Solve simultaneous systems of congruences $x\equiv 10 \pmod{60}$ and $x\equiv 80 \pmod{350}$
How does one solve this using CRT? Because it has duplicate primes in its factorization? I got
$$350=5*5*2*7$$
$$60=2*2*3*5$$
So I tried CRT with
$x\equiv 1\pmod3, x\equiv0\pmod5, x\equiv0 \pmod2, x\equiv3 \pmod7$ and got the wrong answer.
or should it be
$x\equiv 1\pmod3, x\equiv5\pmod{25}, x\equiv2\pmod4, x\equiv3\pmod7$
Best Answer
Your work will be greatly simplify using the formula for the solutions of a system of two simultaneous congruences with coprime moduli $a$ and $b$ based on Bézout's identity $$ua+vb=1,\qquad(u,v\in\mathbf Z):$$ $$\begin{cases} x\equiv \alpha\mod a\\x\equiv\beta\mod b \end{cases}\iff x\equiv\beta\,ua+\alpha\,vb\mod ab.$$ Now your congruences are equivalent to $\begin{cases}x'\equiv 1\mod 6\\x'\equiv8\mod 35 \end{cases}$ and $x\equiv 10x'\mod60\cdot 350$.
Added:
The given congruences mean there exist $k,l \in\mathbf Z$ such that $x=10+60k=80+350l$, i.e. $x=10(1+6k)=10(8+35l)$, so $1+6k=8+35l$, which I denote $x'$, whence the simplified system of congruences and the relation between $x$ and $x'$. congruences