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 a little more straightforward by Eric)

That 111111 is divisible by 13 is an excellent observation. So the 50-digit number is the sum of multiples of 111111, i.e. multiples of 13, plus (10*x+1)*10^24 where x is the unknown digit. If 10*x+1 is a multiple of 13, we have a solution. Now 91=13*7. so we can use x=9. Also 10^24 is congruent to 1 mod 13 (10^4 is congruent to 3), so any multiplier of 10^24 must be congruent to 0 mod 13, so I think that 9 has to be the only possible value for x.

Posted by Richard on 2003-11-19 00:10:42