iamcste wrote:

mavesum wrote:
But what i have figured out is , the result of 2^3 + 3^3 would definately be a multiple of 2+3 i.e 5

anyways in this case remainder is 0

Ans is D

dude, this worked for a^3+b^3 as a^3+b^3=(a+b)(a^2+...) hence a^3+b^3 was divisble by a+b

this wont work for other cases. Also, whats the source for the problem

It actually works for any odd power. For example:

x^7 + y^7 = (x + y)(x^6 - (x^5)y + (x^4)(y^2) - (x^3)(y^3) + (x^2)(y^4) - x(y^5) + y^6)

That's not a factorization you'll ever need on the GMAT, but you can use it for this question:

13^7 + 14^7 + 15^7 + 16^7
= 13^7 + 16^7 + 14^7 + 15^7
= (13 + 16)*(a lot of terms) + (14 + 15)*(a lot of terms)
= 29*(a lot of terms plus a lot of terms)

So 13^7 + 14^7 + 15^7 + 16^7 is a multiple of 29, and since it's even (odd + even + odd + even = even), it's a multiple of 58, and the remainder will be zero when it's divided by 58.

It's not the kind of question you'll see on the GMAT, though.