Difference between revisions of "2012 AMC 8 Problems/Problem 12"
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<math> \textbf{(A)}\hspace{.05in}1\qquad\textbf{(B)}\hspace{.05in}3\qquad\textbf{(C)}\hspace{.05in}5\qquad\textbf{(D)}\hspace{.05in}7\qquad\textbf{(E)}\hspace{.05in}9 </math> | <math> \textbf{(A)}\hspace{.05in}1\qquad\textbf{(B)}\hspace{.05in}3\qquad\textbf{(C)}\hspace{.05in}5\qquad\textbf{(D)}\hspace{.05in}7\qquad\textbf{(E)}\hspace{.05in}9 </math> | ||
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+ | ==Video Solution== | ||
+ | https://youtu.be/7an5wU9Q5hk?t=1186 | ||
==Solution== | ==Solution== | ||
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<math> 3^5 \implies 3 </math> | <math> 3^5 \implies 3 </math> | ||
− | We observe that there is a pattern for the units digit which recurs every four powers of three. Using this pattern, we | + | 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. <math>2011</math> divided by <math>4</math> leaves a remainder of <math>3</math>, so the answer is the units digit of <math>3^{3+1}</math>, or <math>3^4</math>. Thus, we find that the units digit of <math> 13^{2012} </math> is |
<math> \boxed{{\textbf{(A)}\ 1}} </math>. | <math> \boxed{{\textbf{(A)}\ 1}} </math>. | ||
Latest revision as of 16:15, 11 November 2020
Contents
Problem
What is the units digit of ?
Video Solution
https://youtu.be/7an5wU9Q5hk?t=1186
Solution
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 .
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.