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