Finding $9202 \pmod {15}$

Question

I was trying to calculate this using this formula:

$$ac \cong bc \pmod{mc}$$

If I make c = 5 so that mc = 3 * 5, I have to divide 9202 by 5 to get a. The problem is that this number isn't an integer. I tried the other way around with c = 3 and I had the same issue. Am I using the formula wrong or does it just not work in some cases?

Answer 1

The simplest approach is just to divide $9202$ by $15$ and find the remainder. In this case $9202=15\cdot 613+7$ so $9202 \equiv 7 \pmod {15}$

Your approach does not strictly work because if $c=5$ you cannot have $ac=9202$ because it is not a multiple of $5$. You could reduce $9202$ modulo $3$ and $5$ and use the Chinese Remainder Theorem to combine the results.

Answer 2

The result can be applied only if $\text{gcd}(9202,15)>1$.

Answer 3

Instead consider that $9000$ mod $15$ is $0$ since $9000$ is clearly a multiple of $15$. Since $180$ is a multiple of $15$, we know $9180$ mod $15$ is $0$. So $9202$ mod $15=22$ mod $15=7$

Answer 4

$9202\equiv1\bmod3$ because $9+2+0+2\equiv1\bmod3$.

$9202\equiv2\bmod5$ because $9200\equiv0\bmod5$.

Now use the Chinese remainder theorem

(or just figure $9202\equiv2\bmod5\implies 9202\equiv 2, 7, $ or $ 12\bmod15$,

and of those choices only $7$ is consistent with $9202\equiv1\bmod 3$).