If x ≠ 0, does x have an odd number of factors?

Question

(1) √x is an integer.

(2) x^2 is an integer.

KrishnakumarKA1 · Accepted Answer

Hi,

This question, if you know below properties of perfect square numbers then its just a twenty second problem.

1). A perfect square will always have an odd number of factors

2). A square of a prime number will always have exactly 3 factors

3). A perfect square will have an odd number of odd factors and even number of even factors.

Let’s say if one student could not recollect this during the exam,then its always good to try some numbers.

Let’s say for an example,

If x = 2,3 then it has even number of factors. Infact all the prime numbers as even number of factors.

If x is an odd number (not prime), let’s say x = 15, then x has 1,3,5,15 four factors.

If x is an even number (not prime), let’s say x = 20, then x has 1,2,4,5,10 and 20 six factors.

Now lets x is 4 or 9, then the factors are 1,2, 4 and for 9 we have 1,3,9.

We can see that, perfect square numbers have odd number of factors.

This is infact because,

When we prime factorize as perfect square number, always the power of primes is even.

Then the number of factors would be odd.

For example, 36 = 2^2 * 3^2

So, the number of factors is equal to (2+1) * (2+1) = 9 factors.

During the exam, it won’t take more than 40 or 50 secs to check the possibilities(But I would still recommend to know the properties as these are commonly asked questions in GMAT).

Statement I is sufficient:

√x is an integer.

So, x is a perfect square number.

So sufficient.

Statement II is insufficient:

x^2 is an integer.

First thing is “x” here may or may not be an integer.

Let’s say x = √2 is not an integer. But x^2 is not an integer.

Even if we consider “x” is an integer, we are not still sure whether “x” is a perfect square number or not.

So not sufficient.

So the answer is A here.

chetan2u · Answer

If x ≠ 0, does x have an odd number of factors?

ODD number of factors :- Only perfect square have odd factors  
so the question is :-   Is x a perfect square?

(1) √x is an integer. 
let √x=y, where y is an integer
so  x=y^2 , thus x is a perfect square
sufficient

(2) x^2 is an integer. 
if x is an integer, ans can be YES
but if x is not an integer, for example  x^2=3......x=\sqrt{3} ... ans is NO
insuff

A

NidSha · Answer

chetan2u

I had a doubt in the part where you mention - 
(2) x^2 is an integer.
if x is an integer, ans is YES
but if x is not an integer, or example x2=3......x=3√x2=3......x=3... ans is NO
insuff

if i assume = x = 16 i.e. integer and because 16 is a perfect sq. we can conclude = odd no. of integers
However if i assume = x = 2 i.e. integer but because 2 is not a perfect sq. we can conclude = no odd no. of integers,

hence unless we know what the exact value of x is (irrespective of integer or not) = would it not be impossible to state whether or not x having odd # of factors?

if i am missing out on a concept, please guide .
TIA

chetan2u · Answer

I think you mean x^2 =16 then x=4... X is an integer so x^2 has odd factors
And x^2 =2 then x=√2, here we do not take X as an integer, that is why it is not perfect square.

If you meant X is integer..
X=16, so x^2 =16^2, thus 16^2 is square
X=2 so x^2=2^2=4, again a perfect square

NidSha · Answer

chetan2u

sorry, i couldn't express concrete,

Ask = does x have odd no. of factors?

i am assuming that i the concept here is right => perfect squares have odd no. of factors | non perfect squares we cant say unless we know the exact number

Hence,
1. if x^sq = 256 ; hence x = 16 and integer; and yes, going by the conceptual assumption because x = perfect square (as 4*4 = 16) => x has odd no. of factors
2. However, if x^sq = 4 ; hence x = 2 and integer; and again, going by the conceptual assumption because x = not a perfect square => x having odd no. of factors is not clear

by the above two example, shouldn't Statement (2) of the question be insufficient irrespective of x being an integer or not?

