OG Quant #163

[DCS80]

[(8^2)(3^3)(2^4)]/(96^2)

i understand the concept of breaking down 96^2 in terms of the numerator, but how do you factor it appropriately? maybe I'm missing a simple concept, but the answer is vague in the explanations section.

thx

[vkb001]

I don't have OG with me, so can't look at answers given there.

And, I don't understand your concern. You seem to be fine with breaking down 96 into factors. What are you looking for?

[lunarpower]

[(8^2)(3^3)(2^4)]/(96^2)

i understand the concept of breaking down 96^2 in terms of the numerator, but how do you factor it appropriately? maybe I'm missing a simple concept, but the answer is vague in the explanations section.

thx

when it comes to breaking down powers of integers -- if you don't immediately see any tricks or shortcuts, just break all the integers down into primes. (in fact, this should almost be a knee-jerk reaction -- if you see powers of specific whole numbers, your brain should yell, "primes!")

so, here, that would include not only breaking down the 96 in the denominator, but also breaking the 8 (in the numerator) down into more 2's.