0
$\begingroup$

I'm trying to answer this question which has a hint: think about $\mathbb Z_{3600}$.

I tried to set up a linear equation,$\mod{3600},$ without any success. Not even the factorization of $3600$ gives me any ideas on how to set the problem.

Any help?

$\endgroup$
0
2
$\begingroup$

Note that$$\gcd(3600,x)=9\Longrightarrow x=9y\quad,\quad y\in \Bbb Z_{400}\to \gcd(400,y)=1$$therefore the number of solutions of $\gcd(3600,x)=9$ on $\Bbb Z_{3600}$ is equal to number of solution of $\gcd(400,y)=1$ on $\Bbb Z_{400}$ which is $\phi(400)$.

$\endgroup$
0
0
$\begingroup$

A (different) hint: notice that

$$ \gcd(3600,x) = 9 \iff \gcd(400,x/9) = 1.$$

$\endgroup$
2
  • $\begingroup$ Not true. You mean $\gcd(400,x/9)=1$ (and $x$ has to be a multiple of $9$). $\endgroup$
    – TonyK
    Jan 18 '19 at 16:17
  • $\begingroup$ @TonyK Edited, thanks. $\endgroup$
    – SvanN
    Jan 18 '19 at 16:17
0
$\begingroup$

Hint: Such a number must be of the form $9k$, where $k\le 400$ and $\gcd(400,k)=1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.