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605-655 Level|   Number Properties|         
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amitjash
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If n = p/q where p and q are nonzero integers, is n an integer ? 21 Sep 2010, 21:43
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If n = p/q where p and q are nonzero integers, is n an integer ?

(1) n^2 is an integer.
(2) (2n + 4)/2 is an integer.

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Bunuel
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If n = p/q where p and q are nonzero integers, is n an integer ? 21 Sep 2010, 22:35
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If n = p/q where p and q are nonzero integers, is n an integer ?

(1) n^2 is an integer --> \(n^2\) to be an integer \(n\) must be either an integer or an irrational number (for example: \(\sqrt{3}\)), (note that \(n\) cannot be reduced fraction, for example \(\frac{2}{3}\) or \(\frac{11}{3}\) as in this case \(n^2\) won't be an integer). But as \(n\) can be expressed as the ratio of 2 integers, \(n=\frac{p}{q}\), then it cannot be irrational number (definition of irrational number: an irrational number is any real number which cannot be expressed as a fraction a/b, where a and b are integers), so only one option is left: \(n\) is an integer. Sufficient.

(2) (2n + 4)/2 is an integer --> \(\frac{2n+4}{2}=n+2=integer\) --> \(n=integer\). Sufficient.

Answer: D.
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amitjash
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If n = p/q where p and q are nonzero integers, is n an integer ? 21 Sep 2010, 23:50
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OK now i understood the fundamentals about this question. I was thinking what if n=141/100? in that case n^2 will be 2 but n will not be integer.
Thanks bunuel
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If n = p/q where p and q are nonzero integers, is n an integer ? 22 Sep 2010, 18:31
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Bunuel
If n=p/q (p and q are nonzero integers), is n an integer?

(1) n^2 is an integer --> \(n^2\) to be an integer \(n\) must be either an integer or irrational number (for example: \(\sqrt{3}\)), (note that \(n\) can not be reduced fraction, for example \(\frac{2}{3}\) or \(\frac{11}{3}\) as in this case \(n^2\) won't be an integer). But as \(n\) can be expressed as the ratio of 2 integers, \(n=\frac{p}{q}\), then it can not be irrational number (definition of irrational number: an irrational number is any real number which cannot be expressed as a fraction a/b, where a and b are integers), so only one option is left: \(n\) is an integer. Sufficient.

(2) (2n+4)/2 is an integer --> \(\frac{2n+4}{2}=n+2=integer\) --> \(n=integer\). Sufficient.

Answer: D.
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Hi Bunuel,

for (1), what if p and q are 1732 and 1000, in that casem n^2 will be an integer but not n. Isnt it possible.. 1.732^2 doesnt exactly lead to 3 but we can have such integers giving exact value of sqrt of an integer for n.
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If n = p/q where p and q are nonzero integers, is n an integer ? 22 Sep 2010, 23:37
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BalakumaranP
Bunuel
If n=p/q (p and q are nonzero integers), is n an integer?

(1) n^2 is an integer --> \(n^2\) to be an integer \(n\) must be either an integer or irrational number (for example: \(\sqrt{3}\)), (note that \(n\) cannot be reduced fraction, for example \(\frac{2}{3}\) or \(\frac{11}{3}\) as in this case \(n^2\) won't be an integer). But as \(n\) can be expressed as the ratio of 2 integers, \(n=\frac{p}{q}\), then it can not be irrational number (definition of irrational number: an irrational number is any real number which cannot be expressed as a fraction a/b, where a and b are integers), so only one option is left: \(n\) is an integer. Sufficient.

(2) (2n+4)/2 is an integer --> \(\frac{2n+4}{2}=n+2=integer\) --> \(n=integer\). Sufficient.

Answer: D.

Hi Bunuel,

for (1), what if p and q are 1732 and 1000, in that casem n^2 will be an integer but not n. Isnt it possible.. 1.732^2 doesnt exactly lead to 3 but we can have such integers giving exact value of sqrt of an integer for n.
Show more

That's not true. No reduced fraction when squared can equal to an integer.

