Question

When $1999^{2000}$ is divided by $5,$ the remainder is

$4$

$3$

$2$

$1$

$0$

Solution:

Note that to find the remainder when divided by $5,$ we only care about the units digit.

This means we only have to observe how the powers of the units digit work, namely the powers of $9.$

Then, looking at powers of $9,$ we see that the units digit alternates between $9$ and $1:$ $9, 81, 729, \cdots$ This means that $1999^{2000}$ ends in a $1$ since the power is even.

The remainder when divided by $5$ is then $1,$ since it is $1$ more than a multiple of $10.$

Thus, D is the correct answer.

1999 AMC 8 Problem 24

Solution:

Problem 24 in Other Years