Problem 1

Prove that among any 39 consecutive natural numbers it is always possible to find one whose sum of digits is divisible by 11.

Suppose that the smallest number in your sequence of 39 consecutive natural numbers is n. Let k be the smallest number in your sequence that has a zero for the units digit. k is at most n+9. Your sequence then has three blocks of ten numbers each k,k+1,...,k+9; k+10,k+11,...,k+19; and k+20,k+21,...,k+29. The first number in each block (k, k+10 and k+20) has a zero for the units digit. Let m be the smallest of these three numbers that does not have a nine for the tens digit. m is either k or k+10. There are thus 20 consecutive numbers in your sequence, m,m+1,...,m+19 where the units digit of m is zero and the tens digit of m is not 9.

Let d be the sum of the digits of m. The sum of the digits of m,m+1,...,m+19 are then k,k+1,k+2,...,k+9,k+1,k+2,...,k+10 respectively. One of the eleven consecutive integers, k,k+1,k+2,...,k+10 is divisible by eleven.