# A simple doubt in number theory problem for an even number, [duplicate]

I considered an even number $$n\geq 9$$, where it is divisible by some positive integer $$k$$.

Also, $$k$$ does not divide $$\frac{n}{2}\$$ (negation of the hypothesis in my prior question).

Let $$n = kq$$.

Can we say that $$q$$ is even in this case? Can anyone help to get it theoretically? Thanks a lot for the help.

My attempt:

Let $$n = kq$$.

Since $$k$$ does not divide $$\frac{n}{2}$$, we have

$$n/2 = k.q_1 + r$$ for $$0.

How to conclude the even or odd property of $$k$$ and $$q$$ here?

## marked as duplicate by lulu, Bill Dubuque elementary-number-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Apr 15 at 13:51

• This is not clear. Say $k=n$. Then of course $k$ divides $n$, $k$ does not divide $\frac n2$ but $q=1$ is odd. Is that a counterexample to your claim? – lulu Apr 15 at 11:17
• Try a couple of examples. What if $n=10$? Or $12$? What can $k$ possibly be? What can $q$ possibly be? – Arthur Apr 15 at 11:17
• Strongly related: math.stackexchange.com/questions/3184984/… – Dirk Apr 15 at 11:19
• @monalisa Can you clarify your question? As stated there is no difficulty coming up with counterexamples. I assume that you intended something else? – lulu Apr 15 at 11:21
• Follows immediately from my answer in your prior question, namely by divisor $\rm\color{#c00}{recip}\color{#0a0}{rocity}\ % divisor reciprocity$ $${\rm If} \ \,\rm\ \large 2,K\mid N\ \ \ {\rm then} \ \ \rm\,\ \color{#c00}K\ {\LARGE \mid} {\large \frac{N}{\color{#0a0}2}}\! \iff{\large \color{#0a0}2}\ {\LARGE \mid} {\large \frac{N}{\color{#c00}K}}\!=\!Q\,\ [\rm\!\!\iff\!\ 2K\mid N\,]$$ Thus $\,\rm Q\,$ is even $\!\rm \iff\! K\mid N/2,\,$ which by hypothesis is false here, and true in your prior question. This is just the negation of your prior question. – Bill Dubuque Apr 15 at 14:07

As a counter example, take $$n = 14$$ and $$k = 2$$. Then $$2$$ does not divide $$14/2 = 7$$ and $$q = 7$$ is not an even number.

• I tried numerically but theoretically, I am unable to prove it. – monalisa Apr 15 at 11:32
• @monalisa: You can't prove it because it's false. You must disprove it, which means giving a (numerical) counterexample. – Cameron Buie Apr 15 at 11:48
• @CameronBuie Actually we can do better than a single counterexample, we can prove that it is always false, i.e. $q$ is always odd - see my comment on the question. – Bill Dubuque Apr 15 at 14:09
• @Bill: Fair point – Cameron Buie Apr 15 at 17:10