What is the remainder when 32^(32^32) is divided by 7?
A. 5
B. 4
C. 2
D. 0
E. 1
[spoiler]OA:B[/spoiler]
A. 5
B. 4
C. 2
D. 0
E. 1
[spoiler]OA:B[/spoiler]
Remainder when 32^(32^32) OR (28+4)^(32^32) is divided by 7 is the same as remainder when 4^(32^32) is divided by 7. This is because all the terms in the expansion will be multiples of 7 except the last one, which is 4^(32^32).
Also 32^32 when divided by 3 gives remainder (-1)^32 OR 1=>32^32 = 3k+1
4^(32^32)= 4^(3k+1)by 7 = 4* (4^3)^k by 7 = 4*1 = 4 by 7.
Hence the required remainder is 4.
Alternate solution
Also if p is prime and N^(p-1) is divided by p, the remainder is 1 (as long as N is not a multiple of p). So, N^6 will always leave remainder 1 when divided by 7 (if N is not a multiple of 7). And so will N*6k, where k is a positive integer.
So, let's explore the remainder when 32^32 by 6 = 2^32 by 6= 2* (2^31 by 3) = 2 (2 by 3) = 4 when divided by 6
Now we have 4^(32^32) by 7 = 4^(6k+4) by 7= 4^4 by 7 (by using the consideration above)= 256 by 7 = 4 when divided by 7
Hence the required remainder is 4.
