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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]

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New post 26 Jan 2011, 05:17
4
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A
B
C
D
E

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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

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If y is a positive integer is root(y) an integer?  [#permalink]

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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 --> \(y\) cannot 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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Re: Airthmetic DS  [#permalink]

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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]

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New post 26 Jan 2011, 06:08
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]

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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]

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New post Updated on: 15 May 2013, 07:25
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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.

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

Originally posted by yezz on 15 May 2013, 07:19.
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]

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New post 05 Jun 2013, 02:55
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

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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Re: Airthmetic DS  [#permalink]

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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 \(y = 1\)then 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]

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New post 03 Feb 2014, 11:26
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(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]

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