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*10^2410^24+10^48+10^49 mod 13 =0(From the problem)
=> x1+1+10 mod 13=0
=> x = 10 mod 13
=> x= 3.
Edited on September 11, 2007, 8:26 am

Posted by Praneeth
on 20070911 08:20:38 