# Difference between revisions of "2000 AMC 8 Problems/Problem 14"

## Problem

What is the units digit of $19^{19} + 99^{99}$?

$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9$

## Solution

Finding a pattern for each half of the sum, even powers of $19$ have a units digit of $1$, and odd powers of $19$ have a units digit of $9$. So, $19^{19}$ has a units digit of $9$.

Powers of $99$ have the exact same property, so $99^{99}$ also has a units digit of $9$. $9+9=18$ which has a units digit of $8$, so the answer is $\boxed{D}$.

## Solution 2

Using modular arithmetic: $$99 \equiv 9 \equiv -1 \pmod{10}$$

Similarly, $$19 \equiv 9 \equiv -1 \pmod{10}$$

We have $$(-1)^{19} + (-1)^{99} = -1 + -1 \equiv \boxed{(\textbf{D}) \ 8} \pmod{10}$$

-ryjs

## Solution 3

We find a pattern: both of our numbers have units digit $9.$

Experimentation gives that powers of $9$ alternate between units digits $9$ (for odd powers) and $1$ (for even powers).

Since both of our powers ($19$ and $99$) are odd, we are left with $9+9=18,$ which has units digit $\boxed{(\textbf{D}) \ 8}.$

-ryjs