Difference between revisions of "2016 AMC 10B Problems/Problem 8"
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So the answer is <math>\textbf{(A)}\ 0</math>. | So the answer is <math>\textbf{(A)}\ 0</math>. | ||
− | + | ==Solution 2== | |
+ | |||
+ | We can also solve this problem by using patterns. | ||
+ | Notice that when the powers of <math>5</math> are listed out, you get <math>25</math>, <math>125</math> , <math>625</math>. | ||
+ | This proves that the last <math>2</math> digits of <math>2015^{2016}</math> have to be <math>25</math>. | ||
+ | Now, looking at <math>2017</math>, we see that only the <math>17</math> matters. Subtracting, | ||
+ | we get our answer <math>08</math>. | ||
+ | |||
+ | Therefore, our answer is | ||
+ | <math>\textbf{(A)}\ 0</math>. | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2016|ab=B|num-b=7|num-a=9}} | {{AMC10 box|year=2016|ab=B|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 12:46, 21 February 2016
Contents
Problem
What is the tens digit of
Solution
Notice that is congruent to when is even and when is odd. (Check for yourself). Since is even, and .
So the answer is .
Solution 2
We can also solve this problem by using patterns. Notice that when the powers of are listed out, you get , , . This proves that the last digits of have to be . Now, looking at , we see that only the matters. Subtracting, we get our answer .
Therefore, our answer is .
See Also
2016 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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 AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.