Final answer:
For a set of 7 integers each raised to the power of n and added together to be divisible by 7, n must be of the form 6k, corresponding to even values of n. This is consistent with Fermat's Little Theorem, as 7 is a prime.
Step-by-step explanation:
The scenario presented involves raising a set of 7 integers to the power of n and determining for which values of n the sum of these numbers is divisible by 7. According to Fermat's Little Theorem, if we have a prime number p, and an integer a such that a is not divisible by p, then ap-1 ≡ 1 (mod p). In this context, if we consider 7 as our prime p, and raise any integer that is not a multiple of 7 to the 6th power, the result modulo 7 will be 1.
Thus, the sum of seven such terms, each raised to an exponent of n that is one less than a multiple of 7 (i.e., of the form 6k where k is an integer), will be divisible by 7. For these values of n, each term contributes 1 to the sum modulo 7, and the total sum 7*1 = 7 is clearly divisible by 7.
Therefore, for a set of 7 integers raised to a power n that results in the sum being divisible by 7, n must be of the form 6k, which corresponds to the even numbers that are one less than multiples of 7. So the answer is c) Only even values of n.