Difference between revisions of "2000 AMC 8 Problems/Problem 14"
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We have | We have | ||
<cmath>-1^{19} + -1^{99} = -1 + -1 \equiv \boxed{(\textbf{D}) \ 8} \pmod{10}</cmath> | <cmath>-1^{19} + -1^{99} = -1 + -1 \equiv \boxed{(\textbf{D}) \ 8} \pmod{10}</cmath> | ||
| + | |||
| + | -ryjs | ||
| + | |||
| + | ==Solution 3== | ||
| + | We find a pattern: both of our numbers have units digit <math>9.</math> | ||
| + | |||
| + | 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}. | ||
| + | |||
| + | -ryjs | ||
==See Also== | ==See Also== | ||
Revision as of 00:18, 12 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 and are odd, we are left with 
which has units digit \boxed{(\textbf{D}) \ 8}.
-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. 
