Question

The solution of \(7x ≡ 3(mod 5)\) is:

• A. \(x ≡ 5(mod 5)\)
• B. \(x ≡ 4(mod 5)\)
• C. \(x ≡ 1(mod 5)\)
• D. \(x ≡ 2(mod 5)\)

Answer 1

The Correct Option is B

To solve the congruence equation \(7x ≡ 3 \pmod{5}\), we first need to find the multiplicative inverse of 7 modulo 5. Two numbers \(a\) and \(b\) are multiplicative inverses modulo \(m\) if \(a \cdot b ≡ 1 \pmod{m}\). We need \(7 \cdot y ≡ 1 \pmod{5}\) for some integer \(y\).

Testing values for \(y\):

• For \(y = 1\), \(7 \cdot 1 = 7\), and \(7 \mod 5 = 2\).
• For \(y = 2\), \(7 \cdot 2 = 14\), and \(14 \mod 5 = 4\).
• For \(y = 3\), \(7 \cdot 3 = 21\), and \(21 \mod 5 = 1\).

The multiplicative inverse of 7 modulo 5 is 3 because \(7 \cdot 3 ≡ 1 \pmod{5}\).

Multiply both sides of the original congruence by 3:

\(3 \cdot 7x ≡ 3 \cdot 3 \pmod{5}\)

This simplifies to:

\(21x ≡ 9 \pmod{5}\)

Since \(21 \mod 5 = 1\), the equation simplifies to:

\(x ≡ 9 \pmod{5}\)

Finally, calculate the remainder of 9 divided by 5:

\(9 \mod 5 = 4\)

Thus, the solution is:

\(x ≡ 4 \pmod{5}\)