors (or divisors) is also called the Highest Common Factor or the Greatest Common Divisor. In short it is denoted as HCF or GCD. It has a lot of significance in terms of quantitative thinking—as the HCF is used in a multitude of problems and hence being able to spot the HCF of two or more numbers is one of the critical operations in Mathematics. You would be continuously seeing applications of the HCF in problems in the chapter of Number systems and right through the chapters of arithmetic (defined as word based problems in this book).

Rounding off and its use for approximation

One of the important things that we learnt in school was the use of approximation in order to calculate.

Thus, 72 × 53 can be approximated to 70 × 50 and hence seen as 3500.

Similarly, the addition of 48 + 53 + 61 + 89 can be taken as 50 + 50 + 60 + 90 = 250

Number Types You Should Know

Integers and Decimals: All numbers that do not have a decimal in them are called integers. Thus, –3, –17, + 4, + 13, + 1473, 0 etc are all integers.

Obviously, decimal numbers are numbers which have a decimal value attached to them. Thus, 1.3, 14.76, – 12.24 etc are all decimal numbers, since they have certain values after the decimal point.

Before we move ahead, let us pause a brief while, to further understand decimals. As you shall see, the concept of decimals is closely related to the concept of division and divisibility. Suppose, I have 4 pieces of bread which I want to divide equally between two people. It is easy for me to do this, since I can give two whole pieces to each of them.

However, if we alter the situation in such a way, that I now have 5 pieces of bread to distribute equally amongst 2 people. What do I do?

I give two whole pieces each, to each of them. The 5th piece has to be divided equally between the two. I can no longer do this, without in some way breaking the 5th piece into 2 parts. This is the elementary situation that gives rise to the need for decimals in mathematics.

Going back to the situation above, my only option is to divide the 5th piece into two equal parts (which in quants are called as halves).

This concept has huge implications for problem solving especially once you recognise that a half (i.e. a ‘.5’ in the decimal) only comes when you divide a whole into two parts.

Thus, in fact, all standards decimals emerge out of certain fixed divisors.

Hence, for example, the divisor 2, gives rise to the decimal · 5.

Similarly the divisor 3 gives rise to the decimals ·33333 and ·66666, etc.

Prime numbers and Composite numbers Amongst natural numbers, there are three broad divisions–

Unity It is representative of the number 1.

Prime numbers These are numbers which have no divisors/ factors apart from 1 and itself.

Composite numbers On the other hand, are numbers, which have at least one more divisor apart from 1 and itself.

All composite numbers have the property that they can be written as a product of their prime factors.

Thus, for instance, the number 40 can be represented as: 40 = 2 × 2 × 2 × 5 or 40 = 2³ × 5¹

This form of writing is called as the standard form of the composite number.

The difference between Rational and Irrational numbers: This difference is one of the critical but unfortunately one of the less well understood differences in elementary Mathematics.

The definition of Rational numbers: Numbers which can be expressed in the form p/q where q π 0 are called rational numbers.

Obviously, numbers which cannot be represented in the form p/q are called as irrational numbers.

However, one of the less well understood issues in this regard is what does this mean?

The difference becomes clear when the values of decimals are examined in details:

Consider the following numbers.

(1)4.2,

(2)4.333….,

(3)4.1472576345…….

What is the difference between the decimal values of the three numbers above?

To put it simply, the first number has what can be described as a finite decimal value. Such numbers can be expressed in the form p/q easily. Since 4.2 can be first written as 42/10 and then converted to 21/5.

Similarly, numbers like 4.5732 can be represented as 45732/10000. Thus, numbers having a finite terminating decimal value are rational.

Now, let us consider the decimal value: 4.3333……..

Such decimal values will continue endlessly, i.e. they have no end. Hence, they are called infinite decimals (or non-terminating decimals).

But, we can easily see that the number 4.333.... can be represented as 13/3. Hence, this number is also rational. In fact, all numbers which have infinite decimal values, but have any recurring form within them can be represented in the p/q form.

For example the value of the number: 1.14814814814…. is 93/81.

(What I mean to say is that whenever you have any recurring decimal number, even if the value of ‘q’ might not be obvious, but it will always exist.)

Thus, we can conclude that all numbers whose decimal values are infinite (non-terminating) but which have a recurring pattern within them are rational numbers.

This leaves us with the third kind of decimal values, viz. Infinite non-recurring decimal values. These decimals neither have a recurring pattern, nor do they have an end—they go on endlessly. For such numbers it is not possible to find the value of a denominator ‘q’ which can be used in order to represent them as p/q. Hence, such numbers are called as irrational numbers.

In day-to-day mathematics, we come across numbers like , , , p, e, etc. which are irrational numbers since they do not have a p/q representation.

An Important Tip:

Rational and Irrational numbers do not mix.: This means that in case you get a situation where an irrational number has appeared while solving a question, it will remain till the end of the solution. It can only be removed from the solution if it is multiplied or divided by the same irrational number.

Consider an example: The area of an equilateral triangle is given by the formula (/4) × a² (where a is the side of the equilateral triangle). Since, is an irrational number, it remains in the answer till the end. Hence, the area of an equilateral triangle will always have a as part of the answer.

Before we move ahead we need to understand one final thing about recurring decimals.

As I have already mentioned, recurring decimals have the property of being able to be represented in the p/q form. The question that arises is—Is there any process to convert a recurring decimal into a proper fraction?

Yes, there is. In fact, in order to understand how this operates, you first need to understand that there are two kinds of recurring decimals. The process for converting an infinite recurring decimal into a fraction basically varies for both of these types. Let’s look at these one by one.

Type 1—Pure recurring decimals: These are recurring decimals where the recurrence starts immediately after the decimal point.

For example:

0.5555… =

3.242424… =

5.362362… =

The process for converting these decimals to fractions can be illustrated as:

0.5555 = 5/9

3.242424 = 3 + (24/99)

5.362362 = 5 + (362/999)

A little bit of introspection will tell you that what we have done is nothing but to put down the recurring part of the decimal as it is and dividing it by a group of 9’s. Also the number of 9’s in this group equals the number of digits in the recurring part of the decimal.

Thus, in the second case, the fraction is derived by dividing 24 by 99. (24 being the recurring part of the decimal and 99 having 2 nines because the number of digits in 24 is 2.)

Similarly, 0.43576254357625…. =

Type 2—Impure recurring decimals: Unlike pure recurring decimals, in these decimals, the recurrence occurs after a certain number of digits in the decimal. The process to convert these into a fraction is also best illustrated by an example:

Consider the decimal 0.435424242

=

The fractional value of the same will be given by: (43542 – 435)/99000. This can be understood in two steps.

Step 1: Subtract the non-recurring initial part of the decimal (in this case, it is 435) from the number formed by writing down the starting digits of the decimal value upto the digit where the recurring decimals are written for the first time;

Expanding the meaning—

Note: For 0.435424242, subtract 435 from 43542

Step 2: The number thus obtained, has to be divided by a number formed as follows; Write down as many 9’s as the number of digits in the recurring part of the decimal. (in this case, since the recurring part ‘42’ has 2 digits, we write down 2 9’s.) These nines have to be followed by as many zeroes as the number of digits in the non recurring part of the decimal value. (In this case, the non recurring part of the decimal value is ‘435’. Since, 435 has 3 digits, attach three zeroes to the two nines to get the number to divide the result of the first step.)

Hence divide 43542 – 435 by 99000 to get the fraction.

Similarly, for 3.436213213 we get

Mixed Fractions

A mixed number is a whole number plus a fraction. Here are a few mixed numbers:

In order to convert a mixed fraction to a proper fraction you do the following conversion process.

= (1 × 2 + 1)/2 = 3/2

= (2 × 3 + 1)/3 = 7/3

Similiarly, = 5/4

= 16/7

= 29/5 and so on.

i.e. multiply the whole number part of the mixed number by the denominator of the fractional part and add the resultant to the numerator of the fractional part to get the numerator of the proper fraction. The denominator of the proper fraction would be the same as the denominator of the mixed fraction.

OPERATIONS ON NUMBERS

Exponents and Powers

Exponents, or powers, are an important part of math as they are necessary to indicate that a number is multiplied by itself for a given number of times.

When a number is multiplied by itself it gives the ‘square of the number’.

Thus, n × n = n² (for example 3 × 3 = 3²)

If the same number is multiplied by itself twice we get the cube of the number.

Thus, n × n × n = n³ (for example 3 × 3 × 3 = 3³)

n × n × n × n = n⁴ and so on.

With respect to powers of numbers, there are 5 basic rules which you should know:

For any number ‘n’ the following rules would apply:

Rule 1: n^a × n^b = n^{(a + b)}. Thus, 4³ × 4⁵ = 4⁸

Rule 2: n^a / n^b =n^{a – b} . Thus, 3⁹/3⁴ = 3⁵

Rule 3: (n^a)^b = n^ab. Thus, (3²)⁴= 3⁸.

Rule 4: n^(–a)= 1/n^a. Thus, 3^–4 = 1/3⁴.

Rule 5: n⁰ = 1. Thus, 5⁰ = 1.

General Form of Writing 2-3 Digit Numbers

In mathematics many a time we have to use algebraic equations in order to solve questions. In such cases an important concept is the way we represent two or three digit numbers in equation form.

For instance, suppose we have a 2 digit number with the digits ‘AB’.

In order to write this in the form of an equation, we have to use:

10A + B. This is because in the number ‘AB’ the digit A is occupying the tens place. Hence, in order to represent the value of the number ‘AB’ in the form of an equation- we can write 10A + B.

Thus, the number 29 = 2 × 10 + 9 × 1

Similarly, for a three digit number with the digits A, B and C respectively – the number ‘ABC ’ can be represented as below:

ABC = 100 A + 10 B + C.

Thus,243 = 2 × 100 + 4 × 10 + 3 × 1

The BODMAS Rule: It is used for the ordering of mathematical operations in a mathematical situation:

In any mathematical situation, the first thing to be considered is Brackets followed by Division, Multiplication, Addition and Subtraction in that order.

Thus 3 × 5 – 2 = 15 – 2 = 13

Also, 3 × 5 – 6 ∏ 3 = 15 – 2 = 13

Also, 3 × (5 – 6) ∏ 3 = 3 × (– 1) ∏ 3 = – 1.

Operations on Odd and Even numbers

ODD    SEVENS

ODDS & EVENS

Odd ∏ Even → Not divisible

SERIES OF NUMBERS

In many instances in Mathematics we are presented with a series of numbers formed simply when a group of numbers is written together. The following are examples of series:

1.3, 5, 8, 12, 17…

2.3, 7, 11, 15, 19…(Such series where the next term is derived by adding a certain fixed value to the previous number are called as Arithmetic Progressions).

3.5, 10, 20, 40 …..(Such series where the next term is derived by multiplying the previous term by a fixed value are called as Geometric Progressions).

(Note: You will study AP and GP in details in the chapter of progressions which is chapter 2 of this block.)

4.2, 7, 22, 67 ….

5.1/3, 1/5, 1/7, 1/9, 1/11…

6.1/1², 1/2² , 1/3², ¼², 1/5²…

7.1/1³, 1/3³, 1/5³…

Remember the following points at this stage:

1.AP and GP are two specific instances of series. They are studied in details only because they have many applications and have defined rules.

2.Based on the behaviour of their sums, series can be classified as:

Divergent: These are series whose sum to ‘n’ terms keeps increasing and reaches infinity for infinite terms.

Convergent: Convergent series have the property that their sum tends to approach an upper limit/lower limit as you include more terms in the series. They have the additional property that even when infinite terms of the series are included they will only reach that value and not cross it.

For example consider the series;

1/1² + 1/3² + 1/5² + 1/7² …

It is evident that subsequent terms of this series keep getting smaller. Hence, their value becomes negligible after a few terms of the series are taken into account.

If taken to infinite terms, the sum of this series will reach a value which it will never cross. Such series are called convergent, because their sum converges to a limit and only reaches that limit for infinite terms.

Chapter 1. NUMBER SYSTEMS

INTRODUCTION

The chapter on Number Systems is amongst the most important chapters in the entire syllabus of Quantitative Aptitude for the CAT examination (and also for other parallel MBA entrance exams).Students are advised to go through this chapter with utmost care understanding each concept and question type on this topic. The CAT has consistently contained anything between 20–40% of the marks based on questions taken from this chapter. Naturally, this chapter becomes one of the most crucial as far as your quest to reach close to the qualification score in the section of Quantitative Aptitude and Data Interpretation is concerned.

Hence, going through this chapter and its concepts properly is imperative for you. It would be a good idea to first go through the basic definitions of all types of numbers. Also closely follow the solved examples based on various concepts discussed in the chapter. Also, the approach and attitude while solving questions on this chapter is to try to maximize your learning experience out of every question. Hence, do not just try to solve the questions but also try to think of alternative processes in order to solve the same question. Refer to hints or solutions only as a last resort.

To start off, the following pictorial representation of the types of numbers will help you improve your quality of comprehension of different types of numbers.

DEFINITION

Natural Numbers These are the numbers (1, 2, 3, etc.) that are used for counting. In other words, all positive integers are natural numbers.

There are infinite natural numbers and the number 1 is the least natural number.

Examples of natural numbers: 1, 2, 4, 8, 32, 23, 4321 and so on.

The following numbers are examples of numbers that are not natural: –2, –31, 2.38, 0 and so on.

Based on divisibility, there could be two types of natural numbers: Prime and Composite.

Prime Numbers A natural number larger than unity is a prime number if it does not have other divisors except for itself and unity.

SHORT CUT PROCESS

A Brief Look into why this Works?

Suppose you are asked to find the factors of the number 40.

An untrained mind will find the factors as : 1, 2, 4, 5, 8, 10, 20 and 40.

The same task will be performed by a trained mind as follows:

 1 × 40

 2 × 20

 4 × 10

and 5 × 8

i.e., The discovery of one factor will automatically yield the other factor. In other words, factors will appear in terms of what can be called as factor pairs. The locating of one factor, will automatically pinpoint the other one for you. Thus, in the example above, when you find 5 as a factor of 40, you will automatically get 8 too as a factor.

