Divisibility again....

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.

http://www.google.com/cse?cx=partner-pub-5929292452585068:i4gzq1rjo88&ie=ISO-8859-1&cof=FORID:1&q=Prove+that+among+39+sequential+natural+numbers+there+always+is+a+number+with+the+sum+of+its+digits+divisible+by+11&sa=Search

see the results u get on googling it :P