M is a 2003-digit positive decimal integer having the form: 12.....21
where, the first digit is 1, followed by the digit 2 repeated precisely 2001 times, and the last digit is 1.

Determine the highest power of 11 that evenly divides M.

12222...22221 = 11 x 11111...11111 (2002 1's)     

11111...11111 = 11 x 10101....010101 (2001 digits)

Using the divisibility test for a number divisible by 11, taking the difference of the sum of the even digits from the odd digits and checking if the sum is divisible by 11, 10101...010101's odd digits total 1001 and the even digits total 0. The sum, 1001 is divisible by 11, therefore there is at least a 3rd power of 11 that M is divisible by.  

(I realized that my 'solution' was incorrect and incomplete on 2nd reflection.  See Charlie's comments for a solution.)

Edited on August 1, 2009, 4:58 pm

Posted by Dej Mar on 2009-08-01 15:00:11