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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, 06:17
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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
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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, 06:37
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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: If y is a positive integer is root(y) an integer?  [#permalink]

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New post 03 Feb 2014, 12:26
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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: Airthmetic DS  [#permalink]

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New post 26 Jan 2011, 06:47
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, 07:08
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Re: Airthmetic DS  [#permalink]

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New post 27 Jan 2011, 02:49
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, 08:25
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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, 08:19.
Last edited by yezz on 15 May 2013, 08:25, 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, 03:55
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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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, 16:36
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 26 Feb 2019, 21:18
Bunuel wrote:
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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IMO Answer should be A

S1 root( 4* y) is not an integer
Thus root(y) not an Integer.

S2 \(\sqrt{5*y}\) is an integer
Root(y) can be a perfect square if Y=0
Root (y) not a perfect square if Y = 5
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Re: If y is a positive integer is root(y) an integer?  [#permalink]

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New post 26 Feb 2019, 21:21
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roysaurabhkr wrote:
Bunuel wrote:
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.

Similar questions:
http://gmatclub.com/forum/if-x-is-a-pos ... 88994.html
http://gmatclub.com/forum/value-of-x-107195.html
http://gmatclub.com/forum/number-prop-ds-106886.html
http://gmatclub.com/forum/number-system-106606.html
http://gmatclub.com/forum/odd-vs-even-t ... 06562.html
http://gmatclub.com/forum/quant-review- ... 04421.html
http://gmatclub.com/forum/algebra-ds-101464.html
http://gmatclub.com/forum/i-cant-unders ... 01475.html


IMO Answer should be A

S1 root( 4* y) is not an integer
Thus root(y) not an Integer.

S2 \(\sqrt{5*y}\) is an integer
Root(y) can be a perfect square if Y=0
Root (y) not a perfect square if Y = 5


Check the highlighted part. 0 is NOT a positive integer.
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Re: If y is a positive integer is root(y) an integer?  [#permalink]

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New post 20 May 2019, 23:08
Bunuel answer should be A, for the second statement is insufficient

if y = 20 then root( 5* y) root(100) is still an integer BUT root(20) is not an integer so i chose A
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Re: If y is a positive integer is root(y) an integer?  [#permalink]

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New post 21 May 2019, 00:12
dine5207 wrote:
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.


Bunuel answer should be A, for the second statement is insufficient

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


Your logic is not clear. If y = 20, then (2) is satisfied \(\sqrt{5*y}=\sqrt{100}=10=integer\) and \(\sqrt{y}\) is NOT an integer (so you have a NO answer to the question). But what value of y gives an YES answer to the question?
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Re: If y is a positive integer is root(y) an integer?  [#permalink]

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New post 21 May 2019, 02:15
rxs0005 wrote:
If y is a positive integer is \(\sqrt{y}\) an integer?

(1) \(\sqrt{4y}\) is not an integer
(2) \(\sqrt{5y}\) is an integer


#1
\(\sqrt{4y}\)
2√y
y is not an integer
#2
\(\sqrt{5y}\)
y has to be 5
and √y is not integer
sufficient
IMO D
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Re: If y is a positive integer is root(y) an integer?   [#permalink] 21 May 2019, 02:15
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