elementary number theory - What is the last digit of $2003^{2003}$?

What is the last digit of this number?

$$2003^{2003}$$

Thanks in advance.

I don't have any kind of idea how to solve this. Except that $3^3$ is $27$.

Answer

When calculating the last digit of $a\cdot b$ you only need to find the product of the last digit of $a$ and the last digit of $b$.

so $2003^{2003}$ ends in the same digit as $3^{2003}$

$3^1$ ends in $3$

$3^2$ ends in $9$

$3^3$ ends in $7$

$3^4$ ends in $1$

$3^5$ ends in $3$

$3^6$ ends in $9$

$3^7$ ends in $7$

$3^8$ ends in $1$.

So $3^{4k}$ ends in $1$.

From here $3^{2000}$ ends in $1$ (since $2000$ is $4\cdot 500$).

Therefore $3^{2001}$ ends in $3$, $2^{2002}$ ends in $9$ and $3^{2003}$ ends in $7$