Question

Solve the following systems of congruences. xequiv 4(mod 5) xequiv 3(mod 8) xequiv 2(mod 3)

Congruence
ANSWERED
asked 2021-02-21
Solve the following systems of congruences.
\(x\equiv 4(mod\ 5)\)
\(x\equiv 3(mod\ 8)\)
\(x\equiv 2(mod\ 3)\)

Expert Answers (1)

2021-02-22
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
\(x\equiv (mod\ m)\)
\(x \equiv b(mod\ n)\)
Furthermore, any two solutions x and y are congruent modulo mn.
2) Theorem: Addition and Multiplication Properties:
If \(a = b(mod\ n)\) and x is any integer, then \(a + x \equiv b + x(mod\ n)\ and\ ax \equiv bx (mod\ n)\).
3) Theorem: Cancellation Law:
If \(ax \equiv ay (mod\ n)\ and\ (a, n) = 1,\ then\ x \equiv y(mod\ n)\).
Explanation:
Consider the system of congruences
\(x \equiv 4(mod\ 5)\)
\(x \equiv 2(mod\ 3)\)
Since 5 and 3 are relatively prime, then \((5, 3) = 1\).
Then, by using theorem there exists an integer x that satisfies the system of congruences.
From the first congruence \(x = 4 + 5k\) for some integer k and substitute this expression for x into the second congruence.
\(4 + 5k \equiv 2 (mod\ 3)\)
By using addition property,
\(4+5k + (—4) \equiv 2+(—4)(mod\ 3)\)
\(\Rightarrow 5k =—2(mod\ 3)\)
Since \(5 \equiv 2(mod\ 3)\),
\(\Rightarrow 2k \equiv -2(mod\ 3)\)
Since \((2, 3) = 1\) then by using cancellation law,
\(\Rightarrow k\equiv -1(mod\ 3)\)
Now, \(-1 \equiv 2(mod\ 3)\),
Therefore, \(k = 2 (mod\ 3)\)
Thus, \(x = 4 + 5(2) = 14\) satisfies the system and \(x = 14 (mod\ 5 * 3)\) or \(x = 14 (mod\ 15)\) gives all solutions to the given system of congruences.
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