finding smallest natural number $n$ such that $n \equiv 1023 \bmod 2015$ and $n \equiv 1302 \bmod 2016$ I had the following exercise:
\begin{equation}
\begin{cases}
n \equiv 1023 \bmod 2015\\ 
n \equiv 1302 \bmod 2016
\end{cases}
\end{equation}
Since $2015$ and $2016$ are coprime, we could use the Chinese Remainder theorem to compute $n$ and then add or substract multiples of $2015 \cdot 2016$ in order to find $n$. However, this seemed a quite large computation to me, so I computed that $\text{gcd}(2015, 1023) = 31$ and $\text{gcd}(2016, 1302) = 42$. Since $2015$ divides $1023 - n$, we must have that $31$ divides $1023 - n$, which would result in $n \equiv 1023 \equiv 0 \bmod 31$, since $31$ divides $1023$. In the same way, I would get that $n \equiv 0 \bmod 42$. But this would give as a solution that $n = 0$ (or $31 \cdot 42$ depending on whether we treat $0$ as a natural number), but this is clearly not correct...
I have no idea where I made an error, so my questions are:
[1] Where did I make a mistake
[2] This method I used is clearly not right, so how could I simplify the given system of congruences to something which is easier to compute?
Thank you in advance
 A: It's an easy inverse case of CRT, i.e. one modulus is $\equiv 1\,$ mod the other, so the $\rm\color{#c00}{inverse}$ in the CRT formula is trivial to compute, e.g. specializing $ $ Easy CRT $ $  with $\rm\,\color{#c00}m\equiv \pm1\pmod{\! n}\,$ yields
$\rm {\rm if}\ \ m\equiv \pm1\pmod{\!n}\,\ \ {\rm then}\ \ \begin{align}&\rm x\equiv a\!\!\pmod{\!m}\\  &\rm x\equiv b\!\!\pmod{\! n}\end{align}$ $\!\iff \rm  x^{\phantom{|^{|}}}\!\!\! \equiv a \pm m\,(\color{#0a0}{b-a})\,\ \pmod{\!mn}$
Theorem $ $ (Easy CRT) $\rm\ \ $ If $\rm\ m,\,n\:$ are coprime integers then
$\ \ \ \qquad\qquad\qquad\quad\qquad\qquad\displaystyle\begin{align}&\rm x\equiv a\!\!\pmod{\!m}\\  &\rm x\equiv b\!\!\pmod{\! n}\end{align}$ $\displaystyle\! \iff\rm   x \equiv a + m \bigg[\frac{\color{#0a0}{b-a}}{\color{#c00}m}\ mod\ n\bigg]\!\!\!\pmod{\!mn}$
Proof $\rm\,\ m,n\,$ coprime $\:\rm\Rightarrow\, \color{#c00}{{\large\frac{1}m} = m^{-1}}\!\pmod{\! n}\, $ exists, by Bezout or Euler's $\phi$ Theorem.
$\rm\ (\Leftarrow)\ \ \ mod\ m\!:\,\ x \equiv a + m\left[\cdots\right] \equiv a,\ $ and $\,\rm\ mod\,\ n\!:\,\ x \equiv a + m\,\color{#c00}{\large\frac{1}m}\,(b-a) \equiv b$
$\rm (\Rightarrow)\ \ $ The solution is unique $\!\rm\pmod{\!mn} $ since if $\rm\ x',\,x\ $ are solutions then $\rm\ x'\equiv x\ $ mod $\rm\:m,n\:$ hence $\rm\ m,n\mid  x'-x\ \Rightarrow\ m\,n\mid x'-x\  $ since $\rm\ m,\,n\:$ coprime $\rm\:\Rightarrow\ lcm(m,n) = m\,n$.

Hence we conclude that  $\ \begin{align}&\rm x\equiv a\!\!\pmod{\!2016}\\  &\rm x\equiv b\!\!\pmod{\! 2015}\end{align}\!\iff \rm  x \equiv a + 2016\,(b\!-\!a)\,\ \pmod{\!2016\cdot 2015}$
OP is the special case $\rm\,a = 1023,\ b = 1302.\,$
A: We search numbers $s,t$ with $$n=2015s+1023=2016t+1302$$
Taking this modulo $2015$ gives
$$1023=t+1302$$ immediately giving $t=-279\equiv 1736\mod 2015$
Hence, we have $n=2016\cdot 1736+1302=3501078$
