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
$\endgroup$ 12 Answers
$\begingroup$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.
$\endgroup$ 5 $\begingroup$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).
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