Advanced Quant Ch 9, Q 133 If n is an integer and n^4

[LP338]

If n is an integer and n^4 is divisible by 32, which of the following could be the remainder when n is divided by 32?
A) 2
B) 4
C) 5
D) 6
E) 10

My question is a tangent off of the provided solution: With the fact that n has an additional factor of 2 "left over" once you account for the 4 2's that are part of n^4 (because 32 is 2^5), does this mean that n^4 will also be divisible by 2^6, 2^7, and 2^8? They are apparently necessary because of that 5th 2, so for any number n^4, there must be multiples of 4 of all the prime factors, and n^4 will be divisible by any combination of numbers taken from its prime factorization box? Thanks!

[tim]

That is correct. Please let us know if you have any additional questions on this one.