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

Question

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

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

Bunuel · Accepted Answer

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} eq{integer} --> \sqrt{y} eq{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} eq{integer} . Sufficient. Answer: D. Similar questions: https://gmatclub.com/forum/if-x-is-a-pos ... 88994.html https://gmatclub.com/forum/value-of-x-107195.html https://gmatclub.com/forum/number-prop-ds-106886.html https://gmatclub.com/forum/number-system-106606.html https://gmatclub.com/forum/odd-vs-even-t ... 06562.html https://gmatclub.com/forum/quant-review- ... 04421.html https://gmatclub.com/forum/algebra-ds-101464.html https://gmatclub.com/forum/i-cant-unders ... 01475.html

gmatprav · Answer

(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.

rxs0005 · Answer

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

Bunuel · Answer

\sqrt{125}=5\sqrt{5}\approx{11.18}
eq{integer} .

metallicafan · Answer

Thank you Bunuel. I arrived to the same conclusion. However, I have a doubt:
Ok, we know that  x\sqrt{5} .
But how can we be so sure that x is not for example a a huge number like 10*10^10000000.... to make  x\sqrt{5}  a integer?
Maybe, I am forgeting a concept. Please, your help.

Bunuel · Answer

The point is that  \sqrt{5}  is an irrational number, and  the decimal representation of an irrational number never repeats or  terminates   (irrational numbers are not terminating decimals). So,  integer*irrational
eq{integer}  (no matter how large x is,  x\sqrt{5}  will never be an integer).

Hope it's clear.

metallicafan · Answer

Thank you Bunuel! An additional question: all the square roots (or roots in general) of prime numbers are irrational?
Thanks!

Bunuel · Answer

Since no prime can be a perfect square (or perfect cube, ...), then  \sqrt{prime}=irrational .

yezz · Answer

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

Mascarfi · Answer

Bunuel

What if y=125?

125 is an integer.

sqrt of 5*125 = sqrt of 625 = 25

This would make B insufficient.

Am I missing something here?

ENGRTOMBA2018 · Answer

No, your calculations are fine and are consistent with the question. If you have y = 125,  \sqrt{125}   
eq  Integer and thus get an unambiguous "no" making statement 2 sufficient. The original question is

" \sqrt{y}  an integer?" So getting an unambiguous yes or no will be sufficient. Per Bunuel's solution, it is shown that y  can not  be a perfect square leading to  \sqrt{y}
eq Integer

Hope this helps.

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

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