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

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Director
Joined: 07 Jun 2004
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If y is a positive integer is root(y) an integer?  [#permalink]

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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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26 Jan 2011, 06:37
4
7
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.

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

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03 Feb 2014, 12:26
6
(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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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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26 Jan 2011, 07:08
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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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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Updated on: 15 May 2013, 08:25
3
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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05 Jun 2013, 03:55
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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Re: Airthmetic DS  [#permalink]

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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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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.

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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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26 Feb 2019, 21:21
1
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.

Similar questions:
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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
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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

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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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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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.

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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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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