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If is an integer and xy2 = 36, how many values are possible
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30 Aug 2018, 07:52
30 Aug 2018, 07:52
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Question Stats:
19% (01:43) correct
80% (01:35) wrong based on 76 sessions
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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
(A) Two
(B) Three
(C) Four
(D) Six
(E) Eight
Re: If is an integer and xy2 = 36, how many values are possible
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07 Sep 2018, 19:19
07 Sep 2018, 19:19
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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
(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.***
The best answer is E.
Re: If is an integer and xy2 = 36, how many values are possible
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20 Jun 2023, 01:50
20 Jun 2023, 01:50
1
Given:
x is a perfect square, and y is an integer.
x^2 must therefore divide 36, leaving us with:
x^2=1,4,9,36
x=1,2,3,6
Now just list the cases:
If x=1, y^2=36, y=+-6
If x=4, y^2=9, y=+-3
If x=9, y^2=4, y=+-2
If x=36, y^2=1, y=+-1
For a total of 8 possible values of Y
E.
x is a perfect square, and y is an integer.
x^2 must therefore divide 36, leaving us with:
x^2=1,4,9,36
x=1,2,3,6
Now just list the cases:
If x=1, y^2=36, y=+-6
If x=4, y^2=9, y=+-3
If x=9, y^2=4, y=+-2
If x=36, y^2=1, y=+-1
For a total of 8 possible values of Y
E.
