Source: http://www.math.udel.edu/~lazebnik/papers
Problem:
Consider any positive integer N whose (decimal) digits read from left to right are in
non-decreasing order, but the last two digits (tens and ones) are in increasing order.
Prove that the sum of digits of 9N is always exactly 9.
From the solution of author herself::
ReplyDeleteIt was hard to believe, since N could be really large. For example, if a = 1778, b = 2344459,
and c = 12225557779, then
9a = 16002, 9b = 21100131, 9c = 110030020011,
and the sum of digits in each case is 9. The proof is easy. If you find it for for 3- or 4-digit
numbers, the generalization is trivial. This problem was communicated to me about three years
ago by Valery Kanevsky, a friend and an applied mathematician.