Solve $40x + 13\equiv 78 \pmod {105}$.

Question

Solve $40x + 13\equiv 78 \pmod {105}$.

Having a bit of trouble with this problem, I have done other conditional congruence problems but can't quite solve this one. Any help is appreciated.

Answer 1

Our congruence is equivalent to $$40x\equiv 65\pmod{105},$$ which is equivalent to $$8x\equiv 13\pmod{21},$$ which is equivalent to $$8x\equiv -8\pmod{21},$$ which is equivalent to $$x\equiv -1\pmod{21}.$$ If we want to give the answer modulo $105$, the solutions are $x\equiv 20\pmod{105}$, $x\equiv 41\pmod{105}$, $x\equiv 62\pmod{105}$, $x\equiv 83\pmod{105}$, and $x\equiv 104\pmod{105}$.

I prefer to give the answer modulo $21$. Less typing.

Answer 2

$$40x+13\equiv 78\pmod{105}$$

$$\iff 40x\equiv 65\equiv -40\pmod{105}$$

$$\stackrel{:40}\iff x\equiv -1\pmod{\frac{105}{\gcd(105,40)}=21}$$