Can anyone please help with this problem. We covered the Chinese Remainder Theorem in class, but I'm very confused so if anyone could be please explain each step simply I'd be very grateful.
Find the solution to the simultaneous congruences
x ≡ 17 (mod 37)
x ≡ 9 (mod 17)
x ≡ 6 (mod 7)
${\rm mod}\ 37\!:\ x \equiv 17\iff x = 17\!+\!37j\ $ for some integer $\,j$.
${\rm mod}\ 17\!:\ 9\equiv x = 17\!+\!37j\equiv 3j\iff j\equiv 3\iff \color{#c00}{j = 3\!+\!17k}\,$ for some integer $\,k$.
Hence $\ x = 17\!+\!37\color{#c00}j = 17\!+\!37(\color{#c00}{3\!+\!17k}) = 128 + 17\cdot 37\, k$
Therefore $\ x\equiv 128\pmod{27\cdot 37}\ $ is equivalent to the first two congruences.
Next solve $\ x\equiv 6\pmod 7\ $ combined with the prior, in the same way as above, i.e.
${\rm mod}\ 7\!:\,\ 6\equiv x\equiv 128 + 17\cdot 37\, k\equiv 2-k \iff k\equiv \,\ldots$
Remark $ $ This method of iteratively replacing two congruences by a single equivalent congruence works for any number of congruence - see here for a proof.
