Quick way to find product of large numbers / by large number

[jenniferdpitts]

Hi -

I am looking for a quick way to calculate the following that was on a GMAT practice exam:
(8^2)(3^3)(2^4) / 96^2

I tried using the units method, but the answer was incorrect.

Thanks

- sorry this should be in the GMAT prep forum

[esledge]

Hi Jennifer,

No problem; I'll move this to the GMAT Prep folder.

Solution Option 1:
Factor all non-prime bases down to their prime factors.
8 = 2^3
96 = (3)(2^5)

Then substitute into the original, and employ exponent rules to simplify.
(8^2)(3^3)(2^4) / 96^2
= ([2^3]^2)(3^3)(2^4) / [(3)(2^5)]^2
= (2^6)(3^3)(2^4) / (3^2)(2^10) {a power raised to a power-->multiply exponents}
= (2^10)(3^3) / (3^2)(2^10) {bases of 2 on top: multiplying-->add exponents}
= (3^3)/(3^2) {cancel 2^10 from top and bottom}
= 3

Solution Option 2 (a bit faster in my opinion):
Factor big numbers down to manageable numbers you know will cancel with other stuff.
96 = 8*12 = 8*3*4

Then substitute into the original, and employ exponent rules to simplify.
(8^2)(3^3)(2^4) / 96^2
= (8^2)(3^3)(2^4) / [(8)(3)(4)]^2
= (8^2)(3^3)(2^4) / (8^2)(3^2)(4^2)
= (3)(2^4) / (4^2) {recognize: 2^4 = 16 = 4^2}
= 3