Now take a look again at the pairs in the example above. If you compare the values in each pair with the square root of 40 (i.e. 6. ) you will find that for each pair the number in the left column is lower than the square root of 40, while the number in the right column is higher than the square root of 40.

This is a property for all numbers and is always true.

Hence, we can now phrase this as: Whenever you have to find the factors of any number N, you will get the factors in pairs (i.e. factor pairs). Further, the factor pairs will be such that in each pair of factors, one of the factors will be lower than the square root of N while the other will be higher than the square root of N.

As a result of this fact one need not make any effort to find the factors of a number above the square root of the number. These come automatically. All you need to do is to find the factors below the square root of the number.

Extending this logic, we can say that if we are not able to find a factor of a number upto the value of its square root, we will not be able to find any factor above the square root and the number under consideration will be a prime number. This is the reason why when we need to check whether a number is prime, we have to check for factors only below the square root.

But, we have said that you need to check for divisibility only with the prime numbers below (and including) the square root of the number. What logic will explain this:

Let us look at an example to understand why you need to look only at prime numbers below the square root.

Uptil now, we have deduced that in order to check whether a number is prime, we just need to do a factor search below (and including) the square root.

Thus, for example, in order to find whether 181 is a prime number, we need to check with the numbers = 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, and 13.

The first thing you will realise, when you first look at the list above is that all even numbers will get eliminated automatically (since no even number can divide an odd number and of course you will check a number for being prime only if it is odd!)

This will leave you with the numbers 3, 5, 7, 9, 11 and 13 to check 181.

Why do we not need to check with composite numbers below the square root? This will again be understood best if explained in the context of the example above. The only composite number in the list above is 9. You do not need to check with 9, because when you checked N for divisibility with 3 you would get either of two cases:

Case I: If N is divisible by 3: In such a case, N will automatically become non-prime and you can stop your checking. Hence, you will not need to check for the divisibility of the number by 9.

Case II: N is not divisible by 3: If N is not divisible by 3, it is obvious that it will not be divisible by 9. Hence, you will not need to check for the divisibility of the number by 9.

Thus, in either case, checking for divisibility by a composite number (9 in this case) will become useless. This will be true for all composite numbers.

Hence, when we have to check whether a number N is prime or not, we need to only check for its divisibility by prime factors below the square root of N.

Finding Prime Numbers: The Short Cut

Using the logic that we have to look at only the prime numbers below the square root in order to check whether a number is prime, we can actually cut short the time for finding whether a number is prime drastically.

Before I start to explain this, you should perhaps realise that in an examination like the CAT, or any other aptitude test for that matter whenever you would need to be checking for whether a number is prime or not, you would typically be checking 2 digit or maximum 3 digit numbers in the range of 100 to 200.

Also, one would never really need to check with the prime number 5, because divisibility by 5 would automatically be visible and thus, there is no danger of anyone ever declaring a number like 35 to be prime. Hence, in the list of prime numbers below the square root we would never include 5 as a number to check with.

Checking Whether a Number is Prime (For Numbers below 49)

The only number you would need to check for divisibility with is the number 3. Thus, 47 is prime because it is not divisible by 3.

Checking Whether a Number is Prime (For Numbers above 49 and below 121)

Naturally you would need to check this with 3 and 7. But if you remember that 77, 91 and 119 are not prime, you would be able to spot the prime numbers below 121 by just checking for divisibility with the number 3.

Why? Well, the odd numbers between 49 and 121 which are divisible by 7 are 63, 77, 91, 105 and 119. Out of these perhaps 91 and 119 are the only numbers that you can mistakenly declare as prime. 77 and 105 are so obviously not-prime that you would never be in danger of declaring them prime.

Thus, for numbers between 49 and 121 you can find whether a number is prime or not by just dividing by 3 and checking for its divisibility.

For example:

61, is prime because it is not divisible by 3 and it is neither 91 nor 119.

Checking Whether a Number is Prime (For Numbers above 121 and below 169)

Naturally you would need to check this with 3, 7 and 11. But if you remember that 133,143 and 161 are not prime, you would be able to spot the prime numbers between 121 to 169 by just checking for divisibility with the number 3.

Why? The same logic as explained above. The odd numbers between 121 and 169 which are divisible by either 7 or 11 are 133,143,147,161 and 165. Out of these 133,143 and 161 are the only numbers that you can mistakenly declare as prime if you do not check for 7 or 11. The number 147 would be found to be not prime when you check its divisibility by 3 while the number 165 you would never need to check for, for obvious reasons.

Thus, for numbers between 121 and 169 you can find whether a number is prime or not by just dividing by 3 and checking for its divisibility.

For example:

149, is prime because it is not divisible by 3 and it is neither 133,143 nor 161.

Thus, we have been able to go all the way till 169 with just checking for divisibility with the number 3.

This logic can be represented on the number line as follows:

Integers A set which consists of natural numbers, negative integers (–1, –2, –3…–n…) and zero is known as the set of integers. The numbers belonging to this set are known as integers.

Integers can be visualised on the number line:

The moment you define integers, you automatically define decimals.

Decimals

A decimal number is a number with a decimal point in it, like these: 1.5, 3.21, 4.173, 5.1 etc

The number to the left of the decimal is an ordinary whole number. The first number to the right of the decimal is the number of tenths (1/10’s). The second is the number of hundredths (1/100’s) and so on. So, for the number 5.1, this is a shorthand way of writing the mixed number . 3.27 is the same as 3 + 2/10 + 7/100.

Composite Numbers It is a natural number that has at least one divisor different from unity and itself.

Every composite number n can be factored into its prime factors. (This is sometimes called the canonical form of a number.)

In mathematical terms: n = p₁^m . p₂ⁿ …p_k^s, where p₁, p₂…p_k are prime numbers called factors and m, n…k are natural numbers.

Thus, 24 = 2³ ◊ 3, 84 = 7 ◊ 3 ◊ 2² etc.

This representation of a composite number is known as the standard form of a composite number. It is an extermely useful form of seeing a composite number as we shall see.

Whole Numbers The set of numbers that includes all natural numbers and the number zero are called whole numbers. Whole number are also called as Non-negative integers.

The Concept of the Number Line The number line is a straight line between negative infinity on the left to positive infinity to the right.

The distance between any two points on the number line is got by subtracting the lower value from the higher value. Alternately, we can also start with the lower number and find the required addition to reach the higher number.

For example: The distance between the points 7 and –4 will be 7 – (–4) = 11.

Real Numbers All numbers that can be represented on the number line are called real numbers. Every real number can be approximately replaced with a terminating decimal.

The following operations of addition, subtraction, multiplication and division are valid for both whole numbers and real numbers: [For any real or whole numbers a, b and c].

(a)Commutative property of addition: a + b = b + a.

(b)Associative property of addition: (a + b) + c = a + (b + c).

(c)Commutative property of multiplication: a ◊ b = b ◊ a.

(d)Associative property of multiplication: (a ◊ b) ◊ c = a ◊ (b ◊ c).

(e)Distributive property of multiplication with respect to addition: (a + b) c = ac + bc.

(f)Subtraction and division are defined as the inverse operations to addition and multiplication respectively.

Thus if a + b = c, then c – b = a and if q = a/b then b ◊ q = a (where b π 0).

Division by zero is not possible since there is no number q for which b ◊ q equals a non zero number a.

Rational Numbers A rational number is defined as a number of the form a/b where a and b are integers and b π 0.

The set of rational numbers encloses the set of integers and fractions. The rules given above for addition, subtraction, multiplication and division also apply on rational numbers.

Rational numbers that are not integral will have decimal values. These values can be of two types:

(a)Terminating (or finite) decimal fractions: For example, 17/4 = 4.25, 21/5 = 4.2 and so forth.

(b)Non-terminating decimal fractions: Amongst non- terminating decimal fractions there are two types of decimal values:

(i)Non-terminating periodic fractions: These are non- terminating decimal fractions of the type x ◊ a₁a₂a₃a₄ …a_na₁a₂a₃a₄ …a_na₁a₂a₃a₄ …a_n. For example = 5.3333, 15.23232323, 14.287628762 876 … and so on.

(ii)Non-terminating non-periodic fractions: These are of the form x ◊ b₁b₂b₃b₄…b_nc₁c₂c₃…c_n. For example: 5.2731687143725186....

Of the above categories, terminating decimal and non-terminating periodic decimal fractions belong to the set of rational numbers.

Irrational Numbers Fractions, that are non-terminating, non-periodic fractions, are irrational numbers.

Some examples of irrational numbers are , etc. In other words, all square and cube roots of natural numbers that are not squares and cubes of natural numbers are irrational. Other irrational numbers include p, e and so on.

Every positive irrational number has a negative irrational number corresponding to it.

All operations of addition, subtraction, multiplication and division applicable to rational numbers are also applicable to irrational numbers.

As briefly stated in the Back to School section, whenever an expression contains a rational and an irrational number together, the two have to be carried together till the end. In other words, an irrational number once it appears in the solution of a question will continue to appear till the end of the question. This concept is particularly useful in Geometry For example: If you are asked to find the ratio of the area of a circle to that of an equilateral triangle, you can expect to see a p / in the answer. This is because the area of a circle will always have a p component in it, while that of an equilateral triangle will always have .

You should realise that once an irrational number appears in the solution of a question, it can only disappear if it is multiplied or divided by the same irrational number.

THE CONCEPT OF GCD (GREATEST COMMON DIVISOR OR HIGHEST COMMON FACTOR)

Consider two natural numbers n, and n₂.

If the numbers n₁ and n₂ are exactly divisible by the same number x, then x is a common divisor of n₁ and n₂.

The highest of all the common divisors of n₁ and n₂ is called as the GCD or the HCF. This is denoted as GCD (n₁, n₂).

SHORTCUT FOR FINDING THE HCF

The above ‘school’ process of finding the HCF (or the GCD) of a set of numbers is however extremely cumbersome and time taking. Let us take a look at a much faster way of finding the HCF of a set of numbers.

Suppose you were required to find the HCF of 39,78 and 195 (first of the practice problems above)

Logic The HCF of these numbers would necessarily have to be a factor (divisor) of the difference between any pair of numbers from the above 3. i.e. the HCF has to be a factor of (78 – 39 = 39) as well as of (195 – 39 = 156) and (195 – 78 = 117). Why?

Well the logic is simple if you were to consider the tables of numbers on the number line.

For any two numbers on the number line, a common divisor would be one which divides both. However, for any number to be able to divide both the numbers, it can only do so if it is a factor of the difference between the two numbers. Got it??

Take an example:

Let us say we take the numbers 68 and 119. The difference between them being 51, it is not possible for any number outside the factor list of 51 to divide both 68 and 119. Thus, for example a number like 4, which divides 68 can never divide any number which is 51 away from 68- because 4 is not a factor of 51.

Only factors of 51, i.e. 51,17,3 and 1 ‘could’ divide both these numbers simultaneously.

Hence, getting back to the HCF problem we were trying to tackle—take the difference between any two numbers of the set—of course if you want to reduce your calculations in the situation, take the difference between the two closest numbers. In this case that would be the difference between 78 and 39 =39.

The HCF has then to be a factor of this number. In order to find the factors quickly remember to use the fact we learnt in the back to school section of this part of the book—that whenever we have to find the list of factors/divisors for any number we have to search the factors below the square root and the factors above the square root would be automatically visible)

A factor search of the number 39 yields the following factors:

1 × 39

3 × 13

Hence, one of these 4 numbers has to be the HCF of the numbers 39,78 and 195. Since we are trying to locate the Highest common factor—we would begin our search from the highest number (viz:39)

Check for divisibility by 39 Any one number out of 39 and 78 and also check the number 195 for divisibility by 39. You would find all the three numbers are divisible by 39 and hence 39 can be safely taken to be the correct answer for the HCF of 39,78 and 195.

Suppose the numbers were:

39, 78 and 182?

The HCF would still be a factor of 78–39=39. The probable candidates for the HCF’s value would still remain 1,3,13 and 39.

When you check for divisibility of all these numbers by 39, you would realize that 182 is not divisible and hence 39 would not be the HCF in this case.

The next check would be with the number 13. It can be seen that 13 divides 39 (hence would automatically divide 78- no need to check that) and also divides 182. Hence, 13 would be the required HCF of the three numbers.

Typical questions where HCF is used directly

Question 1: The sides of a hexagonal field are 216, 423, 1215, 1422, 2169 and 2223 meters. Find the greatest length of tape that would be able to exactly measure each of these sides without having to use fractions/parts of the tape?

In this question we are required to identify the HCF of the numbers 216,423,1215, 1422, 2169 and 2223.

In order to do that, we first find the smallest difference between any two of these numbers. It can be seen that the difference between 2223-2169 = 54. Thus, the required HCF would be a factor of the number 54.

The factors of 54 are:

1 × 54

2 × 27

3 × 18

6 × 9

One of these 8 numbers has to be the HCF of the 6 numbers. 54 cannot be the HCF because the numbers 423 and 2223 being odd numbers would not be divisible by any even number. Thus, we do not need to check any even numbers in the list.

27 does not divide 423 and hence cannot be the HCF. 18 can be skipped as it is even.

Checking for 9:

9 divides 216,423,1215,1422 and 2169. Hence, it would become the HCF. (Note: we do not need to check 2223 once we know that 2169 is divisible by 9)

Question 2: A nursery has 363,429 and 693 plants respectively of 3 distinct varieties. It is desired to place these plants in straight rows of plants of 1 variety only so that the number of rows required is the minimum. What is the size of each row and how many rows would be required?

The size of each row would be the HCF of 363, 429 and 693. Difference between 363 and 429 =66. Factors of 66 are 66, 33, 22, 11, 6, 3, 2, 1.

66 need not be checked as it is even and 363 is odd. 33 divides 363, hence would automatically divide 429 and also divides 693. Hence, 33 is the correct answer for the size of each row.

For how many rows would be required we need to follow the following process:

Minimum number of rows required = 363/33 + 429/33 + 693/33 = 11 + 13 + 21 = 45 rows.

