Solve the following linear congruence equations:
(a) $2x ≡ 25 \pmod 7$
(b) $17x ≡ 3 \pmod 5$
(c) $27x ≡ 33 \pmod {10}$
(d) $128x ≡ 1 \pmod {5}$
(e) $32x ≡ 23 \pmod {21}$
(f) $54x ≡ 52 \pmod {53}$
(a) $g = \gcd(7,2)=1$ and $g \mid 25$ so a solution exists, and there is only one.
We first recognise that $2x \equiv 25 \equiv 4 \pmod 7$. By inspection we can see $x=2$ is a solution.
So the general solution is $$x \equiv 2 \pmod 7$$
(b) $g = \gcd(5,17)=1$ and $g \mid 3$ so a solution exists, and there is only one.
We can simplify $17x \equiv 2x \equiv 3 \pmod 5$. By inspection $x=4$ is a solution.
So the general solution is $$x \equiv 4 \pmod 5$$
(c) $g = \gcd(10,27)=1$ and $g \mid 33$ so a solution exists, and there is only one.
We simplify $27x \equiv 33 \pmod {10}$ to $7x \equiv 3 \pmod {10}$. By inspection $x=9$ is a solution.
So the general solution is $$x \equiv 9 \pmod {10}$$
(d) $g = \gcd(5,128)=1$ and $g \mid 1$ so a solution exists, and there is only one.
We can simplify $128x \equiv 1 \pmod 5$ to $3x \equiv 1 \pmod 5$. By inspection $x=2$ is a solution.
So the general solution is $$x \equiv 2 \pmod 5$$
(e) $g = \gcd(21,32)=1$ and $g \mid 23$ so a solution exists, and there is only one.
We can simplify $32x \equiv 23 \pmod {21}$ to $11x \equiv 2 \pmod {21}$. By inspection $x=4$ is a solution.
So the general solution is $$x \equiv 4 \pmod {21}$$
(f) $g = \gcd(53,54)=1$ and $g \mid 52$ so a solution exists, and there is only one.
We can simplify $54x \equiv 52 \pmod {53}$ to $x \equiv 52 \pmod {53}$. By inspection $x=52$ is a solution.
So the general solution is $$x \equiv 52 \pmod {53}$$