Lucas's theorem

Theorem. For prime $p$, write $n=\sum n_ip^i$, $k=\sum k_ip^i$ with $0\le n_i,k_i<p$. Then $\binom{n}{k}\equiv\prod_i\binom{n_i}{k_i}\pmod p$.

Compute. $n=10=(101)_3$, $k=3=(10)_3$.

Digits. $n_0=1$, $n_1=0$, $n_2=1$; $k_0=0$, $k_1=1$, $k_2=0$.

Product. $\binom{1}{0}\binom{0}{1}\binom{1}{0}=1\cdot0\cdot1=0$.

Conclusion. $\binom{10}{3}\equiv0\pmod3$.

Check. $\binom{10}{3}=120=3\cdot40$, so $\equiv0$. ✓