## Friday, June 18, 2010

### Mathemagical black holes

Here’s an early kickoff to the weekend from the e-book Mathemagic: Advanced Math Puzzles, available for free download on Scribd. I am shamelessly lifting material from Michael W. Ecker’s paper “Number Play, Calculators, and Card Tricks: Mathemagical Black Holes.”

First, the Sisyphus string: 123. “Suppose we start with any natural number, regarded as a string, such as 9,288,759. Count the number of even digits, the number of odd digits, and the total number of digits. These are 3 (three evens), 4 (four odds), and 7 (seven is the total number of digits), respectively. So, use these digits to form the next string or number, 347. Now repeat with 347, counting events, odds, total number, to get 1, 2, 3, so write down 123. If we repeat with 123, we get 123 again. The number 123 with respect to this process and the universe of numbers is a mathemagical black hole. All numbers in this universe are drawn to 123 by this process, never to escape.” (p. 41)

Second, words to numbers: 4. “Take any whole number and write out its numeral in English, such as FIVE for the usual 5. Count the number of characters in the spelling. In this case, it is 4—or FOUR. So, work now with the 4 or FOUR. Repeat with 4 to get 4 again. As another instance, try 163. To avoid ambiguity, I’ll arbitrarily say that we will include spaces and hyphens in our count. Then, 163 appears as ONE HUNDRED SIXTY-THREE for a total count of 23. In turn, this gives 12, then 6, then 3, then 5, and finally 4.” (p. 44)

Third, narcissistic numbers: 153. This one’s a little tougher. “It is well known that, other than the trivial examples of 0 and 1, the only natural numbers that equal the sum of the cubes of their digits are 153, 370, 371, and 407. Of these, just one has a black-hole property. . . . We start with any positive whole number that is a multiple of 3. Recall that there is a special shortcut to test whether you have a multiple of 3. Just add up the digits and see whether that sum is a multiple of 3. . . . Write down your multiple of 3. One at a time, take the cube of each digit. Add up the cubes to form a new number. Now repeat the process. You must reach 153. And once you reach 153, one more iteration just gets you 153 again. Let’s test just one initial instance. Using the sum of the cubes of the digits, if we start with 432—a multiple of 3—we get 99, which leads to 1458, then 702, which yields 351, finally leading to 153, at which point future iterations keep producing 153. Note also that this operation or process preserves divisibility by 3 in the successive numbers.” (pp. 44-45)

Fourth, Kaprekar’s constant: 6174. “Take any four-digit number except an integral multiple of 1111 (i.e., don’t take one of the nine numbers with four identical digits). Rearrange the digits of your number to form the largest and smallest strings possible. That is, write down the largest permutation of the number, the smallest permutation (allowing initial zeros as digits), and subtract. Apply this same process to the difference just obtained. Within the total of seven steps, you always reach 6174. At that point, further iteration with 6174 is pointless: 7641-1467 = 6174. Example: Start with 8028. The largest permutation is 8820, the smallest is 0288, and the difference is 8532. Repeat with 8532 to calculate 8532-2358 = 6174.”

Ecker gives more examples of mathemagical black holes and provides mathematical explanations, so math puzzlers can indulge themselves.