To solve the congruence \(3x \equiv 8 \pmod{5}\), we start by reducing the congruence modulo \(5\):
\[
3x \equiv 8 \pmod{5}.
\]
Since \(8 \equiv 3 \pmod{5}\), the congruence simplifies to:
\[
3x \equiv 3 \pmod{5}.
\]
Next, we can divide both sides of the congruence by \(3\). However, in modulo arithmetic, dividing by a number is equivalent to multiplying by its multiplicative inverse modulo \(5\).
First, we identify the multiplicative inverse of \(3\) modulo \(5\). Let's find \(y\) such that:
\[
3y \equiv 1 \pmod{5}.
\]
Testing different values of \(y\) from \(0\) to \(4\):
\[
3 \cdot 1 = 3 \equiv 3 \pmod{5},
\]
\[
3 \cdot 2 = 6 \equiv 1 \pmod{5}.
\]
We see that \(2\) is the multiplicative inverse of \(3\) modulo \(5\). Thus, we multiply both sides of \(3x \equiv 3 \pmod{5}\) by \(2\):
\[
2 \cdot 3x \equiv 2 \cdot 3 \pmod{5}.
\]
This simplifies to:
\[
6x \equiv 6 \pmod{5}.
\]
Since \(6 \equiv 1 \pmod{5}\), the equation further reduces to:
\[
x \equiv 1 \pmod{5}.
\]
Therefore, \(x\) must be of the form:
\[
x = 1 + 5k \quad \text{for integer } k.
\]
Next, we list these solutions for \(x\) within the residues \(0, 1, 2, \ldots, 11\):
\[
k = 0 \implies x = 1,
\]
\[
k = 1 \implies x = 6,
\]
\[
k = 2 \implies x = 11.
\]
So the residues modulo \(12\) that satisfy the congruence \(3x \equiv 8 \pmod{5}\) are:
\[
\boxed{1, 6, 11}.
\]
Which of the residues $0,$ $1,$ $2,$ $\dots,$ $11$ satisfy the congruence $3x \equiv 8 \pmod{5}?$
Give your answer as a list, separated by commas, in order from least to greatest.
1 answer