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 solution by Charlie)

Your observation about the powers of 10 mod 13 leads to the conclusion that replacing the powers of 10 by their mod 13 equivalents gives 4*0+10*x+1+4*0 as a mod 13 representation of the 50-digit number since 1+10+9+12+3+4=39 is divisble by 13. Now 10*x+1 is divisble by 13 if x=9, and one may readily check by brute force that 91 is the only two-digit number that ends in 1 and is divisible by 13. Hence x=9 uniquely renders the 50-digit number divisble by 13 as desired.

Posted by Richard on 2003-11-20 20:35:49