Or consider this: \(n^2=\frac{p^2}{q^2}=integer\) --> \(n=\frac{p}{q}=\sqrt{integer}\) --> now, square root of an integer is either an integer or an irrational number. But it cannot be an irrational number as we have that \(\frac{p}{q}=\frac{integer}{integer}=\sqrt{integer}\) and we know that an irrational number cannot be expressed as a fraction of two integers, so \(\sqrt{integer}=n=integer\).

Hope it's clear.
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If n = p/q where p and q are nonzero integers, is n an integer ? 20 Oct 2010, 13:54
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Originally answered B, but D's clear now.

Thanks Bunuel!
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If n = p/q where p and q are nonzero integers, is n an integer ? 25 Oct 2010, 13:44
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Tricky I picked B but makes perfect sense now. Thanks!
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If n = p/q where p and q are nonzero integers, is n an integer ? 22 Sep 2015, 08:58
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Bunuel
If n=p/q (p and q are nonzero integers), is n an integer?

(1) n^2 is an integer --> \(n^2\) to be an integer \(n\) must be either an integer or an irrational number (for example: \(\sqrt{3}\)), (note that \(n\) can not be reduced fraction, for example \(\frac{2}{3}\) or \(\frac{11}{3}\) as in this case \(n^2\) won't be an integer). But as \(n\) can be expressed as the ratio of 2 integers, \(n=\frac{p}{q}\), then it can not be irrational number (definition of irrational number: an irrational number is any real number which cannot be expressed as a fraction a/b, where a and b are integers), so only one option is left: \(n\) is an integer. Sufficient.

(2) (2n+4)/2 is an integer --> \(\frac{2n+4}{2}=n+2=integer\) --> \(n=integer\). Sufficient.

Answer: D.
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Hi
I am having difficulty understanding statement1.
If n^2=100 then n=10
If n^2=3 then n=sqrt(3) --> why cant we express it as p/q
sorry It may be a dumbest question, but need to clarify..Thanks
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If n = p/q where p and q are nonzero integers, is n an integer ? 25 Sep 2015, 21:14
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We can not represent irrational numbers as p/q as the irrational numbers are non repeating non terminating decimals and there is no way we can write p/q for such expression.

One more thing: To check if a square of a number is irrational or not, we need to check if its prime factors have "even powers"
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If n = p/q where p and q are nonzero integers, is n an integer ? 13 Jun 2016, 00:07
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Bunuel
If n=p/q (p and q are nonzero integers), is n an integer?

(1) n^2 is an integer --> \(n^2\) to be an integer \(n\) must be either an integer or an irrational number (for example: \(\sqrt{3}\)), (note that \(n\) can not be reduced fraction, for example \(\frac{2}{3}\) or \(\frac{11}{3}\) as in this case \(n^2\) won't be an integer). But as \(n\) can be expressed as the ratio of 2 integers, \(n=\frac{p}{q}\), then it can not be irrational number (definition of irrational number: an irrational number is any real number which cannot be expressed as a fraction a/b, where a and b are integers), so only one option is left: \(n\) is an integer. Sufficient.

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You are concluding from statement 1 that N is an integer, but it is also given that N is of the form P/Q, a fraction. How can an Integer be a fraction?
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If n = p/q where p and q are nonzero integers, is n an integer ? 13 Jun 2016, 00:26
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Bunuel
If n=p/q (p and q are nonzero integers), is n an integer?

(1) n^2 is an integer --> \(n^2\) to be an integer \(n\) must be either an integer or an irrational number (for example: \(\sqrt{3}\)), (note that \(n\) can not be reduced fraction, for example \(\frac{2}{3}\) or \(\frac{11}{3}\) as in this case \(n^2\) won't be an integer). But as \(n\) can be expressed as the ratio of 2 integers, \(n=\frac{p}{q}\), then it can not be irrational number (definition of irrational number: an irrational number is any real number which cannot be expressed as a fraction a/b, where a and b are integers), so only one option is left: \(n\) is an integer. Sufficient.


You are concluding from statement 1 that N is an integer, but it is also given that N is of the form P/Q, a fraction. How can an Integer be a fraction?
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For example, if n = 2/1, or 6/3, or 10/10, ....
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If n = p/q where p and q are nonzero integers, is n an integer ? 19 Apr 2020, 07:11
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By defining ‘n’ as a ratio of integers, the question data eliminates the possibility of ‘n’ being an irrational number (in simpler terms, a root). That’s valuable information, especially when you are interpreting information like the one given in statement 1.

