Solve system of equation by chinese remainder theorem: x≡5(bmod24) x≡17(bmod18)

Step 1 The system of equation x≡5(bmod24) x≡17(bmod18) Solve system of equation by chinese remainder theorem: Chinese Remainder Theorem: Let m1,m2,m3……mr be a collection of pairwise relatively prime integers. Then the system of simultaneous congruences x≡a1(bmodm1) x≡a2(bmodm2) .... x≡ar(bmodmr) has a unique solution modulo M=m1,m2,…….mr for any given integers a1,a2,……ar. Step 2 Here m1=24,m2=18 gcd(24,18)=6 ⇒gcd(24,18)≠q1 Therefore, m1 and m2 are not relatively prime. Hence, the given system x≡5(bmod24) x≡17(bmod18) has no solution.

"Solve system of equation by chinese remainder theorem:..." was answered by our community

High School
Algebra
Algebra II
Polynomial factorization

Jaya Legge

Answered question

2021-04-25

Solve system of equation by chinese remainder theorem:

x \equiv 5 (b mod 24)

x \equiv 17 (b mod 18)

Answer & Explanation

Alannej

Skilled2021-04-27Added 104 answers

Step 1
The system of equation

x \equiv 5 (b mod 24)

x \equiv 17 (b mod 18)

Solve system of equation by chinese remainder theorem:
Chinese Remainder Theorem:
Let

m_{1}, m_{2}, m_{3} \dots \dots m_{r}

be a collection of pairwise relatively prime integers. Then the system of simultaneous congruences

x \equiv a_{1} (b mod m_{1})

x \equiv a_{2} (b mod m_{2})

....

x \equiv a_{r} (b mod m_{r})

has a unique solution modulo

M = m_{1}, m_{2}, \dots \dots . m_{r}

for any given integers

a_{1}, a_{2}, \dots \dots a_{r}

.
Step 2
Here

m_{1} = 24, m_{2} = 18

gcd(24,18)=6

\Rightarrow gcd (24, 18) \neq q 1

Therefore,

m_{1} and m_{2}

are not relatively prime.
Hence, the given system

x \equiv 5 (b mod 24)

x \equiv 17 (b mod 18)

has no solution.

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Algebra II

Didn't find what you were looking for?