Which integers $x$ (general solution) satisfy the following congruences?
(a) $2x ≡ 2 × 1 \pmod 5$
(b) $7x ≡ 7 × 3 \pmod {14})$
(c) $10x ≡ 10 × 12 \pmod {6}$
(d) $8x ≡ 8 × 5 \pmod {48}$
(e) $−3x ≡ 3 × 5 \pmod {21}$
(f) $−12x ≡ 12 × 7 \pmod {108}$
(g) $15x ≡ 0 \pmod 8$
We'll use Proposition (3.10) that if $ac ≡ bc \pmod n$ then $a ≡ b \pmod {\frac{n}{g} }$ where $g= \gcd (c, n)$.
(a) By Proposition (3.10) we have $x \equiv 1 \pmod {\frac{5}{1}}$. That is, $x \equiv 1 \pmod {5}$.
This means $x = 1 + 5t$ for some integer $t$.
(b) By Proposition (3.10) we have $x \equiv 3 \pmod {\frac{14}{7}}$. That is, $x \equiv 1 \pmod {2}$.
This means $x = 1 + 2t$ for some integer $t$.
(c) By Proposition (3.10) we have $x \equiv 12 \pmod {\frac{6}{2}}$. That is, $x \equiv 0 \pmod {3}$.
This means $x = 3t$ for some integer $t$.
(d) By Proposition (3.10) we have $x \equiv 5 \pmod {\frac{48}{8}}$. That is, $x \equiv 5 \pmod {6}$.
This means $x = 5 + 6t$ for some integer $t$.
(e) By Proposition (3.10) we have $-x \equiv 5 \pmod {\frac{21}{3}}$. That is, $-x \equiv 5 \pmod {7}$ or $x \equiv -5 \equiv 2 \pmod {7}$.
This means $x = 2 +7t$ for some integer $t$.
(f) By Proposition (3.10) we have $-x \equiv 7 \pmod {\frac{108}{12}}$. That is, $-x \equiv 7 \pmod {9}$ or $x \equiv -7 \equiv 2 \pmod {9}$.
This means $x = 2 +9t$ for some integer $t$.
This means $x = 8t$ for some integer $t$.