2000 AMC 8 Problems/Problem 14

Revision as of 23:18, 11 April 2020 by Ryjs (talk | contribs) (added solution 3)

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

-ryjs

See Also

2000 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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All AJHSME/AMC 8 Problems and Solutions

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