Question

Solve the following systems of congruences. xequiv 2(mod 5)

Congruence
ANSWERED
asked 2020-10-28
Solve the following systems of congruences.
\(x\equiv 2(mod\ 5)\)

Answers (1)

2020-10-29
Formula used:
1) Theorem: System of congruences:
Let m and n be relatively prime and a and b integers. There exists an integer x that satisfies the system of congruences
\(x \equiv a(mod\ m)\)
\(x \equiv b(mod\ n)\)
Furthermore, any two solutions x and y are congruent modulo mn.
2) Theorem: Addition and Multiplication Properties:
If \(a \equiv b(mod\ n)\) and x is any integer, then \(a + x -= b + x(mod\ n)\ and\ ax \equiv bx (mod\ n)\).
Explanation:
Consider the system of congruences \(x \equiv (mod\ 5)\)
\(x \equiv 3 (mod\ 8)\)
Since 5 and 8 are relatively prime, then \((5, 8) = 1\).
Then, by using a theorem, there exists an integer x that satisfies the system of congruences.
From the first congruence \(x = 2 + 5k\) for some integer k and substitute this expression for x into the second congruence.
\(2+ 5k \equiv 3 (mod\ 8)\)
By using addition property,
\(2+ 5k + (—2) \equiv 3 + (—2)(mod\ 8)\)
\(= 5k = 1 (mod\ 8)\)
By using multiplication property,
\(\Rightarrow 2+5k+(-2) \equiv 3+(-2)(mod\ 8)\)
\(\Rightarrow 5k \equiv 1(mod\ 8)\)
By using multiplication property,
\(5*5k -\equiv 5*1(mod\ 8)\)
\(\Rightarrow 25k \equiv 5(mod\ 8)\)
Since \(25 \equiv 1 (mod\ 8)\),
\(\Rightarrow k \equiv 5 (mod\ 8)\)
Thus, \(x = 2 + 5(5) = 27\) satisfies the system and \(x \equiv 27 (mod\ 5 * 8)\ or\ x \equiv 27 (mod\ 40)\) gives all solutions to the given system of congruences.
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