jmc

Let $a$ be the number of apples that Sophia originally had. Clearly $a=6b$ for some positive integer $b$. Additionally, $a-1\equiv 0(\mathrm{mod}n)⟹6b\equiv 1(\mathrm{mod}n)$. This is solvable for $b$ if and only if $6$ is invertible modulo $n$. In other words, $\gcd (6,n)=1$. The only such $n$ less than $10$ are $1,5,7$, so there are $\boxed{3}$ possibilities for $n$.