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.

By Fermat's theorem, (10^12 -1) mod 13=
=> 1+10+10^2+....+10^11 mod 13=0
Similarly
10^12+10^13+...+10^23 mod 13 =0
and so on
10^36+10^37+...+10^47 mod 13=0
Let x is in 26th place from left, then
A mod 13 = 0
=> x*10^24-10^24+10^48+10^49 mod 13 =0(From the problem)
=> x-1+1+10 mod 13=0
=> x = -10 mod 13
=> x= 3.

Edited on September 11, 2007, 8:26 am

Posted by Praneeth on 2007-09-11 08:20:38