2000 AMC 8 Problems/Problem 14

Problem

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

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

Video Solution by OmegaLearn

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 (like about 3 hours), this is a real brainless bash.

--hefei417

2000 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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