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

https://youtu.be/7an5wU9Q5hk?t=1552

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

See Also

2000 AMC 8 (ProblemsAnswer KeyResources)
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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