1
$\begingroup$

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.

$\endgroup$

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Najib Idrissi, hardmath, Chappers, kjetil b halvorsen, hunter
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 4
    $\begingroup$ $x$ has to divide $319$ $\endgroup$ – Gregory Grant May 17 '15 at 7:22
  • $\begingroup$ ho thanks ! it was easy $\endgroup$ – lopata May 17 '15 at 7:23
  • $\begingroup$ @GregoryGrant not only has, but it is also sufficient for $x$ to divide $319$. i.e. $x$ satisfies the conditions iff $x$ divides $319$. $\endgroup$ – user26486 May 17 '15 at 14:03
3
$\begingroup$

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 $319$

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

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

$\endgroup$
0
$\begingroup$

$320 \equiv 1 \pmod{x} \implies x \mid 320 -1 =319$ 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.

$\endgroup$

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