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

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

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2000 AMC 8 Problems/Problem 14

Contents

Problem

Solution

Solution 2

Solution 3

See Also