[Periodicity and Non-periodicity]  [Decimal Number System]

Find the last four digits of the number $2^{1965}$.
Note that $5^{n+4} \equiv 5^{n}\left(\bmod 10^{4}\right)$ for $n \geq 4$. Indeed, $5^{n+4}-5^{n}$ is divisible by $5^{4}$ and $5^{4} \equiv 9^{2} \equiv 1(\bmod 16)$.

The number 1965 gives a remainder of 1 when divided by 4, hence $5^{1965} \equiv 5^{5}=3125\left(\bmod 10^{4}\right)$.

## Answer

3125.