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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, here, here and here. –  Julian Kuelshammer Nov 11 '12 at 23:31
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2 Answers

up vote 3 down vote accepted

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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x=680k+279: How do I get the 279 ? –  Tim Nov 12 '12 at 9:31
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@Tim: one way to get the 279 is just to try - a spreadsheet makes this very quick-put 1 through 680 in column A, then in B1 put =mod(A1*39,680) and copy down. For larger numbers, look at the Extended Euclidean algorithm –  Ross Millikan Nov 12 '12 at 14:52
    
@RossMillikan - +1 nice suggestion; spreadsheets come in handy for such tasks, and also for getting insight into problems. Also, nice link. –  amWhy Nov 12 '12 at 15:19
    
I still don't get, how you got the 279 ? With a spreadsheet, but how do you do it with one? –  user49170 Nov 12 '12 at 20:43
    
$\ddot\smile~~~~~~$ –  B. S. Aug 24 '13 at 12:02
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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).

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