Difference between revisions of "2016 AMC 10B Problems/Problem 8"
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Notice that, for <math>n\ge 2</math>, <math>2015^n\equiv 15^n</math> is congruent to <math>25\pmod{100}</math> when <math>n</math> is even and <math>75\pmod{100}</math> when <math>n</math> is odd. (Check for yourself). Since <math>2016</math> is even, <math>2015^{2016} \equiv 25\pmod{100}</math> and <math>2015^{2016}-2017 \equiv 25 - 17 \equiv \underline{0}8\pmod{100}</math>. | Notice that, for <math>n\ge 2</math>, <math>2015^n\equiv 15^n</math> is congruent to <math>25\pmod{100}</math> when <math>n</math> is even and <math>75\pmod{100}</math> when <math>n</math> is odd. (Check for yourself). Since <math>2016</math> is even, <math>2015^{2016} \equiv 25\pmod{100}</math> and <math>2015^{2016}-2017 \equiv 25 - 17 \equiv \underline{0}8\pmod{100}</math>. | ||
− | So the answer is <math>\textbf{(A)}\ 0</math>. | + | So the answer is <math>\textbf{(A)}\ \boxed{0}</math>. |
− | |||
==Solution 2== | ==Solution 2== |
Latest revision as of 20:35, 25 January 2021
Problem
What is the tens digit of
Solution 1
Notice that, for , is congruent to when is even and when is odd. (Check for yourself). Since is even, and .
So the answer is .
Solution 2
In a very similar fashion, we find that , which equals . Next, since every power (greater than ) of every number ending in will end in (which can easily be verified), we get . (In this way, we don't have to worry about the exponent very much.) Finally, , and thus , as above.
Video Solution
~savannahsolver
Video Solution
https://youtu.be/zfChnbMGLVQ?t=2683
~ pi_is_3.14
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 |
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