Thanks for your prompt responses  chetan2u

chetan2u · Answer

Yes you are absolutely correct in your observation..
If x^2 is an integer, X may be a square or may not be .
However if it is not an integer, it is always NO.

NidSha · Answer

chetan2u Thanks a lot for your help

fskilnik · Answer

Beautiful problem!

?\,\,\,\,:\,\,\,\,\,\# \,\,\left( {{	ext{positive}}} ight)\,\,{	ext{factors}}\,\,{	ext{odd}}\,\,{	ext{?}}

Important: it is not known (pre-statements) whether x is an integer (this is part of the problem)! 
If x is not an integer, the answer (=focus) is <NO>, because non-integers don´t have factors (by definition, factors are related to integers only)!

Statement (2) will explore that. Let´s start with it!

\left( 2 ight)\,\,\,\left\{ \begin{gathered}
 \,Take\,\,x = \sqrt 2 \,\,\,\,\,\,\left\langle {{	ext{NO}}} ightangle \,\,\,\,\,\, \hfill \
 \,Take\,\,x = 1\,\,\,\,\,\left\langle {{	ext{YES}}} ightangle \hfill \ 
\end{gathered} ight.

\left( 1 ight)\,\,\sqrt x \,\,\,\operatorname{int} \,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\{ \begin{gathered}
 x \geqslant 0\,\,\,\,{	ext{implicitly}}\,\,\,\,\,\,\mathop \Rightarrow \limits^{x\,\, 
e \,\,0} \,\,\,\,\,x > 0 \hfill \
 x\,\, = {\left( {\sqrt x } ight)^2}\,\, = \,\,{\operatorname{int} ^{\,2}} = {	ext{perfect}}\,\,{	ext{square}} \hfill \ 
\end{gathered} ight.\,

x\,\,{	ext{perfect}}\,\,{	ext{square}}\,\,\, \geqslant \,\,\,{	ext{1}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\left\{ \begin{gathered}
 \,\,x = 1\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{	ext{YES}}} ightangle \,\,\, \hfill \
 \,\,x \geqslant 4\,\,\,\,\, \Rightarrow \,\,\,\,\,{	ext{x}}\,{	ext{ = }}\,{	ext{prime}}{{	ext{s}}^{\,{	ext{even}}\,{	ext{powers}}}}\,\,\, \hfill \ 
\end{gathered} ight.

{	ext{x}}\,{	ext{ = }}\,{	ext{prime}}{{	ext{s}}^{\,\boxed{{	ext{even}}\,\,{	ext{powers}}}}}\,\,\,\,\, \Rightarrow \,\,\,\,? = \left( {\boxed{{	ext{even}}}\, + 1} ight)\,\, \cdot \left( {\boxed{{	ext{even}}}\, + 1} ight) \cdot \ldots \cdot \left( {\boxed{{	ext{even}}}\, + 1} ight) = {	ext{odd}}\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{	ext{YES}}} ightangle \,\,

Important: the explanation above shows the reason why every nonzero perfect square has an odd number of positive factors.

The above follows the notations and rationale taught in the GMATH method.

Deepshikha007 · Answer

Hi

I want to know for statement 1, should we not use x=1 as square root of 1 is also 1, an integer. But 1 would have even no of factors. So the statement would not be sufficient. 
If the team could please help. 
Thank you .

Posted from my mobile device

KanishkM · Answer

I am no expert here,

But isn't factor of 1 is just one i.e. = 1

Even if we take the inline logic for calculating factors
1^(1+1) = 1^(2) = 1

It will still be 1, which is still an odd factor.

Let me know if your query got addressed or not.

Deepshikha007 · Answer

I do understand the part that 1 would have just 1 factor hence odd no. of factors.
But if we look at it technically then 1^1+1= 1+1 = 2 , an even number.

So technically its even but logically odd.

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If x ≠ 0, does x have an odd number of factors?

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