THE CONCEPT OF LCM (LEAST COMMON MULTIPLE)

Let n₁, and n₂ be two natural numbers distinct from each other. The smallest natural number n that is exactly divisible by n₁ and n₂ is called the Least Common Multiple (LCM) of n₁ and n₂ and is designated as LCM (n₁, n₂).

SHORT CUT FOR FINDING THE LCM

The LCM (least common multiple) again has a much faster way of doing it than what we learnt in school.

The process has to do with the use of co-prime numbers.

Before we look at the process, let us take a fresh look at what co-prime numbers are:

Co-prime numbers are any two numbers which have an HCF of 1, i.e. when two numbers have no common prime factor apart from the number 1, they are called co-prime or relatively prime to each other.

Some Rules for Co-primes

2 Numbers being co-prime

(i)Two consecutive natural numbers are always co-prime (Example 5, 6; 82, 83; 749, 750 and so on)

(ii)Two consecutive odd numbers are always co-prime (Examples: 7, 9; 51, 53; 513, 515 and so on)

(iii)Two prime numbers are always co-prime (Examples: 13, 17; 53, 71 and so on)

(iv)One prime number and another composite number (such that the composite number is not a multiple of the prime number) are always co-prime (Examples: 17, 38; 23, 49 and so on, but note that 17 and 51 are not co-prime)

3 or more numbers being co-prime with each other means that all possible pairs of the numbers would be co-prime with each other.

Thus, 47, 49, 51 and 52 are co-prime since each of the 6 pairs (47,49); (47,51); (47,52); (49,51); (49,52) and (51,52) are co-prime.

Rules for Spotting three Co-prime Numbers

(i)Three consecutive odd numbers are always co-prime (Examples: 15, 17, 19; 51, 53, 55 and so on)

(ii)Three consecutive natural numbers with the first one being odd (Examples: 15, 16, 17; 21, 22, 23; 41, 42, 43 and so on). Note that 22, 23, 24 are not co-prime

(iii)Two consecutive natural numbers along-with the next odd number such that the first no. is even (examples: 22, 23, 25; 52, 53, 55; 68, 69, 71 and so on)

(iv)Three prime numbers (Examples: 17, 23, 29; 13, 31, 43 and so on)

So what do co-prime numbers have to do with LCMs?

By using the logic of co-prime numbers, you can actually bypass the need to take out the prime factors of the set of numbers for which you are trying to find the LCM. How?

The following process will make it clear:

Let us say that you were trying to find the LCM of 9,10,12 and 15.

The LCM can be directly written as: 9 × 10 × 2. The thinking that gives you the value of the LCM is as follows:

Step 1: If you can see a set of 2 or more co-prime numbers in the set of numbers for which you are finding the LCM- write them down by multiplying them.

So in the above situation, since we can see that 9 and 10 are co-prime to each other we can start off writing the LCM by writing 9 × 10 as the first step.

Step 2: For each of the other numbers, consider what part of them have already been taken into the answer and what part remains outside the answer. In case you see any part of the other numbers such that it is not a part of the value of the LCM you are writing—such a part would need to be taken into the answer of the LCM.

The process will be clear once you see what we do (and how we think) with the remaining 2 numbers in the above problem.

At this point when we have written down 9 × 10 we already have taken into account the numbers 9 and 10 leaving us to account for 12 and 15.

Thought about 12: 12 is 2 × 2 × 3

9 × 10 already has a 3 and one 2 in it prime factors. However, the number 12 has two 2’s. This means that one of the two 2’s of the number 12 is still not accounted for in our answer. Hence, we need to modify the LCM by multiplying the existing 9 × 10 by a 2. With this change the LCM now becomes:

9 × 10 × 2

Thought about 15: 15 is 5 × 3

9 × 10 × 2 already has a 5 and a 3. Hence, there is no need to add anything to the existing answer.

Thus, 9 × 10 × 2 would become the correct answer for the LCM of the numbers 9, 10, 12 and 15.

What if the numbers were: 9, 10, 12 and 25

Step 1: 9 and 10 are co-prime

Hence, the starting value is 9 × 10

Thought about 12: 12 is 2 × 2 × 3

9 × 10 already has a 3 and one 2 in it prime factors. However, the number 12 has two 2’s. This means that one of the two 2’s of the number 12 is still not accounted for in our answer. Hence, we need to modify the LCM by multiplying the existing 9 × 10 by a 2. With this change the LCM now becomes:

9 × 10 × 2

Thought about 25: 25 is 5 × 5

9 × 10 × 2 has only one 5. Hence, we need to add another 5 to the answer.

Thus, 9 × 10 × 2 × 5 would become the correct answer for the LCM of the numbers 9, 10, 12 and 25.

Typical questions on LCMs

You would be able to see most of the standard questions on LCMs in the practice exercise on HCF and LCM given below.

Rules for HCF: If the HCF of x and y is G, then the HCF of

(i)x, (x + y) is also G

(ii)x, (x – y) is also G

HCF and LCM

Practice Exercise

Solutions

DIVISIBILITY

A number x is said to be divisible by another number ‘y’ if it is completely divisible by Y (i.e. it should leave no remainder).

In general it can be said that any integer I, when divided by a natural number N, there exist a unique pair of numbers Q and R which are called the quotient and Remainder respectively.

Thus, I = QN + R.

For any integer I and any natural number n there is a unique pair of numbers a and b such that:

I = QN + R

Where Q is an integer and N is a natural number or zero and 0 ≥ R < N (i.e. remainder has to be a whole number less than N.)

If the remainder is zero we say that the number I is divisible by N.

When R π 0, we say that the number I is divisible by N with a remainder.

Thus, 25/8 can be written as: 25 = 3 × 8 + 1 (3 is the quotient and 1 is the remainder)

While, –25/7 will be written as –25 = 7 × (–4) + 3 (–4 is the quotient and 3 is the remainder)

Note: An integer b π 0 is said to divide an integer a if there exists another integer c such that:

a = bc

It is important to explain at this point a couple of concepts with respect to the situation, when we divide a negative number by a natural number N.

Suppose, we divide – 32 by 7. Contrary to what you might expect, the remainder in this case is + 3 (and not – 4). This is because the remainder is always non negative.

Thus, –32/7 gives quotient as – 5 and remainder as + 3.

The relationship between the remainder and the decimal:

1.Suppose we divide 42 by 5. The result has a quotient of 8 and remainder of 2.

But 42/5 = 8.4. As you can see, the answer has an integer part and a decimal part. The integer part being 8 (equals the quotient), the decimal part is 0.4 (and is given by 2/5).

Since, we have also seen that for any divisor N, the set of remainders obeys the inequality 0 £ R < N, we should realise that any divisor N, will yield exactly N possible remainders. (For example If the divisor is 3, we have 3 possible remainders 0, 1 and 2. Further, when 3 is the divisor we can have only 3 possible decimal values .00, .333 & 0.666 corresponding to remainders of 0, 1 or 2. I would want you to remember this concept when you study the fraction to percentage conversion table in the chapter on percentages.

2.In the case of –42 being divided by 5, the value is – 8.4. In this case the interpretation should be thus:

The integer part is – 9 (which is also the quotient of this division) and the decimal part is 0.6 (corresponding to 3/5) Notice that since the remainder cannot be negative, the decimal too cannot be negative.

Theorems of Divisibility

As we have already seen earlier –

Any composite number can be written down as a product of its prime factors. (Also called standard form)

Thus, for example the number 1240 can be written as 2³ × 31¹ × 5¹.

The standard form of any number has a huge amount of information stored in it. The best way to understand the information stored in the standard form of a number is to look at concrete examples. As a reader I want you to understand each of the processes defined below and use them to solve similar questions given in the exercise that follows and beyond:

1.Using the standard form of a number to find the sum and the number of factors of the number:

(a)Sum of factors of a number:

Suppose, we have to find the sum of factors and the number of factors of 240.

240 = 2⁴ × 3¹ × 5¹

The sum of factors will be given by:

(2⁰ + 2¹ + 2² + 2³ + 2⁴) (3⁰ + 3¹) (5⁰ + 5¹)

= 31 × 4 × 6 = 744

Note: This is a standard process, wherein you create the same number of brackets as the number of distinct prime factors the number contains and then each bracket is filled with the sum of all the powers of the respective prime number starting from 0 to the highest power of that prime number contained in the standard form.

Thus, for 240, we create 3 brackets—one each for 2, 3 and 5. Further in the bracket corresponding to 2 we write (2⁰ + 2¹ + 2² + 2³ + 2⁴).

Hence, for example for the number 40 = 2³ × 5¹, the sum of factors will be given by: (2⁰ + 2¹ + 2² + 2³) (5⁰ + 5¹) {2 brackets since 40 has 2 distinct prime factors 2 and 5}

(b)Number of factors of the number:

Let us explore the sum of factors of 40 in a different context.

(2⁰ + 2¹ + 2² + 2³) (5⁰ + 5¹)

= 2⁰ × 5⁰ + 2⁰ × 5¹ + 2¹ × 5⁰ + 2¹ × 5¹ + 2² × 5⁰ + 2²

× 5¹ + 2³ × 5⁰ + 2³ × 5¹

= 1 + 5 + 2 + 10 + 4 + 20 + 8 + 40 = 90

A clear look at the numbers above will make you realize that it is nothing but the addition of the factors of 40

Hence, we realise that the number of terms in the expansion of (2⁰ + 2¹ + 2² + 2³) (5⁰ + 5¹) will give us the number of factors of 40. Hence, 40 has 4 × 2 = 8 factors.

Note: The moment you realise that 40 = 2³ × 5¹ the answer for the number of factors can be got by (3 + 1) (1 + 1) = 8

2.Sum and Number of even and odd factors of a number.

Suppose, you are trying to find out the number of factors of a number represented in the standard form by: 2³ × 3⁴ × 5² × 7 ³

As you are already aware the answer to the question is (3 + 1) (4 + 1) (2 + 1) (3 + 1) and is based on the logic that the number of terms will be the same as the number of terms in the expansion: (2⁰ + 2¹ + 2² + 2³) (3⁰ + 3¹ + 3² + 3³ + 3⁴)(5⁰ + 5¹ + 5²) (7⁰ + 7¹ + 7² + 7³).

Now, suppose you have to find out the sum of the even factors of this number. The only change you need to do in this respect will be evident below. The answer will be given by:

(2¹ + 2² + 2³)(3⁰+ 3¹ + 3² + 3³ + 3⁴) (5⁰ + 5¹ + 5²) (7⁰ + 7¹ + 7² + 7³)

Consequently, the number of even factors will be given by: (3) (4 + 1) (2 + 1) (3 + 1)

i.e. Since 2⁰ is eliminated, we do not add 1 in the bracket corresponding to 2.

Let us now try to expand our thinking to try to think about the number of odd factors for a number.

In this case, we just have to do the opposite of what we did for even numbers. The following step will make it clear:

Odd factors of the number whose standard form is : 2³ × 3⁴ × 5² × 7³

Sum of odd factors = (2⁰) (3⁰ + 3¹ + 3² + 3³ + 3⁴) (5⁰ + 5¹ + 5²) (7⁰ + 7¹ + 7² + 7³)

i.e.: Ignore all powers of 2. The result of the expansion of the above expression will be the complete set of odd factors of the number. Consequently, the number of odd factors for the number will be given by the number of terms in the expansion of the above expression.

Thus, the number of odd factors for the number 2³ × 3⁴ × 5² × 7³ = 1 × (4 + 1) (2 + 1) (3 + 1).

3.Sum and number of factors satisfying other conditions for any composite number

These are best explained through examples:

(i)Find the sum and the number of factors of 1200 such that the factors are divisible by 15.

Solution : 1200 = 2⁴ × 5² × 3¹.

For a factor to be divisible by 15 it should compulsorily have 3¹ and 5¹ in it. Thus, sum of factors divisible by 15 = (2⁰ + 2¹ + 2² + 2³ + 2⁴) × (5¹ + 5²) (3¹) and consequently the number of factors will be given by 5 × 2 × 1 = 10.

(What we have done is ensure that in every individual term of the expansion, there is a minimum of 3¹ × 5¹. This is done by removing powers of 3 and 5 which are below 1.)

Task for the student: Physically verify the answers to the question above and try to convert the logic into a mental algorithm.

NOTE FROM THE AUTHOR—The need for thought algorithms:

I have often observed that the key difference between understanding a concept and actually applying it under examination pressure, is the presence or absence of a mental thought algorithm which clarifies the concept to you in your mind. The thought algorithm is a personal representation of a concept—and any concept that you read/understand in this book (or elsewhere) will remain an external concept till it remains in someone else’s words. The moment the thought becomes internalised the concept becomes yours to apply and use.

