# 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

Experimentation gives $$\text{any number ending with }9^{\text{something even}} = \text{has units digit }1$$

$$\text{any number ending with }9^{\text{something odd}} = \text{has units digit }9$$

Using this we have $$19^{19} + 99^{99}$$ $$9^{19} + 9^{99}$$

Both $19$ and $99$ are odd, so we are left with $$9+9=18,$$ which has units digit $\boxed{(\textbf{D}) \ 8}.$ -ryjs

## Solution 4

$19^{19}+99^{99}=$36972963764972677265718790562880544059566876428174110243025997242355257045527752342141065001012823272794097888 9548326540119429996769494359451621570193644014418071060667659303363419435659472789623878 which ends in 8. :-) If you are really stupid and have time, this is a real brainless bash.

--hefei417