Show that x = (66B − 65a) mod 143.

Question

For each natural number $m$ we define $J_m = \{0, 1, . . . , m − 1\}$, the set of all possible residues modulo $m$.

Let $x \in J_{143}$.

Define $a \equiv x \pmod{11}$, $B \equiv x \pmod{13}$

Show that $x = (66B − 65a) \pmod{143}$.

I have no idea how to get started on this problem any tips or solutions would be much appreciated.

Answer 1

The Chinese Remainder Theorem guarantees that your congruences have a unique solution modulo $143$. So, just check that the given formula for $x$ works. For example, $$66b-65a\equiv0b-(-1)a\equiv a\pmod{11}\ .$$