Problem
NT-B2-M11-P017 Blocks and a Lower Bound for Digit Sum
Let \(k\ge1\). Split the decimal representation of a positive integer \(M\) from right to left into blocks of \(k\) digits, and let \(T(M)\) be the sum of these blocks as ordinary integers. Prove that \(M\equiv T(M)\pmod{10^k-1}\). Then prove: if \(10^k-1\mid M\), then the digit sum of \(M\) is at least \(9k\).
The first part follows from \(10^k\equiv1\). For the second part, repeat the operation \(M\mapsto T(M)\).
If \(M=B_0+B_1 10^k+B_2 10^{2k}+\cdots+B_s10^{sk}\), where the \(B_i\) are the blocks, then modulo \(10^k-1\) each power \(10^{ik}\) is congruent to \(1\). Hence \(M\equiv B_0+B_1+\cdots+B_s=T(M)\pmod{10^k-1}\).
Also, the digit sum of \(T(M)\) is not greater than the digit sum of \(M\): carries during addition can only decrease the digit sum. If \(M\ge10^k\), then \(T(M)
A strong problem: it looks like notation technique, but it actually uses descent on the number.