a) Let $n_1,...,n_{39}$ be the consecutive integers in that order. Among the first 20, there are two such that their unit digit is 0. They differ by 10 and at least one, say $n_k$, has its second digit from the right which is not 9 (thus it is at most 8). Then $n_k \leq n_{20}$. Let $s(x)$ be the sum of the digits of $x$. Then $s(n_{k+i}) = s(n_k) + i$ for $i=0,1,...,9$. moreover $n_{k+10}$ has unit digit 0, and the digit before the last is one more than the one of $n_k$, thus $s(n_{k+10}) = s(n_k)+1$, and $n_{k+10} \ leq n_{30}$. It follows that $s(n_{k+10+i}) = s(n_k) + i + 1$ for $i=0,...,9$. It follows that $s(n_j)$ takes each integer value from $s(n_k)$ to $s(n_k)+10$ which are 11 consecutive integers. One of them is divisible by 11.

Again U.S.S.R 1961

Could somebody explain the last line of Pierre's proof to me Thanks

I used the fact that among any 11 consecutive integers, there is one which is divisible by 11, since they give all the residues $0,1,2,...,10$ modulo 11, in some order. Pierre.

Iris Aliaj wrote: a) Prove that it is possible to choose one number out of any 39 consecutive positive integers, having the sum of its digits divisible by 11; b) Find the first 38 consecutive positive integers none of which have the sum of its digits divisible by 11. Solution to b)999981,999982,…,1000018.