Difference between revisions of "2012 AMC 8 Problems/Problem 12"
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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. <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>. | ||
Revision as of 17:55, 5 October 2014
Problem
What is the units digit of 
?
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. 
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. 
