Numbers: Their Tales, Types, and Treasures.
Chapter 8: Special Numbers
8.9.SOME NUMBER PECULIARITIES
Number oddities need not necessarily be restricted to a single number. There are times when these oddities appear with partner numbers. Consider the addition of the two numbers 192 + 384 = 576. You may ask, what is so special about this addition? Look at the outside digits (bold): 192 +384 = 576. They are in numerical sequence left to right (1, 2, 3, 4, 5, 6) and then reversing to get the rest of the nine digits (7, 8, 9). You might have also noticed that the three numbers we used in this addition problem have a strange relationship, as you can see from the following:
192 = 1 × 192,
384 = 2 × 192,
576 = 3 × 192.
The representation of all nine digits often fascinates the observer. Let's consider a number of such situations.
One such unexpected result happens when we subtract the symmetric numbers consisting of the digits in consecutive reverse order and in numerical order: 987,654,321 – 123,456,789 to get 864,197,532. This symmetric subtraction used each of the nine digits exactly once in each of the numbers being subtracted, and, surprisingly, resulted in a difference that also used each of the nine digits exactly once.
Here are a few more such strange calculations—this time using multiplication—where on either side of the equals sign all nine digits are represented exactly once: 291,548,736 = 8 × 92 × 531 × 746, and 124,367,958 = 627 × 198,354 = 9 × 26 × 531,487.
Another example of a calculation where all the digits are used exactly once (not counting the exponent), is 567^{2} = 321,489. This also works for the following: 854^{2} = 729,316. These are, apparently, the only two squares that result in a number that allow all the digits to be represented once.
When we take the square and the cube of the number 69, we get two numbers that together use all the ten digits exactly once. 69^{2} = 4,761, and 69^{3} = 328,509. That is, the two numbers 4,761 and 328,509 together represent all ten digits.
A somewhat convoluted calculation that results in a surprise ending begins with the following: 6,667^{2} = 44,448,889. When this result, 44,448,889, is multiplied by 3 to get 133,346,667, we notice that the last four digits are the same as the four digits of the number we began with, namely, 6,667. We use this example to bring us to a more general number oddity, which occurs when we take the number 625 to any power. We notice that the resulting number will always end with the last three digits being 625 (see figure 8.3).
625^{1} |
= |
625 |
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625^{2} |
= |
390,625 |
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625^{3} |
= |
244,140,625 |
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625^{4} |
= |
152,587,890,625 |
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625^{5} |
= |
95,367,431,640,625 |
||
625^{6} |
= |
59,604,644,775,390,625 |
||
625^{7} |
= |
37,252,902,984,619,140,625 |
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625^{8} |
= |
23,283,064,365,386,962,890,625 |
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625^{9} |
= |
14,551,915,228,366,851,806,640,625 |
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625^{10} |
= |
9,094,947,017,729,282,379,150,390,625 |
||
… |
Table 8.3: Powers of 625.
There are only two such numbers of three digits that have this property. The other is 376, which we can see from the list in table 8.4.
376^{1} |
= |
376 |
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376^{2} |
= |
141,376 |
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376^{3} |
= |
53,157,376 |
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376^{4} |
= |
19,987,173,376 |
||
376^{5} |
= |
7,515,177,189,376 |
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376^{6} |
= |
2,825,706,623,205,376 |
||
376^{7} |
= |
1,062,465,690,325,221,376 |
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376^{8} |
= |
399,487,099,562,283,237,376 |
||
376^{9} |
= |
150,207,149,435,418,497,253,376 |
||
376^{10} |
= |
56,477,888,187,717,354,967,269,376 |
||
… |
Table 8.4: Powers of 376.
If one questions whether there are two-digit numbers that have this property, the answer is clearly yes, and they are 25 and 76.
