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AMC8每日一题(2000年真题#14)

2018-04-28
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2000 AMC 8 竞赛试题/第14题

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 $\boxed{D}$ .