To find the fractions in lowest terms, I need to avoid any common factors between $n$ and $99$. I know that $99=3^2\cdot 11$, so any multiple of $3$ or $11$ will have a common factor with $99$. There are $32$ multiples of $3$ between $1$ and $98$, and $8$ multiples of $11$ in the same range. However, some multiples of $3$ are also multiples of $11$, and I don't want to count them twice. The only multiples of $3$ and $11$ are multiples of $33$, and there are $2$ of them between $1$ and $98$. Therefore, the number of fractions that are not in lowest terms is $32+8-2=38$. The total number of fractions in the form $\frac{n}{99}$ with $0<n<99$ is $98$. Therefore, the number of fractions that are in lowest terms is $98-38=60$.


How many fractions in the form $\frac{n}{99}$, with $0<n<99$, are in lowest terms?