1
$\begingroup$

I'm stuck at the ending part of a math exercise on congruences.

I must solve the following system of congruences $S$:

$x \equiv 2\ (3)$

$x \equiv 3\ (5)$

I was first asked to give the remainders of the division of $3y +2$ by 5, with knowing the remainders of the division of $y$ by 5.

Here's what I did:

-If $y \equiv 0\ (5)$, then $3y +2 \equiv 2\ (5)$

-If $y \equiv 1\ (5)$, then $3y +2 \equiv 0\ (5)$

-If $y \equiv 2\ (5)$, then $3y +2 \equiv 3\ (5)$

-If $y \equiv 3\ (5)$, then $3y +2 \equiv 1\ (5)$

-If $y \equiv 4\ (5)$, then $3y +2 \equiv 4\ (5)$

Here's the part of my exercise I'm stuck with:

I must find the solutions of $3y +2 \equiv 3\ (5)$, knowing that each solution $x$ is like $x$ = $15z + 8$ with $z$ which is an integer. Then, I will need to prove that each integer $x$ like $x = 15z + 8$ is a solution of the system $S$.

I really don't know what to do with this part of my exercise, what shall I do?

Thanks for your answers.

$\endgroup$
1
$\begingroup$

Hint : $15z+8\equiv 8\ (\ mod\ 3\ )$ and $15z+8\equiv 8\ (\ mod\ 5\ )$ because of $3|15$ and $5|15$.

$\endgroup$
  • $\begingroup$ Note that $8$ is a solution, hence $15z+8$ is a solution for every $z\in\mathbb Z$ $\endgroup$ – Peter Dec 2 '15 at 14:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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