RahlowJenkins wrote:

1^x + 2^x will always be divisible by 3 for interger x>0

No, that's not true - 1^2 + 2^2 is not divisible by 3, for example. What you're saying is true when x is odd, and not true when x is even.

I responded to the question in the original post on BTG by rearranging terms and factoring, but I can give a different solution here, using mathematics you definitely don't need to know for the test. But since modular arithmetic will be familiar to some people here, I'll use it on the above question.

Here the \(\equiv\) symbol means "has the same remainder as" and "mod 29" means "when dividing by 29". Modular arithmetic establishes that, if you're interested in finding the remainder of a sum, product, or difference, you can replace one number in that sum, product or difference with any other number that gives the same remainder).

\(13^7 + 14^7 + 15^7 + 16^7 \equiv 13^7 + 14^7 + (-14)^7 + (-13)^7 \equiv 0 \text{ mod 29}\)

So the sum is divisible by 29. Since it's even, it must also be divisible by 58.

Here I'm using the fact that 15 and -14 both have the same remainder when divided by 29, since they're both 15 more than a multiple of 29 (-14 = -29 + 15). Similarly, 16 and -13 have the same remainder when divided by 29.

You certainly don't ever need modular arithmetic on the GMAT, however, and the above question is not a realistic test question.

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