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

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

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

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

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.

chetan2u

Thanks a lot for your help

Beautiful problem!

\(?\,\,\,\,:\,\,\,\,\,\# \,\,\left( {{\text{positive}}} \right)\,\,{\text{factors}}\,\,{\text{odd}}\,\,{\text{?}}\)

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!

\(\left( 2 \right)\,\,\,\left\{ \begin{gathered}\\
\,Take\,\,x = \sqrt 2 \,\,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle \,\,\,\,\,\, \hfill \\\\
\,Take\,\,x = 1\,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \hfill \\ \\
\end{gathered} \right.\)

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

\(x\,\,{\text{perfect}}\,\,{\text{square}}\,\,\, \geqslant \,\,\,{\text{1}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\left\{ \begin{gathered}\\
\,\,x = 1\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \,\,\, \hfill \\\\
\,\,x \geqslant 4\,\,\,\,\, \Rightarrow \,\,\,\,\,{\text{x}}\,{\text{ = }}\,{\text{prime}}{{\text{s}}^{\,{\text{even}}\,{\text{powers}}}}\,\,\, \hfill \\ \\
\end{gathered} \right.\)

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

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.

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

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.

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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