Find all the solutions of the following linear congruences:
(a) $6x ≡ 2 \pmod {4}$
(b) $12x ≡ 6 \pmod {18}$
(c) $15x ≡ 10 \pmod {25}$
(d) $7x ≡ 21 \pmod {1001}$
(a) $g = \gcd(4,6)=2$ and $g \mid 2$ so there are 2 incongruent solutions.
We can simplify $6x \equiv 2 \pmod 4$ as $2x \equiv 2 \pmod 4$. By inspection $x=1$ and $x=3$ are solutions.
So the general solutions is $x \equiv 1 \pmod 4$ and $x \equiv 3 \pmod 4$.
(b) $g = \gcd(18,12)=6$ and $g \mid 6$ so there are 6 incongruent solutions.
By Proposition (3.10) we can simplify $12x \equiv 6 \pmod {18}$ as $2x \equiv 1 \pmod 3$. By inspection $x \equiv 2 \pmod {3}$.
So the 6 incongruent solutions are
$x \equiv 2 \pmod {18}$
$x \equiv 5 \pmod {18}$
$x \equiv 8 \pmod {18}$
$x \equiv 11 \pmod {18}$
$x \equiv 14 \pmod {18}$
$x \equiv 17 \pmod {18}$
(c) $g = \gcd(25,15)=5$ and $g \mid 10$ so there are 5 incongruent solutions.
By Proposition (3.10) we can simplify $15x ≡ 10 \pmod {25}$ as $3x \equiv 2 \pmod 5$. By inspection $x \equiv 4 \pmod 5$.
So the 5 incongruent solutions are
$ x \equiv 4 \pmod {25} $
$ x \equiv 9 \pmod {25} $
$ x \equiv 14 \pmod {25} $
$ x \equiv 19 \pmod {25} $
$ x \equiv 24 \pmod {25} $
(d) $g = \gcd(1001,7)=7$ and $g \mid 21$ so there are 7 incongruent solutions.
By Proposition (3.10) we can simplify $7x ≡ 21 \pmod {1001}$ as $x \equiv 3 \pmod {143}$. By inspection $x \equiv 3 \pmod {143}$.
So the 7 incongruent solutions are
$ x \equiv 3 \pmod {1001} $
$ x \equiv 146 \pmod {1001} $
$ x \equiv 289 \pmod {1001} $
$ x \equiv 432 \pmod {1001} $
$ x \equiv 575 \pmod {1001} $
$ x \equiv 718 \pmod {1001} $
$ x \equiv 861 \pmod {1001} $