Originally Posted by

**nikhil** q 1)what is the remainder when 2^2006 is divided by 7

ans 1) 2^3 con 1 mod 7

or 2^2004 con 1 mod 7.......................1

also 2^2 con -3 mod 7.........................2

by 1 and 2

or 2^2006 con -3 mod 7

hence remainder is 4.

note: con stands for congruence

can anyone use the same procedure to find the remainder when 32^[(32)^32] is divided by 7.

my attempt

32 con 4 mod 7

or 32^32 con 4^32 mod 7

or 32^32 =7a+4^32

therefor

32^[(32)^32]=32^[(7a)+4^32] and here I am gone

in the solved example, problem was easy because of 1(2^3 con 1 mod 7) but in question 2 it is not the case.