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]

11 Dec 2012, 04:23

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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..

[Reveal] Spoiler:

My Answer is B

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Last edited by Bunuel on 11 Dec 2012, 04:28, edited 2 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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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]

11 Dec 2012, 10:47

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MacFauz wrote:

. Here goes..
Source : Me..

[Reveal] Spoiler:

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

Re: If y = (5-x)^1/2 where x and y are integers, what is the [#permalink] 11 Dec 2012, 10:47

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

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