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Algebra Difficulty 4.8 AIME Prove it Croatia

Each digit of a positive integer nn (except the first one) is larger than the digit next left to it. Determine the sum of all digits of the number 9n9n. (Russia 1999)

Solution

Let n=am−1am−2…a1a0‾n = \overline{a_{m-1}a_{m-2}\dots a_1a_0}. We know that a0>a1>⋯>am−1a_0 > a_1 > \dots > a_{m-1}.

Since 9n=10n−n9n = 10n - n, from
am−1am−2am−3…a1a00−\multicolumn1cam−1am−2…a2a1a0bmbm−1bm−2…b2b1b0 \begin{array}{c c c c c c c c c c c c c c c} & a_{m-1} & a_{m-2} & a_{m-3} & \dots & a_1 & a_0 & 0 \\ - & \multicolumn{1}{c}{a_{m-1}} & a_{m-2} & \dots & a_2 & a_1 & a_0 \\ \hline b_m & b_{m-1} & b_{m-2} & \dots & b_2 & b_1 & b_0 \end{array}
we conclude b0=10−a0b_0 = 10 - a_0, b1=a0−(a1+1)b_1 = a_0 - (a_1 + 1), b2=a1−a2b_2 = a_1 - a_2, b3=a2−a3b_3 = a_2 - a_3, ..., bm−1=am−2−am−1b_{m-1} = a_{m-2} - a_{m-1}, bm=am−1b_m = a_{m-1}. Those are the digits of 9n9n, and their sum is am−1+(am−2−am−1)+⋯+(a1−a2)+(a0−a1−1)+(10−a0)=9a_{m-1} + (a_{m-2} - a_{m-1}) + \dots + (a_1 - a_2) + (a_0 - a_1 - 1) + (10 - a_0) = 9.

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