2000 AMC 8 Problems/Problem 14: Difference between revisions
| Line 30: | Line 30: | ||
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 23:23, 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:
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 | ||
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