Number oddities are boundless. Some of these seem a bit far-fetched but nonetheless can be appealing to us from a recreational point of view. For example, consider taking any three-digit number that is multiplied by a five-digit number, all of whose digits are the same. When you add its last five digits to the remaining digits, a number will result where all digits are the same. Here are a few such examples:
237 × 33,333 = 7,899,921, then 78 + 99,921 = 99,999;
357 × 77,777 = 27,766,389, then 277 + 66,389 = 66,666;
789 × 44,444 = 35,066,316, then 350 + 66,316 = 66,666;
159 × 88,888 = 14,133,192, then 141 + 33,192 = 33,333.
These amazing number peculiarities, although entertaining, allow us to exhibit the beauty of mathematics so as to win over those individuals who have not had the experience of seeing mathematics from this point of view. We offer some more of these here to further entice the reader.
Armstrong Numbers
As we continue to expose some of the most celebrated numbers in mathematics, we come to those that are often referred to as Armstrong numbers or narcissistic numbers. In 1966, Michael F. Armstrong, while teaching a course in Fortran and general computing, came across these numbers as an exercise for his students. These numbers were named Armstrong numbers and were popularized in an article by Tim Hartnell in the February 23, 1988, issue of the Australian newspaper; and in the April 19, 1988, edition, the author formally named them “the Armstrong numbers.” The Armstrong numbers have the property that each number is equal to the sum of its digits, when each is taken to the power equal to the number of digits in the original number. For example, we have the three-digit Armstrong number 153, which is equal to the sum of its digits, each taken to the third power as 1^{3} + 5^{3} + 3^{3} = 1 + 125 + 27 = 153.
The nine-digit number 472,335,975 = 4^{9} + 7^{9} + 2^{9} + 3^{9} + 3^{9} + 5^{9} + 9^{9} + 7^{9} + 5^{9} is, therefore, also an Armstrong number. All Armstrong numbers are shown in the appendix, section 6, where we notice that there are no Armstrong numbers for k = 2, 12, 13, 15, 18, 22, 26, 28, 30, and 36 (and k > 39). In fact, there are only eighty-nine Armstrong numbers in the decimal system. The largest Armstrong number is thirty-nine digits long, and it is equal to the sum of its digits, each of which is taken to the thirty-ninth power:
1^{39} + 1^{39} + 5^{39} + 1^{39} + 3^{39} + 2^{39} + 2^{39} + 1^{39} + 9^{39} + 0^{39} + 1^{39} + 8^{39} + 7^{39} + 6^{39} + 3^{39} + 9^{39} + 9^{39} + 2^{39} + 5^{39} + 6^{39} + 5^{39} + 0^{39} + 9^{39} + 5^{39} + 5^{39} + 9^{39} + 7^{39} + 9^{39} + 7^{39} + 3^{39} + 9^{39} + 7^{39} + 1^{39} + 5^{39} + 2^{39} + 2^{39} + 4^{39} + 0^{39} + 1^{39} = 115,132,219,018,763,992,565,095,597,973,971,522,401.
The following is a list of the consecutive Armstrong numbers.
k = 3: 370; 371
k = 8: 24,678,050; 24,678,051
k = 11: 32,164,049,650; 32,164,049,651
k = 16: 4,338,281,769,391,370; 4,338,281,769,391,371
k = 25: 3,706,907,995,955,475,988,644,380;
3,706,907,995,955,475,988,644,381
k = 29: 19,008,174,136,254,279,995,012,734,740;
19,008,174,136,254,279,995,012,734,741
k = 33: 186,709,961,001,538,790,100,634,132,976,990;
186,709,961,001,538,790,100,634,132,976,991
k = 39: 115,132,219,018,763,992,565,095,597,973,971,522,400;
115,132,219,018,763,992,565,095,597,973,971,522,401
Incidentally, our first Armstrong number, 153, has some other amazing properties as well. It is also a triangular number, where
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 = 153.
The number 153 is not only equal to the sum of the cubes of its digits, but it is also a number that can be expressed as the sum of consecutive factorials 1! + 2! + 3! + 4! + 5! = 153.
Can you discover any other properties of this ubiquitous number 153?