Difference between revisions of "2000 AMC 8 Problems/Problem 14"
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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). | 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). | ||
| − | Since both <math>19</math> and <math>99</math> are odd, we are left with <math>9+9=18,</math> which has units digit \boxed{(\textbf{D}) \ 8}. | + | 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> |
-ryjs | -ryjs | ||
Revision as of 23:18, 11 April 2020
Problem
What is the units digit of 
?
Solution
Finding a pattern for each half of the sum, even powers of 
have a units digit of 
, and odd powers of have a units digit of 
. So, 
has a units digit of .
Powers of 
have the exact same property, so 
also has a units digit of . 
which has a units digit of 
, so the answer is 
.
Solution 2
Using modular arithmetic:
![\[99 \equiv 9 \equiv -1 \pmod{10}\]](http://latex.artofproblemsolving.com/7/5/e/75ef9dad0de69599575109d040e7639b9fc39fa3.png)
Similarly,
![\[19 \equiv 9 \equiv -1 \pmod{10}\]](http://latex.artofproblemsolving.com/1/5/c/15c3fc39f1bc9a64c1f573eb133b3078538c5951.png)
We have
![\[-1^{19} + -1^{99} = -1 + -1 \equiv \boxed{(\textbf{D}) \ 8} \pmod{10}\]](http://latex.artofproblemsolving.com/3/b/e/3bef7317510c9ca0743db51825c787b9e5832c9a.png)
-ryjs
Solution 3
We find a pattern: both of our numbers have units digit 
Experimentation gives that powers of alternate between units digits (for odd powers) and (for even powers).
Since both of our powers ( and ) are odd, we are left with 
which has units digit 
-ryjs
See Also
| 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. 
