System of two moduli

Question

Compute which element $[0],[1],\ldots,[2014]$ in $\Bbb{Z}/2015\Bbb{Z}$ under the map of the Chinese remainder theorem is mapped to $([12],[5])\in\Bbb{Z}/31\Bbb{Z} \times \Bbb{Z}/65\Bbb{Z}$. $$ x=12 \pmod{31} \\ x=5 \pmod{65} $$ I used the Euclidean algorithm to find $\gcd(31,65)= 31n+65m=1$ for $n= 21$ and $m= -10$. Now $x=12 \cdot 65 \cdot (-10) + 5 \cdot 31 \cdot 21 = -7800 +3255 =-4545 = [1500]$.

My question is: how do I find that $-4545= [1500]$?

Answer 1

To find the equivalence class of $-4545$ in $\mathbb{Z}/2015\mathbb{Z}$, we must find $x \in \left\{0, 1, ..., 2014\right\}$ for which $k \cdot 2015 = x - 4545$ holds for some $k \in \mathbb{Z}$. By this definition, we have $-4545 \in [x]$.

Clearly, this holds for $k = -3$ and $x = 1500$, so we have $-4545 \in [1500]$.

Answer 2

$-4545=((-3)\times2015)+1500$

$ \Rightarrow -4545\equiv 1500 ($mod $2015)$

Answer 3

$$x=12 \pmod{31} \implies x = 12+31k \implies$$

$$12+31k=5 \pmod{65} \implies k = 48 \implies x = 1500$$