Mathematica competitions in Croatia

Question

Each digit of a positive integer n (except the first one) is larger than the digit next left to it. Determine the sum of all digits of the number 9n. (Russia 1999)

Accepted Answer

Let n = a_m-1a_m-2 a_1a_0. We know that a_0 > a_1 > > a_m-1. Since 9n = 10n - n, from arrayc c c c c c c c c c c c c c c & a_m-1 & a_m-2 & a_m-3 & & a_1 & a_0 & 0 - & 1ca_m-1 & a_m-2 & & a_2 & a_1 & a_0 b_m & b_m-1 & b_m-2 & & b_2 & b_1 & b_0 array we conclude b_0 = 10 - a_0, b_1 = a_0 - (a_1 + 1), b_2 = a_1 - a_2, b_3 = a_2 - a_3, ..., b_m-1 = a_m-2 - a_m-1, b_m = a_m-1. Those are the digits of 9n, and their sum is a_m-1 + (a_m-2 - a_m-1) + + (a_1 - a_2) + (a_0 - a_1 - 1) + (10 - a_0) = 9.