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If is an integer and xy2 = 36, how many values are possible [#permalink]
Expert's post 00:00

Question Stats: 14% (01:32) correct 85% (01:36) wrong based on 34 sessions
If $$\sqrt{x}$$ is an integer and $$xy^2 = 36$$, how many values are possible for the integer y?

(A) Two
(B) Three
(C) Four
(D) Six
(E) Eight
[Reveal] Spoiler: OA

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Sandy
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Re: If is an integer and xy2 = 36, how many values are possible [#permalink]
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sandy wrote:
If $$\sqrt{x}$$ is an integer and $$xy^2 = 36$$, how many values are possible for the integer y?

(A) Two
(B) Three
(C) Four
(D) Six
(E) Eight

$$\sqrt{x}$$ = integer. It means that x is a perfect square.

Given,

$$xy^2 = 36$$

Possible combination:

$$1^26^2 = 36$$

$$2^23^2 = 36$$

$$6^21^2 = 36$$

$$3^22^2 = 36$$

Note: As x and y both are perfect square , we must consider negative values too. Then we have 4 sets of positive and 4 sets of negative values.

Total 8 sets of values. Thus possible values of y is 8. In perfect square negative value becomes positive one.*** Re: If is an integer and xy2 = 36, how many values are possible   [#permalink] 07 Sep 2018, 19:19
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