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.
my solution is 9
the digit is the solution of the following equation:
(10^50 -1)/9 + (x-1)10^24 = 13k (K natural)
using Charlie's remarks regarding modulo 13 and the power series of 10,
10^50 -1 + 9(x-1)10^24 = 0 (mod 13)
9-1+12y=0 (mod 13), where y = x-1
y=8, thus x=9
solution x=9
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Posted by luminita
on 2003-12-31 11:27:06 |