Difference between revisions of "2012 AMC 8 Problems/Problem 12"
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==Solution 2== | ==Solution 2== | ||
− | We find a pattern in the units digit that <math>3^4 \implies 1</math>. We also find <math>2012</math> can be divided by <math>4</math> evenly, meaning <math>2012/4=503</math>. So it gives us the units digit of <math>(3^4) | + | We find a pattern in the units digit that <math>3^4 \implies 1</math>. We also find <math>2012</math> can be divided by <math>4</math> evenly, meaning <math>2012/4=503</math>. So it gives us the units digit of <math>(3^4)</math> power of 503 is the same as <math>(3^4)</math>. Thus, the answer is <math> \boxed{{\textbf{(A)}\ 1}} </math>. ---LarryFlora |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2012|num-b=11|num-a=13}} | {{AMC8 box|year=2012|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:31, 20 August 2021
Problem
What is the units digit of ?
Video Solution
https://youtu.be/7an5wU9Q5hk?t=1186
Solution 1
The problem wants us to find the units digit of , therefore, we can eliminate the tens digit of , because the tens digit will not affect the final result. So our new expression is . Now we need to look for a pattern in the units digit.
We observe that there is a pattern for the units digit which recurs every four powers of three. Using this pattern, we can subtract 1 from 2012 and divide by 4. The remainder is the power of three that we are looking for, plus one. divided by leaves a remainder of , so the answer is the units digit of , or . Thus, we find that the units digit of is .
Solution 2
We find a pattern in the units digit that . We also find can be divided by evenly, meaning . So it gives us the units digit of power of 503 is the same as . Thus, the answer is . ---LarryFlora
See Also
2012 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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.