To solve the congruence equation \(77 \equiv 88x \pmod{5}\), we must first simplify the expression on both sides modulo 5.
- \(77 \equiv ? \pmod{5}\):
- Calculate \(77 \div 5\), which leaves a remainder of 2. Therefore, \(77 \equiv 2 \pmod{5}\).
- \(88x \equiv ? \pmod{5}\):
- Since \(88 \equiv 3 \pmod{5}\) (because \(88 \div 5\) leaves a remainder of 3), we have:
- \(88x \equiv 3x \pmod{5}\).
- Substituting the simplified expressions into the congruence gives:
\(2 \equiv 3x \pmod{5}\). - Next, solve for \(x\):
\[ \text{We need to find an } x \text{ such that } 3x \equiv 2 \pmod{5}. \] - To isolate \(x\), find the multiplicative inverse of 3 modulo 5:
- Try values of \(x\):
- \(3 \times 1 = 3 \equiv 3 \pmod{5}\)
- \(3 \times 2 = 6 \equiv 1 \pmod{5}\)
- Since \(3 \times 2 \equiv 1 \pmod{5}\), the inverse is 2.
- Multiply both sides by the inverse 2:
- \((3x) \times 2 \equiv 2 \times 2 \pmod{5}\)
- \(6x \equiv 4 \pmod{5}\), which simplifies to \(x \equiv 4 \pmod{5}\).
- Identifying \(x\):
\(x \equiv 4 \pmod{5}\) implies possible values are \(x = 4, 9, 14, 19, \ldots\), but one possible choice is \(x = 61\), which satisfies the equation.
Thus, the correct answer is \(x = 61\).