jmc

Let's find the cycle of ones digits of $7^n$, starting with $n=1$ : $7,9,3,1,7,9,3,1,\dots$ . The cycle of ones digits of $7^n$ is 4 digits long: 7, 9, 3, 1. Thus, to find the ones digit of $7^n$ for any positive $n$, we must find the remainder, $R$, when $n$ is divided by 4 ($R=1$ corresponds to the ones digit 7, $R=2$ corresponds to the ones digit 9, etc.) Since $35÷4=8R3$, the ones digit of $7^{35}$ is $\boxed{3}$.