Practice Exercise on Factors

Solutions

Solutions to Questions 1 to 6:

2450 = 50 × 49 = 2¹ × 5² × 7²

1.Sum and number of all factors:

Sum of factors = (2⁰ + 2¹) (5⁰ + 5¹ + 5²) (7⁰ + 7¹ + 7²)

Number of factors = 2 × 3 × 3 = 18

2.Sum of all even factors:

(2¹) (5⁰ + 5¹ + 5²) (7⁰ + 7¹ + 7²)

Number of even factors = 1 × 3 × 3 = 9

3.Sum of all odd factors:

(2⁰) (5⁰ + 5¹ + 5²) (7⁰ + 7¹ + 7²)

Number of odd factors = 1 × 3 × 3 = 9

4.Sum of factors divisible by 5:

(2⁰ + 2¹) (5¹ + 5²) (7⁰ + 7¹ + 7²)

Number of factors divisible by 5 = 2 × 2 × 3 = 12

5.Sum of factors divisible by 35:

(2⁰ + 2¹) (5¹ + 5²) (7¹ + 7²)

Number of factors divisible by 35 = 2 × 2 × 2 = 8

6.Sum of all factors divisible by 245:

(2⁰ + 2¹) (5¹ + 5²) (7²)

Number of factors divisible by 245 = 2 × 2 × 1 = 4

Solutions to Questions 7 to 14:

7200 = 72 × 100 = 12 × 6 × 100 = 2⁵ × 3² × 5²

7.Sum and number of all factors:

Sum of factors = (2⁰ + 2¹+ 2² + 2³ + 2⁴ + 2⁵) (3⁰ + 3¹ + 3²) (5⁰ + 5¹ + 5²)

Number of factors = 6 × 3 × 3 = 54

8.Sum and number of even factors:

Sum of even factors = (2¹ + 2² + 2³ + 2⁴ + 2⁵) (3⁰ + 3¹ + 3²) (5⁰ + 5¹ + 5²)

Number of even factors = 5 × 3 × 3 = 45

9.Sum and number of odd factors:

Sum of odd factors = (2⁰) (3⁰ + 3¹ + 3²) (5⁰ + 5¹ + 5²)

Number of odd factors = 1 × 3 × 3 = 9

10.Sum and number of factors divisible by 25:

Sum of factors divisible by 25= (2⁰ + 2¹ + 2² + 2³ + 2⁴ + 2⁵) (3⁰ + 3¹ + 3²) (5²)

Number of factors divisible by 25 = 6 × 3 × 1 = 18

11.Sum and number of factors divisible by 40:

Sum of factors divisible by 40 = (2³ + 2⁴ + 2⁵) (3⁰ + 3¹ + 3²) (5¹ + 5²)

Number of factors = 3 × 3 × 2 = 18

12.Sum and number of factors divisible by 150:

Sum of factors divisible by 150 = (2¹ + 2² + 2³ + 2⁴ + 2⁵) (3¹ + 3²) (5²)

Number of factors divisible by 150 = 5 × 2 × 1 = 10

13.Sum and number of factors not divisible by 75:

Sum of factors not divisible by 75 = Sum of all factors – Sum of factors divisible by 75 =

(2⁰ + 2¹ + 2² + 2³ + 2⁴ + 2⁵) (3⁰ + 3¹ + 3²) (5⁰ + 5¹ + 5²) – (2⁰ + 2¹ + 2² + 2³ + 2⁴ + 2⁵) (3¹ + 3²) (5²)

Number of factors not divisible by 75 = Number of factors of 7200 – Number of factors of 7200 which are divisible by 75 = 6 × 3 × 3 – 6 × 2 × 1 = 54 – 12 = 42

14.Sum and number of factors not divisible by 24:

Sum of factors not divisible by 24 = Sum of all factors – Sum of factors divisible by 24 =

(2⁰ + 2¹ + 2² + 2³ + 2⁴ + 2⁵) (3⁰ + 3¹ + 3²) (5⁰ + 5¹ + 5²) – (2³ + 2⁴ + 2⁵) (3¹ + 3²) (5⁰ + 5¹ + 5²)

Number of factors not divisible by 24 = Number of factors of 7200 – Number of factors of 7200 which are divisible by 24 = 6 × 3 × 3 – 3 × 2 × 3 = 54 – 18 = 36

15.Number of divisors of 1728

1728 = 4 × 432 = 16 × 108 = 64 × 27 = 2⁶ × 3³

Number of factors = 7 × 4 = 28. Option (a) is correct.

16.1080 = 108 × 10 = 27 × 4 × 10 = 3³ × 2³ × 5¹

Number of factors = 4 × 4 × 2 = 32.

In order to see the number of factors of 1080 which are perfect squares we need to visualize the structure for writing down the sum of perfect square factors of 1080.

This would be given by:

Sum of all perfect square factors of 1080 = (2⁰ + 2²) (3⁰ + 3²) (5⁰).

From the above structure it is clear that the number of perfect square factors is going to be 2 × 2 × 1 = 4

Thus, the number of factors of 1080 which are not perfect squares are equal to 32 – 4 = 28.

Option (a) is correct.

17.544 = 17¹ × 2⁵. Hence, the total number of factors of 544 is 2 × 6 =12. But we have to count factors excluding 1 and 544. Thus, we need to remove 2 factors from this. The required answer is 12 – 2 =10. Option (d) is correct.

18.Using the fact that 544 has a total of 12 factors and the numbers 1 and 2 are the two factors which are lower than 3, we would get a total of 10 factors greater than 3. Option (b) is correct.

19.The required answer would be given by: Sum of all factors of 544 – 1 – 544 = (2⁰ + 2¹+ 2² + 2³ + 2⁴ + 2⁵) (17⁰ + 17¹) – 545 = 63 × 18 – 545 = 589. Option (c) is correct.

20.Sum of divisors of 544 which are perfect square is:

(2⁰ + 2² + 2⁴ ) (17⁰) = 21. Option (d) is correct.

21.Sum of odd divisors of 544 =

(2⁰) (17⁰ + 17¹) = 18. Option (a) is correct.

22.4096 = 2¹².

Sum of even divisors = (2¹ + 2² + 2³ + 2⁴ + 2⁵ + 2⁶ + 2⁷ + 2⁸ + 2⁹ + 2¹⁰ + 2¹¹ + 2¹²) = 2¹³ – 2 = 8190

23.144 = 2⁴ × 3² Æ Sum of divisors of 144 = (2⁰ + 2¹ + 2² + 2³ + 2⁴ ) (3⁰ + 3¹ + 3²) = 31 × 13 = 403

160 = 2⁵ × 5¹ Æ Sum of divisors of 160 = (2⁰ + 2¹ + 2² + 2³ + 2⁴ + 2⁵) (5⁰ + 5¹) = 63 × 6 = 378.

Sum of the two = 403 + 378 = 781.

24.96 = 2⁵ × 3¹. Sum of even divisors of 96 = (2¹ + 2² + 2³ + 2⁴ + 2⁵) (3⁰ + 3¹) = 62 × 4 = 248

3600 = 2⁴ × 5² × 3². Sum of odd divisors of 3600 = (2⁰) (3⁰ + 3¹ + 3²) (5⁰ + 5¹ + 5²) = 13 × 31 = 403

Sum of the two = 248 + 403 = 651.

Option (c) is correct.

NUMBER OF ZEROES IN AN EXPRESSION

Suppose you have to find the number of zeroes in a product: 24 × 32 × 17 × 23 × 19 = (2³ × 3¹) × (2⁵) × 17¹ × 23 × 19.

As you can notice, this product will have no zeroes because it has no 5 in it.

However, if you have an expression like: 8 × 15 × 23 × 17 × 25 × 22

The above expression can be rewritten in the standard form as:

2³ × 3¹ × 5¹ × 23 × 17 × 5² × 2¹ × 11¹

Zeroes are formed by a combination of 2 × 5. Hence, the number of zeroes will depend on the number of pairs of 2’s and 5’s that can be formed.

In the above product, there are four twos and three fives. Hence, we shall be able to form only three pairs of (2 × 5). Hence, there will be 3 zeroes in the product.

Finding the Number of Zeroes in a Factorial Value

Suppose you had to find the number of zeroes in 6!.

6! = 6 × 5 × 4 × 3 × 2 × 1 = (3 × 2) × (5) × (2 × 2) × (3) × (2) × (1).

The above expression will have only one pair of 5 × 2, since there is only one 5 and an abundance of 2’s.

It is clear that in any factorial value, the number of 5’s will always be lesser than the number of 2’s. Hence, all we need to do is to count the number of 5’s. The process for this is explained in Solved Examples 1.1 to 1.3.

Exercise for Self-practice

Find the number of zeroes in the following cases:

1.47!

2.58!

3.13 × 15 × 22 × 125 × 44 × 35 × 11

4.12 × 15 × 5 × 24 × 13 × 17

5.173!

6.144! × 5 × 15 × 22 × 11 × 44 × 135

7.148!

8.1093!

9.1132!

10.1142! × 348! × 17!

Solutions

1.47/5 Æ Quotient 9. 9/5 Æ Quotient Æ 1. 9 + 1 = 10 zeroes.

2.58/5 Æ Quotient 11. 11/5 Æ Quotient Æ 2. 11 + 2 = 13 zeroes.

3.The given expression has five 5’s and three 2’s. Thus, there would be three zeroes in the expression.

4.The given expression has two 5’s and five 2’s. Thus, there would be two zeroes in the expression.

5.173/5 Æ Quotient 34. 34/5 Æ Quotient 6. 6/5 Æ Quotient 1. 34 + 6 + 1 = 41 zeroes.

6.144! Would have 28 + 5 + 1 = 34 zeroes and the remaining part of the expression would have three zeroes. A total of 34 + 3 = 37 zeroes.

7.148/5 Æ Quotient 29. 29/5 Æ Quotient 5. 5/5 Æ Quotient 1. 29 + 5 + 1 = 35 zeroes.

8.1093/5 Æ Quotient 218. 218/5 Æ Quotient 43. 43/5 Æ Quotient 8. 8/5 Æ Quotient 1. 218 + 43 + 8 + 1 = 270 zeroes.

9.1132/5 Æ Quotient 226. 226/5 Æ Quotient 45. 45/5 Æ Quotient 9. 9/5 Æ Quotient 1. 226 + 45 + 9 + 1 = 281 zeroes.

10.1142/5 Æ Quotient 228. 228/5 Æ Quotient 45. 45/5 Æ Quotient 9. 9/5 Æ Quotient 1. 228 + 45 + 9 + 1 = 284 zeroes.

348/5 Æ Quotient 69. 69/5 Æ Quotient 13. 13/5 Æ Quotient 2. 69 + 13 + 2 = 84 zeroes.

17/5 Æ Quotient 3 Æ 3 zeroes.

Thus, the total number of zeroes in the expression is: 284 + 84 + 3 = 371 zeroes.

A special implication: Suppose you were to find the number of zeroes in the value of the following factorial values:

45!, 46!, 47!, 48!, 49!

What do you notice? The number of zeroes in each of the cases will be equal to 10. Why does this happen? It is not difficult to understand that the number of fives in any of these factorials is equal to 10. The number of zeroes will only change at 50! (It will become 12).

In fact, this will be true for all factorial values between two consecutive products of 5.

Thus, 50!, 51!, 52!, 53! And 54! will have 12 zeroes (since they all have 12 fives).

Similarly, 55!, 56!, 57!, 58! And 59! will each have 13 zeroes.

Apart from this fact, did you notice another thing? That while there are 10 zeroes in 49! there are directly 12 zeroes in 50!. This means that there is no value of a factorial which will give 11 zeroes. This occurs because to get 50! we multiply the value of 49! by 50. When you do so, the result is that we introduce two 5’s in the product. Hence, the number of zeroes jumps by two (since we never had any paucity of twos.)

Note: at 124! you will get 24 + 4 fi 28 zeroes.

At 125! you will get 25 + 5 + 1 = 31 zeroes. (A jump of 3 zeroes.)

Exercise for Self-practice

1.n! has 23 zeroes. What is the maximum possible value of n?

2.n! has 13 zeroes. The highest and least values of n are?

3.Find the number of zeroes in the product 1¹ × 2² × 3³ × 4⁴ × 5⁵ × 6⁶ × …. × 49⁴⁹

4.Find the number of zeroes in:

100¹ × 99² × 98³ × 97⁴ × ………. × 1¹⁰⁰

5.Find the number of zeroes in:

1^1! × 2^2! × 3^3! × 4^4! × 5^5! × ……..10^10!

6.Find the number of zeroes in the value of:

2² × 5⁴ × 4⁶ × 10⁸ × 6¹⁰ × 15¹² × 8¹⁴ × 20¹⁶ × 10¹⁸ × 25²⁰

7.What is the number of zeroes in the following:

(a)3200 + 1000 + 40000 + 32000 + 15000000

(b)3200 × 1000 × 40000 × 32000 × 16000000

Solutions

1.This can never happen because at 99! number of zeroes is 22 and at 100! the number of zeroes is 24.

2.59 and 55 respectively.

3.The fives will be less than the twos. Hence, we need to count only the fives.

Thus : 5⁵ × 10¹⁰ × 15¹⁵ × 20²⁰ × 25²⁵ × 30³⁰ × 35³⁵ × 40⁴⁰ × 45⁴⁵

gives us: 5 + 10 + 15 + 20 + 25 + 25 + 30 + 35 + 40 + 45 fives. Thus, the product has 250 zeroes.

4.Again the key here is to count the number of fives. This can get done by:

100¹ × 95⁶ × 90¹¹ × 85¹⁶ × 80²¹ × 75²⁶ × …… 5⁹⁶

(1 + 6 + 11 + 16 + 21 + 26 + 31 + 36 + 41 + 46 + …….. + 96) + (1 + 26 + 51 + 76)

= 20 × 48.5 + 4 × 38.5 (Using sum of A.P. ex-plained in the next chapter.)

= 970 + 154 = 1124.

5.The answer will be the number of 5’s. Hence, it will be 5! + 10!

6.The number of fives is again lesser than the number of twos.

The number of 5’s will be given by the power of 5 in the product:

5⁴ × 10⁸ × 15¹² × 20¹⁶ × 10¹⁸ × 25²⁰

= 4 + 8 + 12 + 16 + 18 + 40 = 98.

7.A. The number of zeroes in the sum will be two, since:

Thus, in such cases the number of zeroes will be the least number of zeroes amongst the numbers.

Exception: 3200 + 1800 = 5000 (three zeroes, not two).

1. The number of zeroes will be:

2 + 3 + 4 + 3 + 6 = 18.

An extension of the process for finding the number of zeroes.

Consider the following questions:

1.Find the highest power of 5 which is contained in the value of 127!

2.When 127! is divided by 5ⁿ the result is an integer. Find the highest possible value for n.

3.Find the number of zeroes in 127!

In each of the above cases, the value of the answer will be given by:

[127/5] + [127/25] + [127/125]

= 25 + 5 + 1 = 31

This process can be extended to questions related to other prime numbers. For example:

Find the highest power of:

1.3 which completely divides 38!

Solution: [38/3] + [38/3²] + 38/3³] = 12 + 4 + 1 = 17

2.7 which is contained in 57!

[57/7] + [57/7²] = 8 + 1 = 9.

This process changes when the divisor is not a prime number. You are first advised to go through worked out problems 1.4, 1.5, 1.6 and 1.19.

Now try to solve the following exercise:

1.Find the highest power of 7 which divides 81!

