Question

# Solve system of equation by chinese remainder theorem: x\equiv 5(\bmod 24) x\equiv 17(\bmod 18)

Polynomial factorization
Solve system of equation by chinese remainder theorem:
$$\displaystyle{x}\equiv{5}{\left({b}\text{mod}{24}\right)}$$
$$\displaystyle{x}\equiv{17}{\left({b}\text{mod}{18}\right)}$$

2021-04-27
Step 1
The system of equation $$\displaystyle{x}\equiv{5}{\left({b}\text{mod}{24}\right)}$$
$$\displaystyle{x}\equiv{17}{\left({b}\text{mod}{18}\right)}$$
Solve system of equation by chinese remainder theorem:
Chinese Remainder Theorem:
Let $$\displaystyle{m}_{{{1}}},{m}_{{{2}}},{m}_{{{3}}}\ldots\ldots{m}_{{{r}}}$$ be a collection of pairwise relatively prime integers. Then the system of simultaneous congruences
$$\displaystyle{x}\equiv{a}_{{{1}}}{\left({b}\text{mod}{m}_{{{1}}}\right)}$$
$$\displaystyle{x}\equiv{a}_{{{2}}}{\left({b}\text{mod}{m}_{{{2}}}\right)}$$
....
$$\displaystyle{x}\equiv{a}_{{{r}}}{\left({b}\text{mod}{m}_{{{r}}}\right)}$$
has a unique solution modulo $$\displaystyle{M}={m}_{{{1}}},{m}_{{{2}}},\ldots\ldots.{m}_{{{r}}}$$ for any given integers $$\displaystyle{a}_{{{1}}},{a}_{{{2}}},\ldots\ldots{a}_{{{r}}}$$.
Step 2
Here $$\displaystyle{m}_{{{1}}}={24},{m}_{{{2}}}={18}$$
gcd(24,18)=6
$$\displaystyle\Rightarrow{\gcd{{\left({24},{18}\right)}}}\ne{q}{1}$$
Therefore, $$\displaystyle{m}_{{{1}}}\ {\quad\text{and}\quad}\ {m}_{{{2}}}$$ are not relatively prime.
Hence, the given system
$$\displaystyle{x}\equiv{5}{\left({b}\text{mod}{24}\right)}$$
$$\displaystyle{x}\equiv{17}{\left({b}\text{mod}{18}\right)}$$ has no solution.