What is the largest whole number n such that [(7^2048)-1] is divisible by 4^n
Answers: 1, 3, 5, 7
So far as I can tell, this is basically a 'difference of squares' question on steroids. Nothing this involved would show up on the GMAT. (Which is to say nothing about having only four answer choices.)
Rewrite [(7^2048)-1] as 7^2048 - 1^2 (anytime you have the difference of two integers raised to even exponents, you have a difference of squares.)
(7^2048) - (1^2) = [(7^1024)+1] * [(7^1024) - 1]
Last term is again a difference of squares. Now we have:
[(7^1024)+1] * [(7^512) +1] * [(7^512)-1]
Again, we end with a difference of squares. Rewrite:
[(7^1024)+1] * [(7^512) +1] * [(7^256)+1] * [7^256)-1]
Same thing:
[(7^1024)+1] * [(7^512) +1] * [(7^256)+1] * [7^128)+1] * [(7^128) - 1]
And so on. This process will repeat over and over. If we keep going, we'll end up with:
[(7^1024)+1] * [(7^512) +1] * [(7^256)+1] * [(7^128) +1] * [(7^64) +1] * [(7^32) +1]* [(7^16) +1]* [(7^8) +1] * [(7^4) +1] * [(7^2) +1] * (7+1) * (7-1)
Note that every term in the product will be even. We have twelve even terms, which means we'll have at least twelve 2's, so we'll have at least six 4's. Therefore, the answer must be '7,' as '5' is too small. {Some additional 2's are included in 7+1, because 8 = 2^3.)
Again, this is well beyond the purview of the GMAT. Learn the rule for difference of squares, then erase this question from your memory banks. Way too complicated for GMAT purposes.