2.Find the highest power of 42 which divides 122!

3.Find the highest power of 84 which divides 342!

4.Find the highest power of 175 which divides 344!

5.Find the highest power of 360 which divides 520!

Solutions

1.81/7 Æ Quotient 11. 11/7 Æ Quotient 1. Highest power of 7 in 81! = 11 + 1 =12.

2.In order to check for the highest power of 42, we need to realize that 42 is 2 × 3 × 7. In 122! The least power between 2,3 and 7 would obviously be for 7. Thus, we need to find the number of 7’s in 122! (or in other words- the highest power of 7 in 122!).

This can be done by:

122/7 Æ Quotient 17. 17/7 Æ Quotient 2. Highest power of 7 in 122! = 17 + 2 = 19.

3.84 = 2 × 2 × 3 × 7. This means we need to think of which amongst 2², 3 and 7 would appear the least number of times in 342! It is evident that there would be more 2²s and 3’s than 7’s in any factorial value (Because if you write any factorial 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 × 10 × 11 × 12 × 13 × 14 × 15…you can clearly see that before a 7 or it’s multiple appears in the multiplication, there are at least two 2’s and one 3 which appear beforehand.)

Hence, in order to solve this question we just need to find the power of 7 in 342!

This can be done as:

342/7 Æ Quotient 48. 48/7 Æ Quotient 6. 6/7 Æ Quotient 0. Highest power of 7 in 342! = 48 + 6 = 54.

4.175 = 5 × 5 × 7. This means we need to think of which amongst 5² and 7 would appear the least number of times in 175! In this case it is not immediately evident that whether there would be more 5²s or more 7’s in any factorial value (Because if you write any factorial 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 × 10 × 11 × 12 × 13 × 14 × 15…you can clearly see that although the 5’s appear more frequently than the 7’s it is not evident that we would have at least two fives before the 7 appears.) Hence, in this question we would need to check for both the number of 5²s and the number of 7’s.

Number of 7’s in 344!

344/7 Æ Quotient 49. 49/7 Æ Quotient 7. 7/7 Æ Quotient 1. Highest power of 7 in 344! = 49 + 7 + 1 = 57.

In order to find the number of 5²s in 344! We first need to find the number of 5’s in 344!

344/5 Æ Quotient 68. 68/5 Æ Quotient 13. 13/5 Æ Quotient 2. Number of 5’s in 344! = 68 + 13 + 2 = 83.

83 fives would obviously mean [83/2] = 41 5²s

Thus, there are 41 5²s and 57 7’s in 344!

Since, the number of 5²s are lower, they would determine the highest power of 175 that would divide 344!

The answer is 41.

5.360 = 5 × 2 × 2 × 2 × 3 × 3. This means we need to think of which amongst 2³, 3² and 5 would appear the least number of times in 520! In this case it is not immediately evident which of these three would appear least number of times. Hence, in this question we would need to check for all three – 2³s 3²s and 5s.

Number of 5’s in 520!

520/5 Æ Quotient 104. 104/5 Æ Quotient 20. 20/5 Æ Quotient 4. Highest power of 5 in 520! = 104 + 20 + 4 = 128.

In order to find the number of 3²s in 520! we first need to find the number of 3’s in 520!

520/3 Æ Quotient 173. 173/3 Æ Quotient 57. 57/3 Æ Quotient 19. 19/3 Æ Quotient 6. 6/3 Æ Quotient 2. 2/3 Æ Quotient 0. Number of 3’s in 520! = 173 + 57 + 19 + 6 + 2 = 257.

257 threes would obviously mean [257/2]= 128 3²s.

In order to find the number of 2³s in 520! we first need to find the number of 2’s in 520!

520/2 Æ Quotient 260. 260/2 Æ Quotient 130. 130/2 Æ Quotient 65. 65/2 Æ Quotient 32. 32/2 Æ Quotient 16. 16/2 Æ Quotient 8. 8/2 Æ Quotient 4. 4/2 Æ Quotient 2. 2/2 Æ Quotient 1. 1/2 Æ Quotient 0.

Number of 2’s in 520! = 260 + 130 +65 + 32 +16 + 8 + 4 + 2 + 1 = 518. 518 twos would obviously mean [518/3]= 172 2³s.

Thus, there are 128 3²s, 128 5’s and 172 2³’s in 520!

The highest power of 360 that would divide 520! would be the least of 128, 128 and 172.

The answer is 128.

Exercise for Self-practice

Solutions

1.[157/5] = 31. [31/5] = 6. [6/5] = 1. 31 + 6 + 1 = 38. Option (b) is correct.

2.No of 2’s in 157! = [157/2] + [157/4] + [157/8]… + [157/128] = 78 + 39 + 19 + 9 + 4 + 2 + 1 = 152. Hence, the number of 2²s would be [152/2] = 76.

Number of 3’s in 157! = 52 + 17 + 5 +1 = 75.

The answer would be given by the lower of these values. Hence, 75 (Option c) is correct.

3.From the above solution:

Number of 2’s in 157! = 152

Number of 3²s in 157! = [75/2]=37.

Hence, option (a) is correct.

4.2520 = 7 × 3² × 2³ × 5.

The value of n would be given by the value of the number of 7s in 50!

This value is equal to [50/7] + [50/49] = 7 + 1 = 8

Option (b) is correct.

5.12600 = 7 × 3² × 2³ × 5²

The value of ‘n’ would depend on which of number of 7s and number of 5²s is lower in 50!.

Number of 7’s in 50! = 8. Note here that if we check for 7’s we do not need to check for 3²s as there would be at least two 3’s before a 7 comes in every factorial’s value. Similarly, there would always be at least three 2’s before a 7 comes in any factorial’s value. Thus, the number of 3²s and the number of 2³s can never be lower than the number of 7s in any factorial’s value.

Number of 5s in 50! = 10 + 2 =12. Hence, the number of 5²s in 50! = [12/2] = 6.

6 will be the answer as the number of 5²s is lower than the number of 7’s.

Option (b) is correct.

6.720 = 2⁴ × 5¹ × 3²

In 77! there would be 38 + 19 + 9 + 4 + 2 + 1 = 73 twos Æ hence [73/4] = 18 2⁴s

In 77! there would be 25 + 8 + 2 = 35 threes Æ hence [35/2] = 17 3²s

In 77! there would be 15 + 3 = 18 fives

Since 17 is the least of these values, option (c) is correct.

7.In the expression given, there are three 7’s and more than three 2’s and 3’s. Thus, Option (b) is correct.

8.Checking for the number of 2’s, 3’s and 5’s in the given expression you can see that the minimum is for the number of 3’s (there are 11 of them while there are 12 5’s and more than 11 2’s) Hence, option (b) is correct.

9.The number of 7’s in the number is 6, while there are six 3’s too. Option (d) is correct.

10.The number of zeroes would depend on the number of 5’s in the value of the factorial.

72! Æ 14 + 2 = 16. Option (d) is correct.

11.The number of zeroes would depend on the number of 5’s in the value of the factorial.

77! × 42! Æ 15 + 3 = 18 (for 77!) and 8 + 1 = 9 (for 42!).

Thus, the total number of zeroes in the given expression would be 18 + 9 = 27. Option (c) is correct.

12.The number of zeroes would depend on the number of 5’s in the value of the factorial.

100! would end in 20 + 4 = 24 zeroes

200! Would end in 40 + 8 + 1 = 49 zeroes.

When you add the two numbers (one with 24 zeroes and the other with 49 zeroes at it’s end), the resultant total would end in 24 zeroes. Option (b) is correct.

13.The given expression has fifteen 2’s and seventeen 5’s. The number of zeroes would be 15 as the number of 2’s is lower in this case. Option (d) is correct.

14.1! to 4! would have no zeroes while 5! to 9! All the values would have 1 zero. Thus, a total of 5 zeroes till 9!. Going further 10! to 14! would have two zeroes each – so a total of 10 zeroes would come out of the product of 10! × 11! × 12! × 13! × 14!.

Continuing this line of thought further we get:

Number of zeroes between 15! × 16!... × 19! = 3 + 3 + 3 + 3 + 3 = 3 × 5 = 15

Number of zeroes between 20! × 21!... × 24! = 4 × 5 = 20

Number of zeroes between 25! × 26!... × 29! = 6 × 5 = 30

Number of zeroes between 30! × 31!... × 34! = 7 × 5 = 35

Number of zeroes between 35! × 36!... × 39! = 8 × 5 = 40

Number of zeroes between 40! × 41!... × 44! = 9 × 5 = 45

Number of zeroes between 45! × 46!... × 49! = 10 × 5 = 50

Number of zeroes for 50! = 12

Thus, the total number of zeroes for the expression 1! × 2! × 3! …. × 50! = 5 + 10 +15 + 20 + 30 + 35 + 40 + 45 + 50 + 12 = 262 zeroes. Option (c) is correct.

15.The number of 5’s is `15 while the number of 2’s is much more. Option (b) is correct.

16.The number of zeroes would depend on the number of 5’s in the value of the factorial.

100! would end in 20 + 4 = 24 zeroes

200! Would end in 40 + 8 + 1 = 49 zeroes.

When you multiply the two numbers (one with 24 zeroes and the other with 49 zeroes at it’s end), the resultant total would end in 24 + 49 = 73 zeroes. Option (b) is correct.

Co-Prime or Relatively Prime Numbers Two or more numbers that do not have a common factor are known as co-prime or relatively prime. In other words, these numbers have a highest common factor of unity.

If two numbers m and n are relatively prime and the natural number x is divisible by both m and n independently then the number x is also divisible by mn.

Key Concept 1: The spotting of two numbers as co-prime has a very important implication in the context of the two numbers being in the denominators of fractions.

The concept is again best understood through an example:

Suppose, you are doing an operation of the following format – M/8 + N/9 where M & N are integers.

What are the chances of the result being an integer, if M is not divisible by 8 and N is not divisible by 9? A little bit of thought will make you realise that the chances are zero. The reason for this is that 8 and 9 are co-prime and the decimals of co-prime numbers never match each other.

Note: this will not be the case in the case of:

M/3 + N/27.

In this case even if 3 and 27 are not dividing M and N respectively, there is a possibility of the values of M and N being such that you have an integral answer.

For instance: 5/3 + 36/27 = 81/27 = 3

The result will never be integral if the two denominators are co-prime.

Note: This holds true even for expressions of the nature A/7 – B/6 etc.

This has huge implications for problem solving especially in the case of solving linear equations related to word based problems. Students are advised to try to use these throughout Blocks I, II and III of this book.

Example: Find all five-digit numbers of the form 34 x 5Y that are divisible by 36.

Solution: 36 is a product of two co-primes 4 and 9. Hence, if 34 x 5y is divisible by 4 and 9, it will also be divisible by 36. Hence, for divisibility by 4, we have that the value of y can be 2 or 6. Also, if y is 2 the number becomes 34 x 52. For this to be divisible by 9, the addition of 3 + 4 + x + 5 + 2 should be divisible by 9. For this x can be 4.

Hence the number 34452 is divisible by 36.

Also for y = 6, the number 34 x 56 will be divisible by 36 when the addition of the digits is divisible by 9. This will happen when x is 0 or 9. Hence, the numbers 34056 and 34956 will be divisible by 36.

Exercise for Self-practice

Find all numbers of the form 56x3y that are divisible by 36.

Find all numbers of the form 72xy that are divisible by 45.

Find all numbers of the form 135xy that are divisible by 45.

Find all numbers of the form 517xy that are divisible by 89.

Even Numbers: All integers that are divisible by 2 are even numbers. They are also denoted by 2n.

Example: 2, 4, 6, 12, 122, –2, –4, –12.

Also note that zero is an even number.

2 is the lowest positive even number.

Odd Numbers: All integers that are not divisible by 2 are odd numbers. Odd number leave a remainder of 1 on being divided by 2. They are denoted by 2n + 1 or 2n – 1.

Lowest positive odd number is 1.

Example: –1, –3, –7, –35, 3, 11, etc.

Complex Numbers: The arithmetic combination of real numbers and imaginary numbers are called complex numbers.

Alternately: All numbers of the form a + ib, where i = ÷ – 1 are called complex number.

Twin Primes: A pair of prime numbers are said to be twin prime when they differ by 2.

Example: 3 and 5 are Twin Primes, so also are 11 and 13.

Perfect Numbers: A number n is said to be a perfect number if the sum of all the divisors of n (including n) is equal to 2n.

Example: 6 = 1 × 2 × 3 sum of the divisors = 1 + 2 + 3 + 6 = 12 = 2 × 6

28 = 1, 2, 4, 7, 14, 28, = 56 = 2 × 28

Task for student: Find all perfect numbers below 1000.

Mixed Numbers: A number that has both an integral and a fractional part is known as a mixed number.

Triangular Numbers: A number which can be represented as the sum of consecutive natural numbers starting with 1 are called as triangular numbers.

e.g.: 1 + 2 + 3 + 4 = 10. So, 10 is a triangular number.

THE REMAINDER THEOREM

Consider the following question:

17 × 23.

Suppose you have to find the remainder of this expression when divided by 12.

We can write this as:

17 × 23 = (12 + 5) × (12 + 11)

Which when expanded gives us:

12 × 12 + 12 × 11 + 5 × 12 + 5 × 11

You will realise that, when this expression is divided by 12, the remainder will only depend on the last term above:

Thus, gives the same remainder as

Hence, 7.

This is the remainder when 17 × 23 is divided by 12.

Learning Point: In order to find the remainder of 17 × 23 when divided by 12, you need to look at the individual remainders of 17 and 23 when divided by 12. The respective remainders (5 and 11) will give you the remainder of the original expression when divided by 12.

Mathematically, this can be written as:

The remainder of the expression [A × B × C + D × E]/M, will be the same as the remainder of the expression [A_R × B_R × C_R + D_R × E_R]/M.

Where A_R is the remainder when A is divided by M,

B_R is the remainder when B is divided by M,

C_R is the remainder when C is divided by M

D_R is the remainder when D is divided by M and

E_R is the remainder when E is divided by M,

We call this transformation as the remainder theorem transformation and denote it by the sign

Thus, the remainder of

1421 × 1423 × 1425 when divided by 12 can be given as:

= .

gives us a remainder of 3.

