I know that two numbers are congruent modulo 7 if they have the same remainder when divided by 7. So I want to find the smallest positive integer $n$ that makes $17n$ have a remainder of 1234 when divided by 7. But 1234 is not a possible remainder, because it is larger than 7. So I need to find the remainder of 1234 when divided by 7, and use that instead. I can use long division or a calculator to find that 1234 divided by 7 has a quotient of 176 and a remainder of 2. So $17n \equiv 1234 \pmod{7}$ is equivalent to $17n \equiv 2 \pmod{7}$. Now I can try different values of $n$ until I find one that works. If $n = 1$, then $17n = 17 \equiv 3 \pmod{7}$, which is not what I want. If $n = 2$, then $17n = 34 \equiv 6 \pmod{7}$, which is also not what I want. If $n = 3$, then $17n = 51 \equiv 2 \pmod{7}$, which is exactly what I want. So $n = 3$ is the smallest positive integer that satisfies the given congruence.


What is the smallest positive integer $n$ such that $17n \equiv 1234 \pmod{7}?$