The last digit (that is, the units digit) of $(2002)^{2002}$ is
(A) 4
(B) 2
(C) 8
(D) 0
(E) 6
The last digit (that is, the units digit) of $(2002)^{2002}$ is
(A) 4
(B) 2
(C) 8
(D) 0
(E) 6

## Solution

The units digit of $(2002)^{2002}$ is the same as the units digit of $2^{2002}$, since the first three digits of 2002 do not affect the units digit.

We write out the first few powers of 2 and check for the units digit.

| $n$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $2^{n}$ | 2 | 4 | 8 | 16 | 32 | 64 | 128 | 256 | 512 |

From this table, we see that the units digits repeat every 4 powers. So the units digit of $2^{2000}$ will be 6 , and thus the units digit of $2^{2002}$ (and so also of $(2002)^{2002}$ ) will be 4 .