3 ms·
Not just the special numbers. Kaprekar's operation always results in a multiple of 9. Given a > b > c > d. Kaprekar's operation gives: 1000a + 100b + 10c
by davnola 15y ago
Not just the special numbers.
Kaprekar's operation always results in a multiple of 9.
Given a > b > c > d. Kaprekar's operation gives:
1000a + 100b + 10c + d - (1000d + 100c + 10b + a)
=999(a - d) + 90(b - c)
=9(111(a - d) + 10(b - c))
And so on for integers with other numbers of digits.
- jimfl 15y agoAny number subtracted from a re-arrangement of the digits results in a multiple of 9.
- sesqu 15y agoWhy?
- jimfl 15y agoA number is the sum of its digits, raised to the appropriate powers Sum(i=0..n, a[i]*10^i) 123, for example is 3 + 2 * 10 + 1 * 100. A power of 10 (10^n) can be re written as (1 + 9 * Sum(i=0..n-1, 10^i). For example: 1000 = 1 + 999 = 1 + 900 + 90 + 9 When you subtract two numbers with the same digits, you end up being able to factor a nine out of these sums fairly easily. I wrote a proof of the number - reverse(number) a while back. It can be found on archive.org http://web.archive.org/web/20050314023901/http://jimfl.tensegrity.net/math/nines/ http://web.archive.org/web/20050314023901/http://jimfl.tense...