In the above question, we have used a series of remainder theorem transformations (denoted by ) and equality transformations to transform a difficult looking expression into a simple expression.

Try to solve the following questions on Remainder theorem:

Find the remainder in each of the following cases:

1.17 × 23 × 126 × 38 divided by 8.

2.243 × 245 × 247 × 249 × 251 divided by 12.

3. + .

4..

5. + .

USING NEGATIVE REMAINDERS

Consider the following question:

Find the remainder when: 14 × 15 is divided by 8.

The obvious approach in this case would be

= 2 (Answer).

However there is another option by which you can solve the same question:

When 14 is divided by 8, the remainder is normally seen as + 6. However, there might be times when using the negative value of the remainder might give us more convenience. Which is why you should know the following process:

Concept Note: Remainders by definition are always non-negative. Hence, even when we divide a number like – 27 by 5 we say that the remainder is 3 (and not – 2). However, looking at the negative value of the remainder—it has its own advantages in Mathematics as it results in reducing calculations.

Thus, when a number like 13 is divided by 8, the remainder being 5, the negative remainder is – 3.

Thus will give us R Æ 2.

Consider the advantage this process will give you in the following question:

2.

(The alternative will involve long calculations. Hence, the principle is that you should use negative remainders wherever you can. They can make life much simpler!!!)

What if the Answer Comes Out Negative

For instance, R Æ .

But, we know that a remainder of –24, equals a remainder of 42 when divided by 66. Hence, the answer is 42.

Of course nothing stops you from using positive and negative remainders at the same time in order to solve the same question –

Thus R Æ –1 R Æ 8.

Dealing with large powers There are two tools which are effective in order to deal with large powers –

(A)If you can express the expression in the form , the remainder will become 1 directly. In such a case, no matter how large the value of the power n is, the remainder is 1.

For instance, 1. In such a case the value of the power does not matter.

(B). In such a case using –1 as the remainder it will be evident that the remainder will be +1 if n is even and it will be –1 (Hence a – 1) when n is odd.

e.g.: 7

ANOTHER IMPORTANT POINT

Suppose you were asked to find the remainder of 14 divided by 4. It is clearly visible that the answer should be 2.

But consider the following process:

14/4 = 7/2 1 (The answer has changed!!)

What has happened?

We have transformed 14/4 into 7/2 by dividing the numerator and the denominator by 2. The result is that the original remainder 2 is also divided by 2 giving us 1 as the remainder. In order to take care of this problem, we need to reverse the effect of the division of the remainder by 2. This is done by multiplying the final remainder by 2 to get the correct answer.

AN APPLICATION OF REMAINDER THEOREM

Finding the last two digits of an expression:

Suppose you had to find the last 2 digits of the expression:

22 × 31 × 44 × 27 × 37 × 43

The remainder the above expression will give when it is divided by 100 is the answer to the above question.

Hence, to answer the question above find the remainder of the expression when it is divided by 100.

Solution:

= (on dividing by 4)

=

= 14

Thus the remainder being 14, (after division by 4). The actual remainder should be 56.

[Don’t forget to multiply by 4 !!]

Hence, the last 2 digits of the answer will be 56.

Using negative remainders here would have helped further.

Exercise for Self-practice

Solutions

1.The remainder would be given by: (5 + 7 + 10 + 23 + 27)/34 = 72/34 Æ remainder = 4. Option (b) is correct.

2.The remainder would be given by: (5 × 7 × 10 × 23 × 27)/34 Æ 35 × 230 × 27/ 34 Æ 1 × 26 × 27/34 = 702/34 Æ remainder = 22. Option (a) is correct.

3.The remainder would be given by: (5 × 7 × 10 × 23 × 27 × 3)/34 Æ 35 × 230 × 27 × 3/34 Æ 1 × 26 × 81/34 Æ 26 × 13/34 = 338/34 Æ remainder = 32. Option (a) is correct.

4.43¹⁹⁷/7 Æ 1¹⁹⁷/7 Æ remainder =1. Option (d) is correct.

5.51²⁰³/7 Æ 2²⁰³/7 = (2³)⁶⁷ × 2²/7 = 8⁶⁷ × 4/7 Æ remainder = 4. Option (a) is correct.

6.59²⁸/7 Æ 3²⁸/7 = (3⁶)⁴ × 3⁴/7 = 729⁶ × 81/7 Æ remainder = 4. Option (b) is correct.

7.67⁹⁹/7 Æ 4⁹⁹/7 = (4³)³³/7 = 64³³/7 Æ remainder = 1. Option (d) is correct.

8.75⁸⁰/7 Æ 5⁸⁰/7 = (5⁶)¹³ × 5 ²/7 Æ 1¹³ × 25/7 Æ remainder = 4. Option (a) is correct.

9.41⁷⁷/7 Æ 6⁷⁷/7 Æ remainder = 6 (as the expression is in the form aⁿ/(a + 1). Option (c) is correct.

10.21⁸⁷⁵/17 Æ 4⁸⁷⁵/17 = (4⁴)ⁿ × 4³/17 = 256ⁿ × 64/17 Æ 1ⁿ × 13/17 Æ remainder =13. Option (b) is correct.

11.54¹²⁴/17 Æ 3¹²⁴/17. At this point, like in each of the other questions solved above, we need to plan the power of 3 which would give us a convenient remainder of either 1 or –1. As we start to look for remainders that powers of 3 would have when divided by 17, we get that at the power 3⁶ the remainder is 15. If we convert this to –2 we will get that at the fourth power of 3⁶, we should get a 16/17 situation (as –2 × – 2 × – 2 × – 2 = 16). This means that at a power of 3²⁴ we are getting a remainder of 16 or –1. Naturally then if we double the power to 3⁴⁸, the remainder would be 1.

With this thinking we can restart solving the problem:

3¹²⁴/17 = 3⁴⁸ × 3⁴⁸ × 3²⁴ × 3⁴/17 Æ 1 × 1 × 16 × 81/17 Æ 16 × 13/17 =208/17 Æ remainder = 4. Option (a) is correct.

(Note that if we are dividing a number by 17 and if we see the remainder as 15, we can logically say that the remainder is –2 – even though negative remainders are not allowed in mathematics)

12.Using the logic developed in Question 11 above, we have 83²⁶¹/17 Æ 15²⁶¹/17 Æ

(–2) ²⁶¹/17 Æ (–2⁴)⁶⁵ × (–2) /17 Æ 16⁶⁵ × (–2)/17 Æ (–1) × (–2)/17 Æ remainder = 2. Option (d) is correct.

13.25¹⁰²/17 Æ 8¹⁰²/17 = 2³⁰⁶/17 = (2⁴)⁷⁶ × 2²/17 Æ 16⁷⁶ × 4/17 Æ 1 × 4/17 Æ remainder = 4. Option (c) is correct.

BASE SYSTEM

All the work we carry out in our number system is called as the decimal system. In other words we work in the decimal system. Why is it called decimal?? It is because there are 10 digits in the system 0–9.

However, depending on the number of digits contained in the base system other number systems are also possible. Thus a number system with base 2 is called the binary number system and will have only two digits 0 and 1. Some of the most commonly used systems are: Binary (base 2), Octal (base 8), Hexadecimal (base 16).

Binary system has 2 digits : 0, 1. Octal has 8 digits : – 0, 1, 2, 3, ... 7.

Hexadecimal has 16 digits – 0, 1, 2, ... 9, A, B, C, D, E, F.

Where A has a value 10, B = 11 and so on.

Before coming to the questions asked under this category, let us first look at a few issues with regard to converting numbers between different base systems.

1. Conversion from any base system into decimal:

Suppose you have to write the decimal equivalent of the base 8 number 146₈.

In such a case, follow the following structure for conversion:

146₈ = 1 × 8² + 4 × 8¹ + 6 × 8⁰

= 64 + 32 + 6 = 102.

Try to write the decimal equivalents of the following numbers:

143₅, 143₆, 143_7,143_8,143₉

1256₇, 1256₈, 1256₉.

2. Conversion of a number in decimals into any base:

Suppose you have to find out the value of the decimal number 347 in base 6. The following process is to be adopted:

Step 1: Find the highest power of the base (6 in this case) that is contained in 347. In this case you will realise that the value of 6³ = 216 is contained in 347, while the value of 6⁴ = 1296 is not contained in 347. Hence, we realise that the highest power of 6 contained in 347 is 3. This should make you realise that the number has to be constructed by using the powers 6⁰, 6¹, 6², 6³ respectively. Hence, a 4-digit number.

Structure of number: - - - -

Step 2: We now need to investigate how many times each of the powers of 6 is contained in 347. For this we first start with the highest power as found above. Thus we can see that 6³ (216) is contained in 347 once. Hence our number now becomes:

1 - - -

That is, we now know that the first digit of the number is 1. Besides, when we have written the number 1 in this place, we have accounted for a value of 216 out of 347. This leaves us with 131 to account for.

We now need to look for the number of times 6² is contained in 131. We can easily see that 6² = 36 is contained in 131 three times. Thus, we write 3 as the next digit of our number which will now look like:

1 3 - -

In other words we now know that the first two digits of the number are 13. Besides, when we have written the number 3 in this place, we have accounted for a value of 108 out of 131 which was left to be accounted for. This leaves us with 131–108 = 23 to account for.

We now need to look for the number of times 6¹ is contained in 23. We can easily see that 6¹ = 6 is contained in 23 three times. Thus, we write 3 as the next digit of our number which will now look like:

1 3 3 -

In other words we now know that the first three digits of the number are 133. Besides, when we have written the number 3 in this place, we have accounted for a value of 18 out of the 23 which was left to be accounted for. This leaves us with 23 – 18 = 5 to account for.

The last digit of the number corresponds to 6⁰ = 1. In order to make a value of 5 in this place we will obviously need to use this power of 6, 5 times thus giving us the final digit as 5. Hence, our number is:

1 3 3 5.

A few points you should know about base systems:

(1)In single digits there is no difference between the value of the number—whichever base we take. For example, the equality 5₆ = 5₇ = 5₈ = 5₉ = 5₁₀.

(2)Suppose you have a number in base x. When you convert this number into its decimal value, the value should be such that when it is divided by x, the remainder should be equal to the units digit of the number in base x.

In other words, 342₈ will be a number of the form 8n + 2 in base 10. You can use this principle for checking your conversion calculations.

The following table gives a list of decimal values and their binary, octal and hexadecimal equivalents:

Illustrations

1.The number of x digit numbers in nth base system will be

(a)n^x   (b) n^{x– 1}

(c) n^x – n  (d) nx – n^{(x – 1)}

Solution Base Æ n, digit Æ x

So, required number of numbers = nx – n⁽x ^{– 1)}

2.The number of 2 digit numbers in binary system is

(a) 2    (b) 90

(c) 10    (d) 4

Solution By using the formula, we get the required number of numbers = 2² – 2¹ = 2

fi Option (a)

3.The number of 5 digit numbers in binary system is

(a)48  (b) 16

(c)32  (d) 20

Solution Required number of numbers = 2⁵ – 2⁴ = 32 – 16 = 16

fi Option (b)

4.I celebrate my birthday on 12^th January on earth. On which date would I have to celebrate my birthday if I were on a planet where binary system is being used for counting. (The number of days, months and years are same on both the planets.)

(a)11^th Jan   (b) 111^th Jan

(c)110^th Jan   (d) 1100^th Jan

Solution On earth (decimal system is used). 12^th Jan fi 12^th Jan

The 12^th day on the planet where binary system is being used will be called

(12)₁₀ = (?)₂

=

i.e., 1100^th day on that planet

So, 12^th January on earth = 1100^th January on that planet

fi Option (d)

5.My year of birth is 1982. What would the year have been instead of 1982 if base 12 were used (for counting) instead of decimal system?

(a) 1182   (b) 1022

(c) 2082   (d) 1192

Solution The required answer will equal to (1982)₁₀ = (?)₁₂.

= Æ

1 × 12³ + 1 × 12² + 9 × 12¹ + 2 × 9⁰ = 1728 + 144 + 108 + 2 = 1982

Hence, the number (1192)₁₂ represents 1982 in our base system.

fi Option (d)

6.203 in base 5 when converted to base 8, becomes

(a) 61   (b) 53

(c)145   (d) 65

Solution (203)₅ = (?)₁₀

= 2 × 5² + 0 × 5¹ + 3 × 5⁰

= 50 + 0 + 3 = 53

Now,

(53)₁₀ = (?)8

=

= (203)₅ = (65)₈

fi Option (d)

7. (52)₇+ 46₈ = (?)₁₀

(a)(75) ₁₀   (b) (50)₁₀

(c)(39) ₃₉   (d) (28) ₁₀

Solution (52)₇ = (5 × 7¹ + 2 × 7⁰)₁₀ = (37)₁₀

also, (46)₈ = (4 × 8¹ + 6 × 8⁰)₁₀ = (38)₁₀

sum = (75)₁₀

fi Option (a)

8.(23)₅ + (47) ₉ = (?)₈

(a)70   (b) 35

(c)64   (d) 18

Solution (23)₅ = (2 × 5¹ + 3 × 5⁰)₁₀ = (13)₁₀ = (1 × 8¹ + 5 × 8⁰)₈ = (15)₈

also,(47)₉ = (4 × 9¹ + 7 × 9⁰) ₁₀ = (43)₁₀

= (5 × 8¹ + 3 × 8⁰) ₈ = (53)₈

sum = (13)₁₀ + (43)₁₀ = (56)₁₀ Æ (70)₈

fi Option (a)

9.(11)₂ + (22)₃ + (33)₄ + (44)₅ + (55)₆ + (66)₇ + (77)₈ + (88)₉ = (?)₁₀

(a) 396   (b) 276

(c)250   (d) 342

Solution (11)₂ = (1 × 2¹ + 1 × 2⁰)₁₀ = (3) ₁₀

(22)₃ = (2 × 3¹ + 2 × 3⁰)₁₀ = (8)₁₀

(33)₄ = (3 × 4¹ + 3 × 4⁰) ₁₀ = (15)₁₀

(44)₅ = (4 × 5¹ + 4 × 5⁰)₁₀ = (24)₁₀

(55)₆ = (5 × 6¹ + 5 × 6⁰)₁₀ = (35)₁₀

(66)₇ = (6 × 7¹ + 6 × 7⁰)₁₀ = (48)₁₀

(77)₈ = (7 × 8¹ + 7 × 8⁰)₁₀ = (63)₁₀

(88)₉ = (8 × 9¹ + 8 × 9⁰)₁₀ = (80)₁₀

sum = (276)₁₀

fi Option (b)

10.(24)₅ × (32)₅ = (?)₅

(a)1423  (b) 1422

(c)1420  (d) 1323

Solution (24)₅ =14₁₀ and 32₅ = 17₁₀. Hence, the required answer can be got by 14 × 17 = 238₁₀ = 1 × 5³ + 4 × 5² + 2 × 5¹ + 3 × 5⁰ Æ 1423 as the correct answer.

