Difference between revisions of "2000 AMC 8 Problems/Problem 14"
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Using this we have | Using this we have | ||
− | + | <cmath>19^{19} + 99^{99}</cmath> | |
− | 19^{19} + 99^{99} | + | <cmath>9^{19} + 9^{99}</cmath> |
− | 9^{19} + 9^{99} | ||
− | |||
Both <math>19</math> and <math>99</math> are odd, so we are left with | Both <math>19</math> and <math>99</math> are odd, so we are left with |
Revision as of 00:23, 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:
Similarly,
We have
-ryjs
Solution 3
Experimentation gives
Using this we have
Both and are odd, so 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.