2000 AMC 8 Problems/Problem 14: Difference between revisions
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==Problem== | |||
What is the units digit of <math>19^{19} + 99^{99}</math>? | |||
<math>\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9</math> | |||
==Solution== | |||
Finding a pattern for each half of the sum, even powers of <math>19</math> have a units digit of <math>1</math>, and odd powers of <math>19</math> have a units digit of <math>9</math>. So, <math>19^{19}</math> has a units digit of <math>9</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>. | |||
==See Also== | |||
{{AMC8 box|year=2000|num-b=13|num-a=15}} | |||
Revision as of 19:15, 30 July 2011
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 
.
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 | ||