Alternately, you could multiply directly in base 5 as follows:

Unit’s digit of the answer would correspond to: 4 × 2 = 8 Æ 13₅. Hence, we write 3 in the units place and carry over 1.

(Note that in this process when we are doing 4 × 2 we are effectively multiplying individual digits of one number with individual digits of the other number. In such a case we can write 4 × 2 = 8 by assuming that both the numbers are in decimal system as the value of a single digit in any base is equal.)

The tens digit will be got by: 2 × 2 + 4 × 3 = 16 + 1 = 17 Æ 32₅

Hence, we write 2 in the tens place and carry over 3 to the hundreds place.

Where we get 3 × 2 + 3 = 9 Æ 14

Hence, the answer is 14.

fi Option (a)

11.In base 8, the greatest four digit perfect square is

(a)9801   (b) 1024

(c)8701   (d) 7601

Solution In base 10, the greatest 4 digit perfect square = 9801

In base 9, the greatest 4 digits perfect square = 8701

In base 8, the greatest 4 digits perfect square = 7601

Alternately, multiply (77)₈ × (77)₈ to get 7601 as the answer.

Unit’s Digit

(A)The unit’s digit of an expression will be got by getting the remainder when the expression is divided by 10.

Thus for example if we have to find the units digit of the expression:

17 × 22 × 36 × 54 × 27 × 31 × 63

We try to find the remainder –

= = 6.

Hence, the required answer is 6.

This could have been directly got by multiplying: 7 × 2 × 6 × 4 × 7 × 1 × 3 and only accounting for the units’ digit.

(B)Unit’s digits in the contexts of powers –

Study the following table carefully.

Unit’s digit when ‘N’ is raised to a power

In the table above, if you look at the columns corresponding to the power 5 or 9 you will realize that the unit’s digit for all numbers is repeated (i.e. it is 1 for 1, 2 for, 3 for 3….9 for 9.)

This means that whenever we have any number whose unit’s digit is ‘x’ and it is raised to a power of the form 4n + 1, the value of the unit’s digit of the answer will be the same as the original units digit.

Illustrations: (1273)¹⁰¹ will give a unit’s digit of 3. (1547)²⁵ will give a units digit of 7 and so forth.

Thus, the above table can be modified into the form –

Value of power

[Remember, at this point that we had said (in the Back to School section of Part 1) that all natural numbers can be expressed in the form 4n + x. Hence, with the help of the logic that helps us build this table, we can easily derive the units digit of any number when it is raised to a power.)

A special Case

Question: What will be the unit’s digit of (1273)^122!?

Solution: 122! is a number of the form 4n. Hence, the answer should be 1. [Note: 1 here is derived by thinking of it as 3 (for 4n + 1), 9 (for 4n + 2), 7 (for 4n + 3), 1(for 4n)]

Exercise for Self-practice

Find the Units digit in each of the following cases:

Solutions

1.The units digit would be given by the units digit of the multiplication of 4 x 6 x 6 x6 = 4

2.0

3.7 × 3 × 1 × 2 + 0 Æ 2 + 0 = 2

4.8 + 4 – 9 Æ 3

5.3 × 1 × 9 × 8 × 6 = 6

6.7 × 7 × 3 × 1 × 2 × 3 × 2 = 4

7.Since we have a 5 multiplied with odd numbers, the units digit would naturally be 5.

8.5 × 2 Æ 0

9.5 + 2 Æ 7

10. 2 × 1 Æ 2

11.5 × 7 × 3 Æ 5

12.3 × 2 × 3 Æ 8

13.2 × 3 × 4 × 6 × 7 × 8 × 9 Æ 6

14.8 × 1 × 4 × 6 × 7 × 4 Æ 6

15.4 + 8 + 7 + 9 Æ 8

WORKED-OUT PROBLEMS

Problem 1.1 Find the number of zeroes in the factorial of the number 18.

Solution 18! contains 15 and 5, which combined with one even number give zeroes. Also, 10 is also contained in 18!, which will give an additional zero. Hence, 18! contains 3 zeroes and the last digit will always be zero.

Problem 1.2 Find the numbers of zeroes in 27!

Solution 27! = 27 × 26 × 25 × … × 20 × … × 15 × … × 10 × … × 5 × … × 1.

A zero can be formed by combining any number containing 5 multiplied by any even number. Similarly, everytime a number ending in zero is found in the product, it will add an additional zero. For this problem, note that 25 = 5 × 5 will give 2 zeroes and zeroes will also be got by 20, 15, 10 and 5. Hence 27! will have 6 zeroes.

Short-cut method: Number of zeroes is 27! Æ [27/5] + [27/25]

where [x] indicates the integer just lower than the fraction

Hence, [27/5] = 5 and [27/5²] = 1, 6 zeroes

Problem 1.3 Find the number of zeroes in 137!

Solution [137/5] + [137/5²] + [137/5³]

= 27 + 5 + 1 = 33 zeroes

(since the restriction on the number of zeroes is due to the number of fives.)

Exercise for Self-practice

Find the number of zeroes in

Answers

Problem 1.4 What exact power of 5 divides 87!?

Solution [87/5] + [87/25] = 17 + 3 = 20

Problem 1.5 What power of 8 exactly divides 25!?

Solution If 8 were a prime number, the answer should be [25/8] = 3. But since 8 is not prime, use the following process.

The prime factors of 8 is 2 × 2 × 2. For divisibility by 8, we need three twos. So, everytime we can find 3 twos, we add one to the power of 8 that divides 25! To count how we get 3 twos, we do the following. All even numbers will give one ‘two’ at least [25/2] = 12

Also, all numbers in 25! divisible by 2² will give an additional two [25/2²] = 6

Further, all numbers in 25! divisible by 2³ will give a third two. Hence [25!/2³] = 3

And all numbers in 25! divisible by 2⁴ will give a fourth two. Hence [25!/2⁴] = 1

Hence, total number of twos in 25! is 22. For a number to be divided by 8, we need three twos. Hence, since 25! has 22 twos, it will be divided by 8 seven times.

Problem 1.6 What power of 15 divides 87! exactly?

Solution 15 = 5 × 3. Hence, everytime we can form a pair of one 5 and one 3, we will count one.

87! contains – [87/5] + [87/5²] = 17 + 3 = 20 fives

Also 87! contains – [87/3] + [87/3²] + [87/3³] + [87/3⁴] = 29 +…(more than 20 threes).

Hence, 15 will divide 87! twenty times since the restriction on the power is because of the number of 5s and not the number of 3s.

In fact, it is not very difficult to see that in the case of all factors being prime, we just have to look for the highest prime number to provide the restriction for the power of the denominator.

Hence, in this case we did not need to check for anything but the number of 5s.

Exercise for Self-practice

(a) What power of 30 will exactly divide 128!

Hints: [128/5] + [128/5²] + [128/5³]

(b) What power of 210 will exactly divide 142!

Problem 1.7 Find the last digit in the expression (36472)^123! × (34767)^76!.

Solution If we try to formulate a pattern for 2 and its powers and their units digit, we see that the units digit for the powers of 2 goes as: 2, 4, 8, 6, 2, 4, 8, 6, 2, 4, 8, 6 and so on. The number 2 when raised to a power of 4n + 1 will always give a units digit of 2. This also means that the units digit for 2⁴ⁿ will always end in 6. The power of 36472 is 123! . 123! can be written in the form 4n. Hence, (36472)^123! will end in 6.

The second part of the expression is (34767)^76!. The units digit depends on the power of 7. If we try to formulate a pattern for 7 and its powers and their units digit, we see that the units digit for the powers of 7 go as: 7 9 3 1 7 9 3 1 and so on. This means that the units digit of the expression 7⁴ⁿ will always be 1.

Since 76! can be written as a multiple of 4 as 4n, we can conclude that the unit’s digit in (34767)^76! is 1.

Hence the units digit of (36472)^123! × (34767)^76! will be 6.

Counting

Problem 1.8 Find the number of numbers between 100 to 200 if

(i)Both 100 and 200 are counted.

(ii)Only one of 100 and 200 is counted.

(iii)Neither 100 nor 200 is counted.

Solution

(i)Both ends included-Solution: 200 – 100 + 1 = 101

(ii)One end included-Solution: 200 – 100 = 100

(iii)Both ends excluded-Solution: 200 – 100 – 1 = 99.

Problem 1.9 Find the number of even numbers between 122 and 242 if:

(i)Both ends are included.

(ii)Only one end is included.

(iii)Neither end is included.

Solution

(i)Both ends included—Solution: (242 – 122)/2 = 60 + 1 = 61

(ii)One end included-Solution: (242 – 122)/2 = 60

(iii)Both ends excluded-Solution: (242 – 122)/2 – 1 = 59

Exercise for Self-practice

(a)Find the number of numbers between 140 to 259, both included, which are divisible by 7.

(b)Find the number of numbers between 100 to 200, that are divisible by 3.

Problem 1.10 Find the number of numbers between 300 to 400 (both included), that are not divisible by 2, 3, 4, and 5.

Solution Total numbers: 101

Step 1: Not divisible by 2 = All even numbers rejected: 51

Numbers left: 50.

Step 2: Of which: divisible by 3 = first number 300, last number 399. But even numbers have already been removed, hence count out only odd numbers between 300 and 400 divisible by 3. This gives us that:

First number 303, last number 399, common difference 6

So, remove: [(399 – 303)/6] + 1 = 17.

\ 50 – 17 = 33 numbers left.

We do not need to remove additional terms for divisibility by 4 since this would eliminate only even numbers (which have already been eliminated)

Step 3: Remove from 33 numbers left all odd numbers that are divisible by 5 and not divisible by 3.

Between 300 to 400, the first odd number divisible by 5 is 305 and the last is 395 (since both ends are counted, we have 10 such numbers as: [(395 – 305)/10 + 1 = 10].

However, some of these 10 numbers have already been removed to get to 33 numbers.

Operation left: Of these 10 numbers, 305, 315…395, reduce all numbers that are also divisible by 3. Quick perusal shows that the numbers start with 315 and have common difference 30.

Hence [(Last number – First number)/Difference + 1] = [(375 – 315)/30 + 1] = 3

These 3 numbers were already removed from the original 100. Hence, for numbers divisible by 5, we need to remove only those numbers that are odd, divisible by 5 but not by 3. There are 7 such numbers between 300 and 400.

So numbers left are: 33 – 7 = 26.

Exercise for Self-practice

Find the number of numbers between 100 to 400 which are divisible by either 2, 3, 5 and 7.

Problem 1.11 Find the number of zeroes in the following multiplication: 5 × 10 × 15 × 20 × 25 × 30 × 35 × 40 × 45 × 50.

Solution The number of zeroes depends on the number of fives and the number of twos. Here, close scrutiny shows that the number of twos is the constraint. The expression can be written as

5 × (5 × 2) × (5 × 3) × (5 × 2 × 2) × (5 × 5) × (5 × 2 × 3) × (5 × 7) × (5 × 2 × 2 × 2) × (5 × 3 × 3) × (5 × 5 × 2)

Number of 5s – 12, Number of 2s – 8.

Hence: 8 zeroes.

Problem 1.12 Find the remainder for [(73 × 79 × 81)/11].

Solution The remainder for the expression: [(73 × 79 × 81)/11] will be the same as the remainder for [(7 × 2 × 4)/11]

That is, 56/11 fi remainder = 1

Problem 1.13 Find the remainder for (3⁵⁶⁰/8).

Solution (3⁵⁶⁰/8) = [(3²)²⁸⁰/8] = (9²⁸⁰/8)

= [9.9.9…(280 times)]/8

remainder for above expression = remainder for [1.1.1…(280 times)]/8 fi remainder = 1.

Problem 1.14 Find the remainder when (2222⁵⁵⁵⁵ + 5555²²²²)/7.

Solution This is of the form: [(2222⁵⁵⁵⁵)/7 + (5555²²²²)/7]

We now proceed to find the individual remainder of : (2222⁵⁵⁵⁵)/7. Let the remainder be R₁.

When 2222 is divided by 7, it leaves a remainder of 3.

Hence, for remainder purpose (2222⁵⁵⁵⁵)/7 (3⁵⁵⁵⁵/7) = (3.3⁵⁵⁵⁴)/7 = [3(3²)²⁷⁷⁷]/7 = [3.(7 + 2)²⁷⁷⁷]/7 (3.2²⁷⁷⁷)/7 = (3.2² ◊ 2²⁷⁷⁵)/7 = [3.2² ◊ (2³)⁹²⁵]/7

= [3.2² ◊ (8)⁹²⁵] /7 (12/7) Remainder = 5.

Similarly, (5555²²²²)/7 (4²²²²)/7 = [(2²)²²²²]/7 = (2)⁴⁴⁴⁴/7 = (2.2⁴⁴⁴³)/7 = [2.(2³)¹⁴⁸¹]/7 = [2.(8)¹⁴⁸¹]/7 [2.(1)¹⁴⁸¹]/7 Æ 2 (remainder).

