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Source: — Problem Solving |

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by Brent@GMATPrepNow » Sat Jun 11, 2016 7:04 am
Here's a nice discussion about this question: https://www.beatthegmat.com/veritas-quan ... 79772.html

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by MartyMurray » Sat Jun 11, 2016 8:01 am
lucas211 wrote:If a and b are integers and (ab)� = 96y, y could be

(a) 5
(b) 9
(c) 27
(d) 81
(e) 125
If a and b are integers, then (ab)� = (all of the prime factors of a and b)�.

So the prime factors in 96y have to all be raised to the fifth power.

96 = 3 x 32 = 3 x 2�.

2 is raised to the fifth power, but 3 is not. So y must include in its prime factors at least 3�.

3� = 81. Seeing that, you can immediately choose answer choice D.
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by [email protected] » Sat Jun 11, 2016 9:54 am
Hi lucas211,

This question is essentially about prime-factorization - the idea that any positive integer greater than 1 is either prime or the product of a bunch of primes.

Here, we're told that (AB)^5 = 96Y, and that A and B are integers, which means...

(AB)(AB)(AB)(AB)(AB) = 96Y

We can rewrite this as....

(A^5)(B^5) = (2^5)(3)(Y)

We're asked for what Y COULD equal. This means that Y could be MORE than one value...so we should start by looking for the smallest value that Y could equal.

Notice how 2^5 could "account for" either A or B, so we need to make sure that the "Y", when combined with the "3" that's already there, could account for the other variable....

If Y = 3^4, then 96Y would = (2^5)(3^5), which gives us two integers raised to the 5th power.

Y COULD = 3^4 = 81

Final Answer: D

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