Difference between revisions of "2000 AMC 8 Problems/Problem 14"

Revision as of 23:22, 11 April 2020

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

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 \begin{align*} 19^{19} + 99^{99} \\ 9^{19} + 9^{99} \\ \end{align*}

Both $19$ and $99$ are odd, so 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
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 24: / Line 24: @@
 ==Solution 3==
-We find a pattern: both of our numbers have units digit <math>9.</math>
+Experimentation gives
+<cmath>\text{any number ending with }9^{\text{something even}} = \text{has units digit }1</cmath>
-Experimentation gives that powers of <math>9</math> alternate between units digits <math>9</math> (for odd powers) and <math>1</math> (for even powers).
+<cmath>\text{any number ending with }9^{\text{something odd}} = \text{has units digit }9</cmath>
-Since both of our powers (<math>19</math> and <math>99</math>) are odd, we are left with <math>9+9=18,</math> which has units digit <math>\boxed{(\textbf{D}) \ 8}.</math>
+Using this we have
+\begin{align*}
+^{19} + 99^{99} \\
+^{19} + 9^{99} \\
+\end{align*}
+Both <math>19</math> and <math>99</math> are odd, so we are left with
+<cmath>9+9=18,</cmath> which has units digit <math>\boxed{(\textbf{D}) \ 8}.</math>
 -ryjs

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