For how many different digits $n$ is the two-digit number $\underline{6}\underline{n}$ divisible by $n$? (The expression $\underline{6}\underline{n}$ should be interpreted as a two-digit integer with tens digit 6 and units digit $n$, not as 6 times $n$.)
This means that $\underline{6}\underline{n}$ must be equal to $n$ times some integer $k$. I can write this as $\underline{6}\underline{n} = kn$, or equivalently, $10 \cdot 6 + n = kn$. Subtracting $n$ from both sides, I get $60 = (k - 1)n$. This means that $n$ must be a factor of 60, and also a digit from 0 to 9. The factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60. Out of these, only 1, 2, 3, 4, 5, and 6 are digits. So there are 6 possible values for $n$ that make $\underline{6}\underline{n}$ divisible by $n$. 