Question

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

Polynomial factorization
ANSWERED
asked 2021-04-25
Solve system of equation by chinese remainder theorem:
\(\displaystyle{x}\equiv{5}{\left({b}\text{mod}{24}\right)}\)
\(\displaystyle{x}\equiv{17}{\left({b}\text{mod}{18}\right)}\)

Answers (1)

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.
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