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

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

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New post Updated on: 20 Jul 2014, 09:49
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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.. :P

My Answer is B

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Originally posted by MacFauz on 11 Dec 2012, 04:23.
Last edited by Bunuel on 20 Jul 2014, 09:49, 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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New post 11 Dec 2012, 04:52
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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.

Answer: 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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New post 11 Dec 2012, 10:47
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.. :P

My Answer is B


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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New post 10 Aug 2013, 07:56
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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New post 22 Sep 2015, 00:43
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

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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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New post 19 Oct 2018, 11:45
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Re: If y = (5-x)^1/2 where x and y are integers, what is the   [#permalink] 19 Oct 2018, 11:45
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