since 7 is prime , u can use fermat's theorem which says than any number a^(p-1) will leave a remainder 1 when divided by p ( obviously a is not a factor of p )

hence 2^6 rem =1 therfore (2^100= (2^6)^16 *2^4)/7 =2^4/7= 2 similarly do for the rest

This problem isn't appropriate for the GMAT, at least by our current understanding of the test: modular arithmetic and Fermat's Little Theorem are well out of scope.

That said, here's how I'd do it.

Simplify first. 4^100 is the same as 2^200. 5 divided by 7 leaves the same remainder as -2 divided by 7, so 5^100 is congruent to (-2)^100, modulo 7. 3 divided by 7 leaves the same remainder as -4 divided by 7, so 3^100 is congruent to (-4)^100, or (-2)^200 modulo 7. That means the question is really asking

"What's the remainder when 2^100 + 2^200 + (-2)^100 + (-2)^200 is divided by 7?"

Simplify further. 2^100 + 2^200 + (-2)^100 + (-2)^200 is the same as 2^100 + 2^200 + 2^100 + 2^200, or 2*(2^100) + 2*(2^200), or 2^101 + 2^201.

Now the question is really "What's the remainder when 2^101 + 2^201 is divided by 7?"

2^201 = (2^3)^67. 2^3 leaves a remainder of 1 when divided by 7, so (2^3)^67 does too.

2^101 = (2^3)^33 * 2^2. The (2^3)^33 leaves a remainder of 1, while 2^2 leaves a remainder of 4.

So in total we have remainders of 1 + 1*4, or 1 + 4, or 5, so the remainder is 5.

It will be cool if the GMAT ever starts testing this, as modular arithmetic is really fun to teach (and hopefully to learn!) - telescoping huge numbers and series is awesome.

But for now, don't lose sleep over it!