If x and y are positive integers, and 4x^2=3y,

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M7MBA: Moderator; Posts: 2058; Joined: Sun Oct 29, 2017 4:24 am; Thanked: 1 times; Followed by:5 members

If x and y are positive integers, and 4x^2=3y,

by M7MBA » Sat May 12, 2018 5:57 am

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If x and y are positive integers, and 4x^2=3y, then which of the following must be a multiple of 9?

I. x^2
II. y^2
III. xy

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III

The OA is the option E.

I am really confused here. Could anyone give me some help, please? I'd be thankful.

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Source: Beat The GMAT — Problem Solving | Sat May 12, 2018 5:57 am

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[email protected]: Elite Legendary Member; Posts: 10392; Joined: Sun Jun 23, 2013 6:38 pm; Location: Palo Alto, CA; Thanked: 2867 times; Followed by:511 members; GMAT Score:800

by [email protected] » Sat May 12, 2018 1:18 pm

Hi M7MBA,

We're told that X and Y are POSITIVE INTEGERS and 4X^2 = 3Y. We're asked which of the following MUST be a multiple of 9. While this question is 'quirky', it can be solved with a little logic and Prime Factorization.

To start, we should prime factor 4X^2 = 3Y...

(2)(2)(X)(X) = (3)(Y)

Since these two calculations are EQUAL to one another, each 'side' has to include the SAME prime factors. The two 2s on the 'left side' must ALSO be on the 'right side' and the 3 on the 'right side' must ALSO be on the 'left side. However, since there are two Xs, there's no way for a factor to appear just once - it would have to appear TWICE. While there are lots of different possible pairs of values for X and Y, since both variables have to be positive integers, the minimum possible values would be...

X = 3, Y = 12
this would give us....
(2)(2)(3)(3) = (3)(3)(2)(2)

The other possible solutions are all MULTIPLES of those minimum values.

I. X^2... X has to be 3 or a multiple of 3, so X^2 will always be 9 or a multiple of 9.
II. Y^2.... Y has to be 12 or a multiple of 12, so Y^2 will always be 144 or a multiple of 144; thus, it will always be a multiple of 9
III. (X)(Y)... (X)(Y) will always be a multiple of 36, so it will always be a multiple of 9

Final Answer: E

GMAT assassins aren't born, they're made,
Rich

Contact Rich at [email protected]

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8086; Joined: Sat Apr 25, 2015 10:56 am; Location: Los Angeles, CA; Thanked: 43 times; Followed by:29 members

by Scott@TargetTestPrep » Tue May 29, 2018 8:05 am

M7MBA wrote:If x and y are positive integers, and 4x^2=3y, then which of the following must be a multiple of 9?

I. x^2
II. y^2
III. xy

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III

Simplifying the equation, we have:

x^2 =3y/4

We see that y must be divisible by 4, and 3y/4 must be a perfect square; thus, the least value of y is 12.

When y = 12 we have:

x^2 =36/4

x^2 = 9

x = 3

Thus, we see that x^2 must be a multiple of 9, since x^2 = 9. We also see that y^2 must be a multiple of 9, since y^2 = 144. Finally, xy is a multiple of 9 since xy = 36.

Answer: E

Scott Woodbury-Stewart
Founder and CEO
[email protected]