# Congruence $320 \equiv 1 (\text{mod }x)$ [closed]

I have the following congruence $320 \equiv 1 (\text{mod }x)$

And the question is : find all the modulos $x$ that make this congruence true.

## closed as off-topic by Najib Idrissi, hardmath, Chappers, kjetil b halvorsen, hunterMay 17 '15 at 14:51

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• $x$ has to divide $319$ – Gregory Grant May 17 '15 at 7:22
• ho thanks ! it was easy – lopata May 17 '15 at 7:23
• @GregoryGrant not only has, but it is also sufficient for $x$ to divide $319$. i.e. $x$ satisfies the conditions iff $x$ divides $319$. – user26486 May 17 '15 at 14:03

Well basically the definition of $$a \equiv b (\text{mod }n)$$ is that n should divide $(a-b)$ So according to your question $x$ should divide $320-1$ that is $319$. It would be $11$ and $29$

Since you want all the $x$'s then it would be $11$, $29$ and $319$

• and 319 too.... – Honza Brabec May 17 '15 at 7:40
• Yes I saw that thanks@HonzaBrabec – user210387 May 17 '15 at 7:41
• And $1$. And, if you allow negative moduli (some people do), four more. – André Nicolas May 17 '15 at 7:43
• @AndréNicolas 320 mod 1 is 0. – Honza Brabec May 17 '15 at 7:45
• @HonzaBrabec 0 is congruent to 1 (mod 1). – wythagoras May 17 '15 at 7:46

$320 \equiv 1 (\text{mod }x) \rightarrow 320-1 \equiv 0 (\text{mod }x) \rightarrow 319 \equiv 0 (\text{mod }x) \rightarrow x=11$ or $29$, since that only works for the divisors.

$320 \equiv 1 \pmod{x} \implies x \mid 320 -1 =319$ and hence there exists an integer $k$ such that $$xk = 319$$

Now should find all pairs $x,k$ that when you multiply them together you get $319$

$1 \times 319$

$29 \times 11$

$-1 \times -319$

$-29 \times -11$

any one of these 8 numbers can work for $x$

Basically you get the prime factorization for $319$ and then you add to your list the negatives of these numbers plus the number itself plus 1, that will give you a complete list for the divisors of $319$ or generally any number.