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Is (x + y)^(1/2) an integer? (1) x^3 = 64 (2) x^2 = y – 3

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HariS1992
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Is (x + y)^(1/2) an integer? (1) x^3 = 64 (2) x^2 = y – 3 [#permalink] New post  Updated on: 03 Oct 2018, 00:10
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Is √(x + y) an integer?

(1) x^3 = 64
(2) x^2 = y – 3
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Originally posted by HariS1992 on 03 Oct 2018, 00:00.
Last edited by Bunuel on 03 Oct 2018, 00:10, edited 1 time in total.
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pushpitkc
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Re: Is (x + y)^(1/2) an integer? (1) x^3 = 64 (2) x^2 = y – 3 [#permalink] New post  03 Oct 2018, 00:07
HariS1992 wrote:
Is √(x + y) an integer?

(1) x^3 = 64
(2) x^2 = y – 3



1. x^3 = 64
We can get the value of x = 4, but without the value of y - we can't
say whether √(x + y) is an integer. (Insufficient)

2. x^2 = y – 3
If x = 1 and y = 4, then √(x + y) is not an integer
If x = 2 and y = 7, then √(x + y) is an integer (Insufficient)

On combining the information in both the statements, we can clearly
get a unique value of x and y and √(x + y) is not an integer (Sufficient - Option C)

Btw, are you sure this is from the OG - 2017(because if it is - it must have definitely been discussed)
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vipulkhandelwal999
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Re: Is (x + y)^(1/2) an integer? (1) x^3 = 64 (2) x^2 = y – 3 [#permalink] New post  03 Oct 2018, 02:59
I think OA should be B

because sqrt(x^2 + x + 3) will never be a perfect square for any integer value of x
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Re: Is (x + y)^(1/2) an integer? (1) x^3 = 64 (2) x^2 = y – 3 [#permalink] New post  03 Oct 2018, 06:49
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vipulkhandelwal999 wrote:
I think OA should be B

because sqrt(x^2 + x + 3) will never be a perfect square for any integer value of x



Take x as -3
U will get your answer
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errorlogger
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Re: Is (x + y)^(1/2) an integer? (1) x^3 = 64 (2) x^2 = y – 3 [#permalink] New post  01 Jan 2022, 08:03
HariS1992 wrote:
Is √(x + y) an integer?

(1) x^3 = 64
(2) x^2 = y – 3


Hi Bunuel,

If x^3 = 64, x should be +4, -4
On Combining (1) (2):
We get √(x + y) = √17 or √(x + y) = 3

It the OA correct?
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IanStewart
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Re: Is (x + y)^(1/2) an integer? (1) x^3 = 64 (2) x^2 = y – 3 [#permalink] New post  01 Jan 2022, 08:42
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errorlogger wrote:
If x^3 = 64, x should be +4, -4


If x^3 = 64, then x can only be equal to 4. It can't be that x = -4, because (-4)^3 = -64, not +64.

You're surely thinking about even exponents: if you see equations like these, with a nonzero even exponent on one side and a positive number on the other, there will always be two solutions (you only have one solution when the right side is zero) :

x^2 = 25 --> x = 5 or -5
x^4 = 81 --> x = 3 or -3
x^20 = 1 --> x = 1 or -1

but if you see a similar equation with an odd exponent on the left side, there is always just one solution:

x^3 = 27 --> x = 3
x^5 = -32 --> x = -2
x^101 = 1 --> x = 1
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Re: Is (x + y)^(1/2) an integer? (1) x^3 = 64 (2) x^2 = y 3 [#permalink] New post  30 Jan 2023, 04:34
Hello from the GMAT Club BumpBot!

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Re: Is (x + y)^(1/2) an integer? (1) x^3 = 64 (2) x^2 = y 3 [#permalink]
30 Jan 2023, 04:34
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