Solve the set of congruences 2x -= 1(mod 5) x -= 3(mod 4)

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Solve the set of congruences 2x -= 1(mod 5) x -= 3(mod 4)

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Solve the system of congruence's
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asked 2021-02-04

Find all solutions to the following system of linear congruences: \(\displaystyle{x}\equiv{1}\text{mod}{2},{x}\equiv{2}\text{mod}{3},{x}\equiv{3}\text{mod}{5},{x}\equiv{4}\text{mod}{7},{x}\equiv{5}\text{mod}{11}\).

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asked 2021-02-21

Use the construction in the proof of the Chinese remainder theorem to find all solutions to the system of congruences \(\displaystyle{x}\equiv{2}{\left(\text{mod}{3}\right)},{x}\equiv{1}{\left(\text{mod}{4}\right)},{\quad\text{and}\quad}{x}\equiv{3}{\left(\text{mod}{5}\right)}\).

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Answer 1

Step 1
Here, we have to solve for congruences
2x=1 mod 5
x=3 mod 4
To solve such questions, we apply chinese remainder theorem.
Chinese Remainder Theorem- Let m, n are relatively prime integers. Then the system of simultaneous congruences \(\displaystyle{x}\equiv{a}_{{1}}{\left(\text{mod}{m}\right)}{x}\equiv{a}_{{2}}{\left(\text{mod}{n}\right)}\) has a unique solution modulo M = mxn, for any given integers \(\displaystyle{a}_{{1}},{a}_{{2}}\).
Step 2
\(\displaystyle{2}{x}={1}\text{mod}{5}\Rightarrow{2}\times{3}={3}\text{mod}{5}\Rightarrow{x}={3}\text{mod}{5}\)
and we have x = 3 mod 4
Here, 4 and 5 are co-prime, therefore By chinese remainder theorem, it has unique solution module 20.
So, only possibilty is x=23 which satisfy the given congruences.
Step 3
Answer:
x=23

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Solve the set of congruences 2x -= 1(mod 5) x -= 3(mod 4)

Solve the set of congruences 2x -= 1(mod 5) x -= 3(mod 4)

Answers (1)

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