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.

The powers of 10 mod 13 repeat in a cycle of 6:
0 1
1 10
2 9
3 12
4 3
5 4
The next power (10^6) is again 1 mod 13 and the cycle starts again.

These total 39, which is 0 mod 13

Fifty ones is the sum of the powers of 10 from 0 to 49, which is [50/6]=8 times through the cycle plus the first two remainder. So the string of fifty ones is (1+10) mod 13 or 11. We need to add two more mod 13 to get to zero mod 13 so that it would be divisible.

The 26th digit in the fifty-digit number is the 25th from the right, representing 10^24. Representing the exponent mod 6 since there is a cycle of six in the mod-13 values, that's 10^0, or 1. We need to add just 2 in that position, making that position a 3, which is the answer.

Posted by Charlie on 2003-11-15 11:32:16