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> | ||
− | ==Solution== | + | |
+ | ==Video Solution by OmegaLearn== | ||
+ | https://youtu.be/7an5wU9Q5hk?t=1186 | ||
+ | |||
+ | ==Video Solution 2== | ||
+ | https://youtu.be/6RGNZj0tt2w ~David | ||
+ | |||
+ | https://youtu.be/6c_s967T7cA ~savannahsolver | ||
+ | |||
+ | ==Solution 1== | ||
The problem wants us to find the units digit of <math> 13^{2012} </math>, therefore, we can eliminate the tens digit of <math> 13 </math>, because the tens digit will not affect the final result. So our new expression is <math> 3^{2012} </math>. Now we need to look for a pattern in the units digit. | The problem wants us to find the units digit of <math> 13^{2012} </math>, therefore, we can eliminate the tens digit of <math> 13 </math>, because the tens digit will not affect the final result. So our new expression is <math> 3^{2012} </math>. Now we need to look for a pattern in the units digit. | ||
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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>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | Ignore the tens digit of <math>13</math>, 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, which is <math>2012/4=503</math>. So <math>3^{2012}</math> = <math>(3^4)^{503}</math>. Because the units digit of <math>3^4 \implies 1</math>,so the units digit <math>1^{503} \implies 1</math>. Thus, the units digit of <math>13^{2012}</math> 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}} |
Latest revision as of 18:59, 15 April 2023
Contents
Problem
What is the units digit of ?
Video Solution by OmegaLearn
https://youtu.be/7an5wU9Q5hk?t=1186
Video Solution 2
https://youtu.be/6RGNZj0tt2w ~David
https://youtu.be/6c_s967T7cA ~savannahsolver
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
Ignore the tens digit of , we find a pattern in the units digit that
. We also find
can be divided by
evenly, which is
. So
=
. Because the units digit of
,so the units digit
. Thus, the units digit of
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.