elementary number theory - Find the last digit of $3^{1999}$

Problem: Find the last digit of $3^{1999}$.

My answer is $3$, but the answer sheet says $7$.

Here is what I did:

• $3^{1999}=(3^9)^{222}\cdot3$


• Using Fermat's Little Theorem: $3^9\equiv1\pmod{10}$


• Therefore, $3^{1999}\equiv(3^9)^{222}\cdot3\equiv1^{222}\cdot3\equiv3\pmod{10}$


• Therefore, the last digit should be $3$

Where did I go wrong?

Answer

Here's a straightforward alternative that does not require Euler's or Fermat's, and only requires noticing that

$$3^2 \equiv -1 \pmod {10}$$ so that
$$\begin{align}3^{1999} &= (3^2)^{999}\cdot3\\&\equiv (-1)^{999}\cdot3\pmod{10}\\&\equiv-3\pmod{10}\\&\equiv{7}\pmod{10}\end{align}$$