For some positive integer $k$, when 60 is divided by $k^2$, the remainder is 6. What is the remainder when 100 is divided by $k$?
I need to find a positive integer $k$ such that 60 divided by $k^2$ gives a quotient of some integer $q$ and a remainder of 6. In other words, I need to solve the equation $60 = qk^2 + 6$ for some integers $q$ and $k$. I can rearrange this equation to get $k^2 = \frac{60 - 6}{q} = \frac{54}{q}$. This means that $k^2$ is a factor of 54, and so is $q$. The factors of 54 are 1, 2, 3, 6, 9, 18, 27, and 54. I can try each of these values for $q$ and see if I get a perfect square for $k^2$. If $q = 1$, then $k^2 = 54$, which is not a perfect square. If $q = 2$, then $k^2 = 27$, which is not a perfect square. If $q = 3$, then $k^2 = 18$, which is not a perfect square. If $q = 6$, then $k^2 = 9$, which is a perfect square, and $k = 3$. If $q > 6$, then $k^2 < 9$, and there are no more positive integers that can be $k$. So the only possible value for $k$ is 3. Now I need to find the remainder when 100 is divided by $k = 3$. I can use long division or a calculator to find that 100 divided by 3 gives a quotient of 33 and a remainder of 1. Therefore, the answer is 1.

