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The integers x and y are positive. Is the sqrt (x^2 +7xy) an integer?

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The integers x and y are positive. Is the sqrt (x^2 +7xy) an integer?  [#permalink]

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The integers x and y are positive. Is the \(\sqrt{x^2 + 7xy}\) an integer?

(1) y = 5x
(2) y - x = 4

Originally posted by neiln413b on 03 Jul 2018, 07:54.
Last edited by Bunuel on 03 Jul 2018, 08:12, edited 1 time in total.
Renamed the topic and edited the question.
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Re: The integers x and y are positive. Is the sqrt (x^2 +7xy) an integer?  [#permalink]

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New post 03 Jul 2018, 08:00
neiln413b wrote:
Hi, any help with this DS question would be greatly appreciated!

The integers x and y are positive. Is the sqrt(x^2 + 7xy) an integer?

1) y=5x
2) y-x=4


IMO A:
option A when substituted gives the answer as 6x, since x is a positive integer 6x will be an integer.

option B when substituted gives an expression which cant be simplified in square root.
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Re: The integers x and y are positive. Is the sqrt (x^2 +7xy) an integer?  [#permalink]

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New post 03 Jul 2018, 11:10
neiln413b wrote:
The integers x and y are positive. Is the \(\sqrt{x^2 + 7xy}\) an integer?

(1) y = 5x
(2) y - x = 4


Square root of x^2 + 7xy will be an integer if x^2 + 7xy is a perfect square.

(1) Substituting y=5x, we get
x^2 + 7xy = x^2 + 7*x*5x = x^2 + 35x^2 = 36x^2.

Now since x is a positive integer, x^2 is a perfect square, and 36 is already a perfect square. So 36x^2 will also be a perfect square. Thus its square root will be an integer. Sufficient.

(2) Substituting y = 4+x we get
x^2 + 7xy = x^2 + 7x(4+x) = x^2 + 28x + 7x^2 = 8x^2 + 28x = 4x(2x+7).

Now say x=1, then 4x(2x+7) = 4*1(2+7) = 4*9 which is a perfect square.
But if say x=2, then 4x(2x+7) = 4*2(4+7) = 8*11 which is not a perfect square.
Thus this statement is Not Sufficient.

Hence A answer
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Re: The integers x and y are positive. Is the sqrt (x^2 +7xy) an integer?  [#permalink]

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New post 03 Jul 2018, 11:13
neiln413b wrote:
The integers x and y are positive. Is the \(\sqrt{x^2 + 7xy}\) an integer?

(1) y = 5x
(2) y - x = 4


1. y = 5x
The expression becomes \(\sqrt{x^2 + 7x(5x)} = \sqrt{36x^2} = 6x\)
Since, integers x and y are positive integers, the expression will be an integer (Sufficient)

2. Here, y = x + 4
The expression becomes \(\sqrt{x^2 + 7x(x+4)} = \sqrt{8x^2 + 28x} = \sqrt{4x(2x + 7)}\)
If x = 1, the expression is an integer whereas if x = 2, it is not an integer (Insufficient) (Option A)
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Re: The integers x and y are positive. Is the sqrt (x^2 +7xy) an integer?  [#permalink]

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New post 04 Jul 2018, 09:21
Statement 1 is sufficient as it is mentioned that x is an integer and this statement simplifies to 6x which has to be an integer (i.e. integer * integer = integer)

Statement 2 is Not sufficient.

Hence the correct answer is A
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Re: The integers x and y are positive. Is the sqrt (x^2 +7xy) an integer? &nbs [#permalink] 04 Jul 2018, 09:21
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