Which of the following congruences equations have no solutions? If any of these have solutions, find them.
(a) $12x ≡ 4 \pmod {18}$
(b) $13x ≡ 5 \pmod {65}$
(c) $18x ≡ 1 \pmod {16}$
(d) $1001x ≡ 121 \pmod {11}$
(e) $15x ≡ 9 \pmod {27}$
(f) $407x ≡ 40 \pmod {666}$
(a) $g = \gcd(18,12)=6$ but $g \not \mid 4$ so there is no solution.
(b) $g = \gcd(65, 13)=13$ and $g \not \mid 5$ so there is no solution.
(c) $g = \gcd(16,18)=2$ and $g \not \mid 1$ so there is no solution.
(d) $g = \gcd(11,1001)=11$ and $g \mid 121$ so there are 11 solutions.
Since we are working with modulo 11, then 11 solutions means $x=1,2, \ldots, 9,1 0 \pmod{11}$.
Alternatively, we can simplify $1001x ≡ 121 \pmod {11}$ as $0x = 0 \pmod {11}$ for which any integer is a solution.
(e) $g = \gcd(27,15)=3$ and $g \mid 9$, so there are 3 solutions.
We can simplify $15x ≡ 9 \pmod {27}$ as $5x \equiv 3 \pmod{9}$. By inspection $x=6$ is a solution.
So the full set of 3 solutions is
$ x \equiv 6 \pmod {27}$
$ x \equiv 15 \pmod {27}$
$ x \equiv 24 \pmod {27}$
(f) $g= \gcd(666,407) = 34$ but $g \not \mid 40$ so there is no solution.