If 22^3+23^3+24^3+....+87^3+88^3 is divided by 110 then the remainder will be
A. 55
B.1
C.33
D.0
E.44
A
A. 55
B.1
C.33
D.0
E.44
A
Proviso: Unless there's a much easier solution out there, I believe that this question is out of scope for the GMAT.nasheen wrote:If 22^3+23^3+24^3+....+87^3+88^3 is divided by 110 then the remainder will be
A. 55
B.1
C.33
D.0
E.44
A
First notice that there are 57 terms in the given expression (since 88-22+1 = 57)
Now notice that we can rewrite the sum as:
(22^3 + 88^3) + (23^3 + 87^3) + (24^3 + 88^3) . . . + (54^3 + 56^3) + 55^3
Why did I do this?
Well, there's a nice factoring technique called sum of cubes factoring. It goes like this:
a^3 + b^3 = (a + b)(a^2 - ab + b^2)
IMPORTANT: You do not need to know the technique of sum of cubes factoring for the GMAT. For this reason, I believe this question is out of scope (unless there's a much easier solution that I've overlooked).
Moving along, we can apply this factoring technique to each of the sum of cubes in our rearranged expression. For example:
22^3 + 88^3 = (22 + 88)(22^2 - (22)(88) + 88^2)
= (110)(some other value)
Notice that our sum of cubes can be rewritten as 110 times some other number.
This means that (22^3 + 88^3) is divisible by 110, which means the remainder will be 0 when we divide (22^3 + 88^3) by 110
Using similar logic, we can show that (23^3 + 87^3) is divisible by 110 (since 23+87=110, which means 110 will be part of the factorization . . . try it)
And we can show that (24^3 + 86^3) is divisible by 110.
And so on.
So, almost all of the terms in (22^3 + 88^3) + (23^3 + 87^3) + (24^3 + 88^3) . . . + (54^3 + 56^3) + 55^3 are divisible by 110.
So, if we were to ignore the 55^3, the sum ((22^3 + 88^3) + (23^3 + 87^3) + (24^3 + 88^3) . . . + (54^3 + 56^3) would have remainder 0 when divided by 110
So, all we need to do now is determine the remainder when 55^3 is divided by 110
Even this step is a bit of work since it requires some modular arithmetic that we don't need for the GMAT.
Alternatively, I suppose one could evaluate 55^3 to get 166,375 and then divide by 110 to get a remainder of 55 (A)
BUT, as I said, this question is out of score for the GMAT (unless I've overlooked a much easier solution).
Cheers,
Brent