From statement 1 alone, \(n^2\) is an integer.

This is where the question data is valuable. Had it not been told to us that n=\(\frac{p}{q}\), p and q being non-zero integers, interpreting statement 1 would take a different turn.

Because, \(n^2\) being an integer does not automatically make ‘n’ an integer; ‘n’ could be a root also.

However, in this question, we know that ‘n’ is a rational number and hence cannot be an irrational root. Therefore, if \(n^2\) is an integer, it necessarily means that n is an integer.
Statement 1 alone is sufficient to answer the question with a definite YES. Answer options B, C and E can be eliminated. Possible answer options are A or D.


From statement 2 alone, \(\frac{(2n+4)}{2}\) is an integer.

Upon simplification, we obtain n+2 as an integer. This is possible only if n is an integer itself.
Statement 2 alone is sufficient to answer the question with a definite YES. Answer option A can be eliminated.

The correct answer option is D.

Hope that helps!
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If n = p/q where p and q are nonzero integers, is n an integer ? 18 Jun 2020, 05:57
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CrackVerbalGMAT
By defining ‘n’ as a ratio of integers, the question data eliminates the possibility of ‘n’ being an irrational number (in simpler terms, a root). That’s valuable information, especially when you are interpreting information like the one given in statement 1.

From statement 1 alone, \(n^2\) is an integer.

This is where the question data is valuable. Had it not been told to us that n=\(\frac{p}{q}\), p and q being non-zero integers, interpreting statement 1 would take a different turn.

Because, \(n^2\) being an integer does not automatically make ‘n’ an integer; ‘n’ could be a root also.

However, in this question, we know that ‘n’ is a rational number and hence cannot be an irrational root. Therefore, if \(n^2\) is an integer, it necessarily means that n is an integer.
Statement 1 alone is sufficient to answer the question with a definite YES. Answer options B, C and E can be eliminated. Possible answer options are A or D.


From statement 2 alone, \(\frac{(2n+4)}{2}\) is an integer.

Upon simplification, we obtain n+2 as an integer. This is possible only if n is an integer itself.
Statement 2 alone is sufficient to answer the question with a definite YES. Answer option A can be eliminated.

The correct answer option is D.

Hope that helps!
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Regarding statement 2:
is it possible that n = 1/3 and the equation becomes (2/3+4)/2? in this case too the result will be an integer but n is a fraction
confused here!
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If n = p/q where p and q are nonzero integers, is n an integer ? 07 Nov 2020, 08:47
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Harps
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By defining ‘n’ as a ratio of integers, the question data eliminates the possibility of ‘n’ being an irrational number (in simpler terms, a root). That’s valuable information, especially when you are interpreting information like the one given in statement 1.

From statement 1 alone, \(n^2\) is an integer.

This is where the question data is valuable. Had it not been told to us that n=\(\frac{p}{q}\), p and q being non-zero integers, interpreting statement 1 would take a different turn.

Because, \(n^2\) being an integer does not automatically make ‘n’ an integer; ‘n’ could be a root also.

However, in this question, we know that ‘n’ is a rational number and hence cannot be an irrational root. Therefore, if \(n^2\) is an integer, it necessarily means that n is an integer.
Statement 1 alone is sufficient to answer the question with a definite YES. Answer options B, C and E can be eliminated. Possible answer options are A or D.


From statement 2 alone, \(\frac{(2n+4)}{2}\) is an integer.

Upon simplification, we obtain n+2 as an integer. This is possible only if n is an integer itself.
Statement 2 alone is sufficient to answer the question with a definite YES. Answer option A can be eliminated.

The correct answer option is D.

Hope that helps!

Regarding statement 2:
is it possible that n = 1/3 and the equation becomes (2/3+4)/2? in this case too the result will be an integer but n is a fraction
confused here!
Show more
14/6 is not an integer, hence, it is not a valid example :)
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If n = p/q where p and q are nonzero integers, is n an integer ? 31 Mar 2025, 19:11
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