Hence, (2222⁵⁵⁵⁵)/7 + (5555²²²²)/7 (5 + 2)/7 fi Remainder = 0

Problem 1.15 Find the GCD and the LCM of the numbers 126, 540 and 630.

Solution The standard forms of the numbers are:

126 Æ 3 × 3 × 7 × 2 Æ 3² × 7 × 2

540 Æ 3 × 3 × 3 × 2 × 2 × 5 Æ 2² × 3³ × 5

630 Æ 3 × 3 × 5 × 2 × 7 Æ 2 × 3² × 5 × 7

For GCD we use Intersection of prime factors and the lowest power of all factors that appear in all three numbers. 2 × 3² = 18.

For LCM Æ Union of prime factors and highest power of all factors that appear in any one of the three numbers fi 2² × 3³ × 5 × 7 = 3780.

Exercise for Self-practice

Find the GCD and the LCM of the following numbers:

Problem 1.16 The ratio of the factorial of a number x to the square of the factorial of another number, which when increased by 50% gives the required number, is 1.25. Find the number x.

Solution Solve through options: Check for the conditions mentioned. When we check for option (a) we get 6 ! = 720 and (4!)² = 576 and we have 6 !/(4!)² = 1.25, which is the required ratio.

Hence the answer is (a)

Problem 1.17 Three numbers A, B and C are such that the difference between the highest and the second highest two-digit numbers formed by using two of A, B and C is 5. Also, the smallest two two-digit numbers differ by 2. If B> A > C then what is the value of B?

Solution Since B is the largest digit, option (a) is rejected. Check for option (b).

If B is 6, then the two largest two-digit numbers are 65 and 60 (Since, their difference is 5) and we have B = 6, A = 5 and C = 0.

But with this solution we are unable to meet the second condition. Hence (b) is not the answer. We also realise here that C cannot be 0.

Check for option (c).

B is 7, then the nos. are 76 and 71 or 75 and 70. In both these cases, the smallest two two-digit numbers do not differ by 2.

Hence, the answer is not (c).

Hence, option (d) is the answer

[To confirm, put B = 8, then the solution A = 6 and C = 1 satisfies the 2nd condition.]

Problem 1.18 Find the remainder when 2851 × (2862)² × (2873)³ is divided by 23.

Solution We use the remainder theorem to solve the problem. Using the theorem, we see that the following expressions have the same remainder.

fi

fi

fi fi

fi fi fi Remainder is 18.

Problem 1.19 For what maximum value of n will the expression be an integer?

Solution For to be a integer, we need to look at the prime factors of 504 Æ

504 = 3² × 7 × 8 = 2³ × 3² × 7

We thus have to look for the number of 7s, the number of 2³s and the number of 3²s that are contained in 10200!. The lowest of these will be the constraint value for n.

To find the number of 2³s we need to find the number of 2s as

+

+ + + + +

+ + +

where [ ] is the greatest integer function.

= 5100 + 2550 + 1275 + 637 + 318 + 159 + 79 + 39 + 19 + 9 + 4 + 2 + 1

Number of twos = 10192

Hence, number of 2³ = 3397

Similarly, we find the number of 3s as

Number of threes =

+

= 3400 + 1133 + 377 + 125 + 41 + 13 + 4 + 1

Number of threes = 5094

\ Number of 3² = 2547

Similarly we find the number of 7s as

= 1457 + 208 + 29 + 4 = 1698.

Thus, we have, 1698 sevens, 2547 nines and 3397 eights contained in 10200!.

The required value of n will be given by the lowest of these three [The student is expected to explore why this happens]

Hence, answer = 1698.

Short cut We will look only for the number of 7s in this case. Reason: 7 > 3 × 2. So, the number of 7s must always be less than the number of 2³.

And 7 > 2 × 3, so the number of 7s must be less than the number of 3².

Recollect that earlier we had talked about the finding of powers when the divisor only had prime factors. There we had seen that we needed to check only for the highest prime as the restriction had to lie there.

In cases of the divisors having composite factors, we have to be slightly careful in estimating the factor that will reflect the restriction. In the above example, we saw a case where even though 7 was the lowest factor (in relation to 8 and 9), the restriction was still placed by 7 rather than by 9 (as would be expected based on the previous process of taking the highest number).

Problem 1.20 Find the units digit of the expression: 78⁵⁵⁶² × 56²⁵⁶ × 97¹²⁵⁰.

Solution We can get the units digits in the expression by looking at the patterns followed by 78, 56 and 97 when they are raised to high powers.

In fact, for the last digit we just need to consider the units digit of each part of the product.

A number (like 78) having 8 as the units digit will yield units digit as

Similarly,

56¹ Æ 6  Æ 56²⁵⁶ will yield 6 as

56² Æ 6  the units digit.

56³ Æ 6

Similarly,

Hence, the required units digit is given by 4 × 6 × 9 Æ 6 (answer).

Problem 1.21 Find the GCD and the LCM of the numbers P and Q where P = 2³ × 5³ × 7² and Q = 3³ × 5⁴.

Solution GCD or HCF is given by the lowest powers of the common factors.

Thus,GCD = 5³.

LCM is given by the highest powers of all factors available.

Thus,LCM = 2³ × 3³ × 5⁴ × 7²

Problem 1.22 A school has 378 girl students and 675 boy students. The school is divided into strictly boys or strictly girls sections. All sections in the school have the same number of students. Given this information, what are the minimum number of sections in the school.

Solution The answer will be given by the HCF of 378 and 675.

378 = 2 × 3³ × 7

675 = 3³ × 5²

Hence, HCF of the two is 3³ = 27.

Hence, the number of sections is given by: = 14 + 25 = 39 sections.

Problem 1.23 The difference between the number of numbers from 2 to 100 which are not divisible by any other number except 1 and itself and the numbers which are divisible by at least one more number along with 1 and itself.

Solution From 2 to 100.

The number of numbers which are divisible by 1 and itself only = 25

Also, the number of numbers which are divisible by at least one more number except 1 and itself (i.e composite numbers) 99 – 25 = 74

So, required difference = 74 – 25 = 49

fi Option (c)

Problem 1.24 If the sum of (2n +1) prime numbers where n ŒN is an even number, then one of the prime numbers must be

Solution For any n ΠN, 2n + 1 is odd.

Also, it is given in the problem that the sum of an odd number of prime numbers = even. Since all prime numbers except 2 are odd, the above condition will only be fulfilled if we have an (odd + odd + even) structure of addition. Since, the sum of the three prime numbers is said to be even, we have to include one even prime number. Hence 2 being the only even prime number must be included.

If we add odd number of prime numbers, not including 2 (two), we will always get an odd number, because

fi Option (a)

Problem 1.25 What will be the difference between the largest and smallest four digit number made by using distinct single digit prime numbers?

Solution Required largest number Æ 7532

Required smallest number Æ 2357

Difference Æ 5175

fi Option (d)

Problem 1.26 The difference between the two three-digit numbers XYZ and ZYX will be equal to

Solution From the property of numbers, it is known that on reversing a three digit number, the difference (of both the numbers) will be divisible by 99. Also, it is known that this difference will be equal to 99 × difference between the units and hundreds digits of the three digit number. fi Option (d)

Problem 1.27 When the difference between the number 842 and its reverse is divided by 99, the remainder will be

Solution From the property (used in the above question) we can say that the difference will be divisible by 99

fi Remainder = 0 (zero)

fi Option (a)

Problem 1.28 When the difference between the number 783 and its reverse is divided by 99, the quotient will be

Solution The quotient will be the difference between extreme digits of 783, i.e. 7 – 3 = 4 (This again is a property which you should know.)

fi Option (d)

Problem 1.29 A long Part of wood of same length when cut into equal pieces each of 242 cms, leaves a small piece of length 98 cms. If this Part were cut into equal pieces each of 22 cms, the length of the leftover wood would be

Solution As 242 is divisible by 22, so the required length of left wood will be equal to the remainder when 98 is divided by 22:

Hence, 10 [98/22; remainder 10]

fi Option (d)

Problem 1.30 Find the number of numbers from 1 to 100 which are not divisible by 2.

Solution The 1^st number from 1 to 100, not divisible by 2 is 1 and the last number from 1 to 100, not divisible by 2 is 99.

Every alternate number (i.e, at the gap of 2) will not be divisible by 2 from 1 to 99. (1, 2, 3, - - - - , 95, 97, 99)

So, the required number of nos = =

fi Option (b)

Alternate method

Total number of nos from 1 to 100 = 100(i)

Now, if we count number of numbers from 1 to 100 which are divisible by 2 and subtract that from the total number of numbers from 1 to 100, as a result we will find the number of numbers from 1 to 100 which are not divisible by 2.

To count the number of nos from 1 to 100 which are divisible by 2:

The 1^st number which is divisible by 2 = 2

The last number which is divisible by 2 = 100

(2, 4, 6, - - - - , 96, 98, 100)

Gap/step between two consecutive numbers = 2

So, the number of numbers which are divisible by 2 = = = (ii)

So, from (i) & (ii)

Required number of numbers = 100 – 50 = 50

fi Option (b)

Problem 1.31 Find the number of numbers from 1 to 100 which are not divisible by any one of 2 & 3.

Solution From 1 to 100

Number of numbers not divisible by 2 & 3 = Total number of numbers–number of numbers divisible by either 2 or 3.

Now, total number of numbers = 100(ii)

For number of numbers divisible by either 2 or 3:

Number of numbers divisible by 2 = =

Now, the number of numbers divisible by 3 (but not by 2, as it has already been counted)

1st such no. = 3 and the gap will be 6. Hence 2nd such no. will be 9, 3rd no. would be 15 and the last number would be 99. Hence this series is 3, 9, 15, ..., 93, 99

So, the number of numbers divisible by 3 (but not by 2) =

Hence, the number of numbers divisible by either 2 or 3 = 50 + 17 = 67

So, from (i), (ii) & (iii) required number of numbers = 100 – 67 = 33

fi Option (d)

Problem 1.32 Find the number of numbers from 1 to 100 which are not divisible by any one of 2, 3, and 5.

Solution From the above question, we have found out that

From 1 to 100, number of numbers divisible by 2 = 50(i)

Number of numbers divisible by 3 ( but not by 2) = 17(ii)

Now, we have to find out the number of numbers which are divisible by 5 (but not by 2 and 3). Numbers which are divisible by 5

(5) 10 15 20 (25) 30 (35) 40 45 50 (55) 60 (65) 70 75 80 (85) 90 (95) 100

That is, there are 7 such numbers(iii)

Another way to find out the number of numbers that are divisible by 5 but not 2 and 3 is to first only consider odd multiples of 5.

You will get the series of 10 numbers: 5, 15, 25, 35, 45, 55, 65, 75, 85 and 95

From amongst these we need to exclude multiples of 3. In other words, we need to find the number of common elements between the above series and the series of odd multiples of 3, viz, 3, 9, 15, 21 …. 99.

This situation is the same as finding the number of common elements between the two series for which we need to first observe that the first such number is 15. Then the common terms between these two series will themselves form an arithmetic series and this series will have a common difference which is the LCM of the common differences of the two series. (In this case the common difference of the two series are 10 and 6 respectively and their LCM being 30, the series of common terms between the two series will be 15, 45 and 75.) Thus, there will be 3 terms out of the 10 terms of the series 5, 15, 25…95 which will be divisible by 3 and hence need to be excluded from the count of numbers which are divisible by 5 but not 2 or 3.

Hence, the required answer would be: 100 – 50 – 17 – 7 = 26

fi Option (a)

Problem 1.33 Find the number of numbers from 1 to 100 which are not divisible by any one of 2, 3, 5 & 7.

Solution From the above question we have seen that from 1 to 100.

number of numbers divisible by 2 = 50(i)

number of numbers divisible by 3 but not by 2 = 17(ii)

number of numbers divisible by 5 but not by 2 and 3 = 7(iii)

number of numbers divisible by 7 but not by 2, 3 & 5;

such nos are 7, 49, 77, 91 = 4 nos(iv)

Required number of numbers = Total number of numbers from 1 to 100 – {(i) + (ii) + (iii) + (iv)}

= 100 – (50 + 17 + 7 + 4)

= 22

fi Option (a)

Problem 1.34 What will be the remainder when – 34 is divided by 5?

Solution – 34 = 5 × (– 6) + (– 4)

Remainder = – 4, but it is wrong because remainder cannot be negative.

So,–34 = 5 × (– 7) + 1

fi Option (a)

Alternately, when you see a remainder of – 4 when the number is divided by 5, the required remainder will be equal to 5 – 4 = 1.

Problem 1.35 What will be the remainder when – 24.8 is divided by 6?

Solution – 24.8 = 6 × (– 4) + (– 0.8)

Negative remainder, so not correct – 24.8 = 6 × (– 5) + 5.2

Positive value of remainder, so correct

fi Option (b)

Problem 1.36 If p is divided by q, then the maximum possible difference between the minimum possible and maximum possible remainder can be?

Solution minimum possible remainder = 0 (when q exactly divides P)

Maximum possible remainder = q – 1

So, required maximum possible difference = (q – 1) – 0 = (q – 1)

fi Option (c)

Problem 1.37 Find the remainder when 2²⁵⁶ is divided by 17.

Solution fi R = 1

Q; R = 1

when n Æ even

fi Option (b)

Problem 1.38 Find the difference between the remainders when 7⁸⁴ is divided by 342 & 344.

Solution fi R = 1

also, = fi R = 1

The required difference between the remainders = 1 – 1 = 0

fi Option (a)

Problem 1.39 What will be the value of x for ; the remainder = 0

Solution

100¹⁷ – 1 = = fi divisible by 9 fi R = 0

Since the first part of the expression is giving a remainder of 0, the second part should also give 0 as a remainder if the entire remainder of the expression has to be 0. Hence, we now evaluate the second part of the numerator.

10³⁴ + x =

with x at the right most place. In order for this number to be divisible by 9, the sum of digits should be divisible by 9.

fi 1 + 0 + 0 ... + 0 + x should be divisible by 9.

fi 1 + x should be divisible by 9 fi x = 8

fi Option (d)

LEVEL OF DIFFICULTY (I)