A number of 50 digits has all its digits equal to 1 except the 26th digit. If the number is divisible by 13, then find the digit in the 26th place.

(In reply to re: a little more straightforward by Tristan)

10^24 is not divisible by 13 by unique factorization into primes as 10^24 factors uniquely as 2^24 * 5^24 and the prime 13 does not appear. The 50-digit number written out in normal number (arabic) notation looks like
111111 111111 111111 111111 x1 111111 111111 111111 111111
where I have put in spaces to make the blocks of 6 ones stand out. Thus the number is the sum of multiples of 111111, which is divisble by 13, and x1*10^24. Thus to make the whole thing divisible by 13, x1*10^24 must also be divisible by 13. Since 10^24 certainly is not divisible by 13, x1 must be. The only two-digit number ending in 1 that is divisble by 13 is 91=13*7, so x=9 is the unique solution.

Posted by Richard on 2003-11-19 01:59:52