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Factor Problem

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Factor Problem

by yvonne0923 » Tue Apr 05, 2011 11:08 am
If x and y are positive integers, is the product xy divisible by 9?

(1)The product xy is divisible by 6
(2)x and y are perfect squares.









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[spoiler]Answer: C[/spoiler]

My steps: factor of 6 = 2,3
factor of 9 = 3,3
LCM = 2*3*3 = 18
but I don't want to plug in all the random numbers into (1) and (2), is there a better way to solve this problem to avoid any missing numbers?


Thanks,
Yvonne
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by Stuart@KaplanGMAT » Tue Apr 05, 2011 12:19 pm
yvonne0923 wrote:If x and y are positive integers, is the product xy divisible by 9?

(1)The product xy is divisible by 6
(2)x and y are perfect squares.

Thanks,
Yvonne
Hi Yvonne,

Let's start by simplifying the question:

When will xy be divisible by 9? When xy contains all the primes contained by 9. So, rephrasing the question:
Does xy contain at least two "3"s?
On to the statements!

(1) xy is divisible by 6, or:

xy contains at least one "2" and and least one "3".

Could it contain two "3"s? Yes. Does it have to contain two "3"s? No. So, (1) is insufficient.

(2) x and y are perfect squares, or:

xy will contain at least 2 of each prime factor by which it's divisible.

[Perfect squares are made up of pairs of primes.]

Do we know which perfect squares xy contains? Nope, no clue: insufficient.

Combined:

from (1), we know that xy has at least one "3"; from (2) we know that if xy has at least one "3", it must have at least two "3"s.

Since xy contains at least two "3"s, it MUST be divisible by 9: together sufficient, apart insufficient, choose (C).
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by MAAJ » Tue Apr 05, 2011 3:22 pm
If x and y are positive integers, is the product xy divisible by 9?
Rephrase to does xy has at least two 3? in its prime factors?
xy -> 3,3,etc ?

(1)The product xy is divisible by 6

xy -> 2,3...?
xy has one 2 and one 3 in its prime factors, but doesn't tell us anything about the other 3 we are looking for. It could have 5,7,11,13,etc... but we are looking for the other "3"
Insufficient

(2)x and y are perfect squares
x = a²
y = b²
But doesn't tell us anything about the prime factors that we are looking for
Insufficient

(1)and (2) combined:
If the prime factors of xy are 2,3...? And if x and y are perfect squares:
Then xy must at least have 2,2,3,3...? in its prime factors because 2 and 3 are not perfect squares (x and y must have at least two equal "prime factors" to make them perfect squares)

And because it has two 3 in its prime factors, then it must be divisible by 9

[spoiler]Correct Answer (C)[/spoiler]
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