Find all solutions of the following congruences:
(a) $10x ≡ 20 \pmod {15}$
(b) $12x ≡ 18 \pmod {48}$
(c) $12x ≡ 48 \pmod {18}$
(a) $g = \gcd(15,10) = 5$ and $5 \mid 20$ so there are 5 solutions.
We simplify $10x ≡ 20 \pmod {15}$ as $x \equiv 2 \pmod {3}$. By inspection $x=2$ is a solution.
So the 5 solutions are
$x \equiv 2 \pmod {15}$
$x \equiv 5 \pmod {15}$
$x \equiv 8 \pmod {15}$
$x \equiv 11 \pmod {15}$
$x \equiv 14 \pmod {15}$
(b) $g = \gcd(48,12)=12$ and $g \not \mid 18$ so there is no solution.
(c) $g = \gcd(18,12)=6$ and $g \mid 48$ so there are 6 solutions.
We simplify $12x \equiv 48 \pmod {18}$ as $12x \equiv 12 \pmod {18}$. This can be further simplified as $x \equiv 1 \pmod 3$. By inspection, $x=1$ is a solution.
So the 6 solutions are
$x \equiv 1 \pmod {18}$
$x \equiv 4 \pmod {18}$
$x \equiv 7 \pmod {18}$
$x \equiv 10 \pmod {18}$
$x \equiv 13 \pmod {18}$
$x \equiv 16 \pmod {18}$