Difference between revisions of "2000 AMC 8 Problems/Problem 14"
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Powers of <math>99</math> have the exact same property, so <math>99^{99}</math> also has a units digit of <math>9</math>. <math>9+9=18</math> which has a units digit of <math>8</math>, so the answer is <math>\boxed{D}</math>. | Powers of <math>99</math> have the exact same property, so <math>99^{99}</math> also has a units digit of <math>9</math>. <math>9+9=18</math> which has a units digit of <math>8</math>, so the answer is <math>\boxed{D}</math>. | ||
| + | |||
| + | ==Solution 2== | ||
| + | Using modular arithmetic: | ||
| + | <cmath>99 \equiv 9 \equiv -1 \pmod{10}</cmath> | ||
| + | |||
| + | Similarly, | ||
| + | <cmath>19 \equiv 9 \equiv -1 \pmod{10}</cmath> | ||
| + | |||
| + | We have | ||
| + | <cmath>-1^{19} + -1^{99} = -1 + -1 \equiv \boxed{(\text{D })8} \pmod{10}</cmath> | ||
==See Also== | ==See Also== | ||
Revision as of 23:15, 11 April 2020
Contents
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
![\[-1^{19} + -1^{99} = -1 + -1 \equiv \boxed{(\text{D })8} \pmod{10}\]](http://latex.artofproblemsolving.com/e/0/a/e0a8012556590852b9a3431362554f9104623e74.png)
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. 
