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 $\boxed{D}$ .

2000 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

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2000 AMC 8 Problems/Problem 14

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