Question:
What is the average of the two smallest positive integer solutions to the congruence $$14u \equiv 46 \pmod{100}~?$$

Answer:
Note that $14$, $46$, and $100$ all have a common factor of $2$, so we can divide it out: the solutions to $$14u \equiv 46 \pmod{100}$$ are identical to the solutions to $$7u \equiv 23 \pmod{50}.$$ Make sure you see why this is the case.

Now we can multiply both sides of the congruence by $7$ to obtain $$49u \equiv 161 \pmod{50},$$ which also has the same solutions as the previous congruence, since we could reverse the step above by multiplying both sides by $7^{-1}$. (We know that $7^{-1}$ exists modulo $50$ because $7$ and $50$ are relatively prime.)

Replacing each side of $49u\equiv 161$ by a $\pmod{50}$ equivalent, we have $$-u \equiv 11\pmod{50},$$ and thus $$u \equiv -11\pmod{50}.$$ This is the set of solutions to our original congruence. The two smallest positive solutions are $-11+50 = 39$ and $-11+2\cdot 50 = 89$. Their average is $\boxed{64}$.