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If y is a positive integer is root(y) an integer?

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If y is a positive integer is root(y) an integer? [#permalink] New post 26 Jan 2011, 05:17
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If y is a positive integer is \sqrt{y} an integer?

(1) \sqrt{4y} is not an integer
(2) \sqrt{5y} is an integer
[Reveal] Spoiler: OA

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Re: Airthmetic DS [#permalink] New post 26 Jan 2011, 05:37
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rxs0005 wrote:
If y is a positive integer is root (y) an integer

S1 root( 4* y) is not an integer

S2 root( 5* y) is an integer

I do not agree with OA


If y is a positive integer is \sqrt{y} an integer?

Note that as y is a positive integer then \sqrt{y} is either a positive integer or an irrational number. Also note that the question basically asks whether y is a perfect square.

(1) \sqrt{4*y} is not an integer --> \sqrt{4*y}=2*\sqrt{y}\neq{integer} --> \sqrt{y}\neq{integer}. Sufficient.

(2) \sqrt{5*y} is an integer --> y can not be a prefect square because if it is, for example if y=x^2 for some positive integer x then \sqrt{5*y}=\sqrt{5*x^2}=x\sqrt{5}\neq{integer}. Sufficient.

Answer: D.

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If y is a positive integer, is root y an integer? [#permalink] New post 15 May 2013, 07:11
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If y is a positive integer, is root y an integer?

(1) Root 4y is not an integer.
(2) Root 5y is an integer.
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Re: Airthmetic DS [#permalink] New post 26 Jan 2011, 05:47
thanks for the explanation the approach i took was

if y = 25 then root( 5* y) root(125) is an integer is valid and root(25) is an integer

if y = 20 then root( 5* y) root(100) is still an integer BUT root(20) is not an integer so i chose A

why is this approach wrong
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Re: Airthmetic DS [#permalink] New post 26 Jan 2011, 06:08
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rxs0005 wrote:
thanks for the explanation the approach i took was

if y = 25 then root( 5* y) root(125) is an integer is valid and root(25) is an integer

if y = 20 then root( 5* y) root(100) is still an integer BUT root(20) is not an integer so i chose A

why is this approach wrong


\sqrt{125}=5\sqrt{5}\approx{11.18}\neq{integer}.
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Re: Airthmetic DS [#permalink] New post 27 Jan 2011, 01:49
nice explanation Bunuel, it was a bit tough for me too.
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Re: If y is a positive integer, is root y an integer? [#permalink] New post 15 May 2013, 07:19
If y is a positive integer, is root y an integer?

(1) Root 4y is not an integer.
(2) Root 5y is an integer.

is y a perfect square?


1) 2 * sqrt y is not an integer .............. therefore sqrt y is not an integer and thus not a perfect square.......suff

2) y could be 1/5 or 5^3 for example still it can never be a perfect square .......suff

D

Last edited by yezz on 15 May 2013, 07:25, edited 1 time in total.
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Re: If y is a positive integer is root(y) an integer? [#permalink] New post 05 Jun 2013, 02:55
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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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Theory on roots problems: math-number-theory-88376.html

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Tough and tricky exponents and roots questions (DS): tough-and-tricky-exponents-and-roots-questions-125967.html
Tough and tricky exponents and roots questions (PS): new-tough-and-tricky-exponents-and-roots-questions-125956.html

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COLLECTION OF QUESTIONS:
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Re: Airthmetic DS [#permalink] New post 26 Nov 2013, 15:36
If y is a positive integer is \sqrt{y} an integer?


1) Means that we have \sqrt{4*y^{odd}} --> \sqrt{y^{odd}} \neq {integer} for y>1 If y = 1then statement is not true.

2) \sqrt{5y} means that y has a 5^{odd} combination in its prime box, which means we have \sqrt{5^{odd}*y}, where y is some integer and \sqrt{5^{odd}*y} \neq {integer}

Both sufficient. D)
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Re: If y is a positive integer is root(y) an integer? [#permalink] New post 03 Feb 2014, 11:26
(1) \sqrt{4y} is not an integer.

That is 2\sqrt{y} is not an integer => \sqrt{y} is not an integer. Sufficient

(2) \sqrt{5y} is an integer

let p be this integer.

\sqrt{5y} = p

then \sqrt{5}\sqrt{y}=p

or \sqrt{y} = \frac{p}{\sqrt{5}}

Now an integer divided by an irrational number cannot be integer. Sufficient

D is the answer.
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Re: If y is a positive integer is root(y) an integer?   [#permalink] 03 Feb 2014, 11:26
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