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 fiftydigit number is the 25th from the right, representing 10^24. Representing the exponent mod 6 since there is a cycle of six in the mod13 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 20031115 11:32:16 