## SAT Test Prep

## CHAPTER 6

WHAT THE SAT MATH IS REALLY TESTING

### Lesson 3: Finding Patterns

**Repeating Patterns**

Finding patterns means looking for simple rules that relate the parts of a problem. One key to simplifying many SAT math problems is exploiting repetition. If something repeats, you usually can *cancel* or *substitute* to simplify.

If , then what is the value of *x*?

This question is much simpler than it looks at first because of the repetition in the equation. If you subtract the repetitive terms from both sides of the equation, it reduces to . Subtracting 4*x*^{2}from both sides then gives , so .

**Patterns in Geometric Figures**

Sometimes you need to play around with the parts of a problem until you find the patterns or relationships. For instance, it often helps to treat geometric figures like jigsaw puzzle pieces.

The figure above shows a circle with radius 3 in which an equilateral triangle has been inscribed. Three diameters have been drawn, each of which intersects a vertex of the triangle. What is the sum of the areas of the shaded regions?

This figure looks very complicated at first. But look closer and notice the *symmetry* in the figure. Notice that the three diameters divide the circle into six congruent parts. Since a circle has 360°, each of the central angles in the circle is . Then notice that the two shaded triangles fit perfectly with the other two shaded regions to form a sector such as this:

Moving the regions is okay because it doesn”t change their areas. Notice that this sector is 1/3 of the entire circle. Now finding the shaded area is easy. The total area of the circle is . So the area of 1/3 of the circle is .

**Patterns in Sequences**

Some SAT questions will ask you to analyze a sequence. When given a sequence question, write out the terms of the sequence until you notice the pattern. Then use whole-number division with remainders to find what the question asks for.

1, 0, –1, 1, 0, –1, 1, 0, –1, …

If the sequence above continues according to the pattern shown, what will be the 200th term of the sequence?

Well, at least you know it”s either 1, 0, or –1, right? Of course, you want a better than a one-in-three guess, so you need to analyze the sequence more deeply. The sequence repeats every 3 terms. In 200 terms, then the pattern repeats itself times with a remainder of 2. This means that the 200th term is the same as the second term, which is 0.

What is the units digit of 27^{40}?

The units digit is the “ones” digit or the last digit. You can”t find it with your calculator because when 27^{40} is expressed as a decimal, it has 58 digits, and your calculator can only show the first 12 or so. To find the units digit, you need to think of 27^{40} as a term in the sequence 27^{1}, 27^{2}, 27^{3}, 27^{4}, …. If you look at these terms in decimal form, you will notice that the units digits follow a pattern: 7, 9, 3, 1, 7, 9, 3, 1, …. The sequence has a repeating pattern of four terms. Every fourth term is 1, so the 40th term is also 1. Therefore, the units digit of 27^{40} is 1.

**Concept Review 3: Finding Patterns**

Solve the following problems by taking advantage of repetition.

__1.__ If 5 less than 28% of *x*^{2} is 10, then what is 15 less than 28% of *x*^{2}?

__2.__ If *m* is the sum of all multiples of 3 between 1 and 100, and *n* is the sum of all multiples of 3 between 5 and 95, what is ?

__3.__ How much greater is the combined surface area of two cylinders each with a height of 4 cm and a radius of 2 cm than the surface area of a single cylinder with a height of 8 cm and a radius of 2 cm?

Solve each of the following problems by analyzing a sequence.

__4.__ What is the units digit of 4^{134}?

__5.__ The first two terms of a sequence are 1 and 2. If every term after the second term is the sum of the previous two, then how many of the first 100 terms are odd?

**SAT Practice 3: Finding Patterns**

**1**__.__ If , then

(A) 5

(B) 15

(C) 20

(D) 40

(E) 60

**2**__.__ Every term of a sequence, except the first, is 6 less than the square of the previous term. If the first term is 3, what is the fifth term of this sequence?

(A) 3

(B) 15

(C) 19

(D) 30

(E) 43

**3**__.__ If the sequence above continues according to the pattern shown, what is the sum of the first 200 terms of the sequence?

(A) –800

(B) –268

(C) –4

(D) 0

(E) 268

**4**__.__ The figure above shows a square with three line segments drawn through the center. What is the total area of the shaded regions?

**5**__.__ What is the units digit of 3^{40}?

(A) 1

(B) 3

(C) 6

(D) 7

(E) 9

**Answer Key 3: Finding Patterns**

**Concept Review 3**

__1.__ Don”t worry about the percent or about finding *x*. Translate: 5 less than 28% of *x*^{2} is 10 means

So 15 less than 28% of *x*^{2} is 0.

__2.__

When you subtract *n* from *m*, all the terms cancel except .

__3.__ Don”t calculate the total surface area. Instead, just notice that the two small cylinders, stacked together, are the same size as the large cylinder. But remember that you are comparing **surface areas**, not **volumes**. The surface areas are almost the same, except that the smaller cylinders have two extra bases. Each base has an area of so the surface area of the smaller cylinders is greater than that of the larger cylinder.

__4.__ Your calculator is no help on this one because 4^{134} is so huge. Instead, think of 4^{134} as a term in the sequence 4^{1}, 4^{2}, 4^{3}, 4^{4}, …. What is the units digit of 4^{134}? If you write out the first few terms, you will see a clear pattern to the units digits: 4, 16, 64, 256, …. Clearly, every odd term ends in a 4 and every even term ends in a 6. So 4^{134} must end in a 6.

__5.__ The first few terms are 1, 2, 3, 5, 8, 13, 21, …. Since we are concerned only about the “evenness” and “oddness” of the numbers, think of the sequence as odd, even, odd, odd, even, odd, odd, even, …. Notice that the sequence repeats every three terms: (odd, even, odd), (odd, even, odd), (odd, even, odd), …. In the first 100 terms, this pattern repeats times. Since each pattern contains 2 odd numbers, the first 33 repetitions contain 66 odd numbers and account for the first 99 terms. The last term must also be odd because each pattern starts with an odd number. Therefore, the total number of odds is .

**SAT Practice 3**

__2.__ **A** If every term is 6 less than the square of the previous term, then the second term must be . The third term, then, is also , and so on. Every term, then, must be 3, including the fifth.

__3.__ **C** The sequence repeats every three terms: (–4, 0, 4), (–4, 0, 4), (–4, 0, 4), …. Each one of the groups has a sum of 0. Since , the first 200 terms contain 67 repetitions of this pattern, plus two extra terms. The 67 repetitions will have a sum of , but the last two terms must be –4 and 0, giving a total sum of –4.

__4.__ **50** Move the shaded regions around, as shown above, to see that they are really half of the square. Since the area of the square is , the area of the shaded region must be half of that, or 50.

__5.__ **A** The number 3^{40} is so big that your calculator is useless for telling you what the last digit is. Instead, think of 3^{40} as being an element in the sequence 3^{1}, 3^{2}, 3^{3}, 3^{4},…. If you write out the first six terms or so, you will see that there is a clear pattern to the units digits: 3, 9, 27, 81, 243, 729, …. So the pattern in the units digits is 3, 9, 7, 1, 3, 9, 7, 1, …. The sequence repeats every four terms. Since 40 is a multiple of 4, the 40th term is the same as the 4th and the 8th and the 12th terms, so the 40th term is 1.