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# If y = (5-x)^1/2 where x and y are integers, what is the

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Joined: 02 Jul 2012
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If y = (5-x)^1/2 where x and y are integers, what is the  [#permalink]

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Updated on: 20 Jul 2014, 08:49
6
7
00:00

Difficulty:

65% (hard)

Question Stats:

54% (01:40) correct 46% (01:54) wrong based on 327 sessions

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If $$y = \sqrt{5-x}$$ where x and y are positive integers, what is the remainder when x is divided by 5?

(1) x < 5

(2) y > 1

Another question from me. No OA as this is my own question. Please let me know if you dont agree with my answer and if so, the reason.. . Here goes..
Source : Me..

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Originally posted by MacFauz on 11 Dec 2012, 03:23.
Last edited by Bunuel on 20 Jul 2014, 08:49, edited 3 times in total.
Renamed the topic and edited the question.
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Re: If y = (5-x)^1/2 where x and y are integers, what is the  [#permalink]

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11 Dec 2012, 03:52
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If $$y = \sqrt{5-x}$$ where x and y are positive integers, what is the remainder when x is divided by 5?

Since square root from a negative number is undefined and $$y$$ is a positive integer, then $$5-x>0$$ --> $$x<5$$. Since also given that $$x$$ is positive, then $$0<x<5$$, thus $$x$$ could be 1, 2, 3 or 4. Now, $$y$$ would be an integer only for $$x=1$$ or $$x=4$$.

If $$x=1$$, then $$y=2$$.
If $$x=4$$, then $$y=1$$.

(1) x < 5. Not sufficient.

(2) y > 1 --> $$y=2$$ --> $$x=1$$. Sufficient.

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Re: If y = (5-x)^1/2 where x and y are integers, what is the  [#permalink]

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11 Dec 2012, 09:47
1
MacFauz wrote:
If $$y = \sqrt{5-x}$$ where x and y are positive integers, what is the remainder when x is divided by 5?

(1) x < 5

(2) y > 1

Another question from me. No OA as this is my own question. Please let me know if you dont agree with my answer and if so, the reason.. . Here goes..
Source : Me..

Sq root of a neg number is not defined, hence 5-x >0 or x<5 now x is a positive integer hence X lies between 1 & 5 .

1) X<5 - Already determined , but not sufficient
2) y>1 - next integer y= 2 ; => x = 1

Hence remainder when X divided by 5 is 1

Ans B
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Re: If y = (5-x)^1/2 where x and y are integers, what is the  [#permalink]

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10 Aug 2013, 06:56
y^2 = 5 -x

x= 5 - y^2

(1).

x=1,4 since x<5

(2).

y>1

x=1 hence REM=1

Hence (B)
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Re: If y = (5-x)^1/2 where x and y are integers, what is the  [#permalink]

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21 Sep 2015, 23:43
y=sqrt*(5-x)
y^2=5-x
y^2+x=5

x/5=q+R , what is R?

St1. x<5, no new information, INSUFF

St2. y>1 means y=2, so x=1. R=1. SUFF

B
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Re: If y = (5-x)^1/2 where x and y are integers, what is the  [#permalink]

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19 Oct 2018, 10:45
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Re: If y = (5-x)^1/2 where x and y are integers, what is the   [#permalink] 19 Oct 2018, 10:45
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