Using Chinese Remainder Theorem for large modulo I'm an undergraduate and currently in a course for abstract algebra. I'm trying to resolve the following problem:
Compute which element of $\mathbb{Z}/2550\mathbb{Z}$ under the map of the Chinese Remainder Theorem is mapped to $(\bar{14},\bar{32})$ in $\mathbb{Z}/50\mathbb{Z} \times\mathbb{Z}/51\mathbb{Z}$.
Now I think I should solve the following system:
$x=14 \mod 50$
$x=32 \mod 50$
$x=14 \mod 51$
$x=32 \mod 51,$
but if I attempt this using the 'general' CRT I have to compute very large multiplications (such as $50 \cdot 50 \cdot 51 \cdot 51$) and since no calculator is used I think I'm on the wrong course.
Any suggestions how to handle this?
 A: All you need to do is to solve the system$$\left\{\begin{array}{l}x\equiv14\pmod{50}\\x\equiv32\pmod{51}.\end{array}\right.$$The numbers $50$ and $51$ are coprime and $1=51-50$. Therefore,$$-18(=14-32)=-18\times51+18\times50$$and so$$32-18\times51=14-18\times51=-866\equiv1664\pmod{\!2550}.$$So, the answer is $\overline{1\,664}$.
A: A very simple  way: $\bmod (50,51)\!:\,\ (14,32)\equiv 16(\color{#0a0}4,\color{#c00}2)\to 16(\color{#90f}{104})=1664\ $   since by Easy CRT 
$\quad\ \ \begin{align}x&\equiv\color{#0a0} 4\!\!\!\pmod{50}\\ x&\equiv \color{#c00}2\!\!\!\pmod{51}\end{align}$ $\iff x\equiv\color{#c00} 2+51\underbrace{\left[\dfrac{\color{#0a0}4-\color{#c00}2}{51}\bmod 50\right]}_{\Large \equiv\ 2/1\ \ \ \ }\!\equiv \color{#90f}{104}\,\pmod{50\cdot 51}$
Remark $ $ Without factoring  $\,  x\equiv 51\cdot\color{#0a0}{14} - 50\cdot \color{#c00}{32}\equiv 714\! -\! 1600\equiv -886\equiv 2550-886\equiv 1664\ $ hence the arithmetic is harder mentally without factoring out $16$ (or $32)$ at the start.
