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Can somone explain me, how I can check if an number is an divisor of a sum with large exponents? Something like this:

Is $5$ a divisor of $3^{2012} - 4^{2011}$?

And how can I calculate something like that:

$39x \bmod 680 = 1$.

Thanks for your help

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    For some basic information about writing math at this site see e.g. [here](http://meta.math.stackexchange.com/questions/5020/), [here](http://meta.stackexchange.com/a/70559/155238), [here](http://meta.math.stackexchange.com/questions/1773/) and [here](http://math.stackexchange.com/editing-help#latex).2012-11-11

2 Answers 2

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For your second question, solving $39x \bmod 680 = 1\tag{1}$

is equivalent to solving the following congruence equation, $\bmod(680)$:

$39x \equiv 1 \pmod{680}.\tag{2}$

There is more than one solution: there are infinitely many solutions for $x$. Every integer $x$ which satisfies the following equation is a solution: $39x = 680k + 1$

Experiment with particular values for $k$ and see what values of $x$ you arrive at. Then try to define the set of all solutions.


ADDED: Solving $(2)$ gives us

$x \equiv 279 \pmod{680}.\tag{3}$

Then assuming you are looking for all integer solutions for $x$ we have, as solutions, all $x$ satisfying

$x = 680k + 279\quad k\in \mathbb{Z}.\tag{4}$

Note that when $k=0$, $x = 279$, which is the least positive solution solving your equation. So the set of all integer solutions satisfying $(1)$is given by $\{x\mid x =279 \pm 680k, k\in \mathbb{Z}\}.$


Please, in the future, if you have more than one sufficiently unrelated questions, post them separately.

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    $\ddot\smile~~~~~~$2013-08-24
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You can look at $3,3^2,3^3,\dots$, reducing each modulo $5$, until you see (and can prove) a pattern; then do the same with $4,4^2,4^3,\dots$.

Your other question is different --- you should make it a separate question (or, better, search the site, because that kind of linear congruence has been discussed here many times).