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Difference between revisions of "2012 AMC 8 Problems/Problem 12"
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==Problem==
 
What is the units digit of  <math>13^{2012}</math>?
 
What is the units digit of  <math>13^{2012}</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>
 
<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 by OmegaLearn==
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https://youtu.be/7an5wU9Q5hk?t=1186
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==Video Solution 2==
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https://youtu.be/6RGNZj0tt2w  ~David
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https://youtu.be/6c_s967T7cA ~savannahsolver
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==Solution 1==
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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> 3^1 \implies 3 </math>
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<math> 3^2 \implies 9 </math>
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<math> 3^3 \implies 7 </math>
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<math> 3^4 \implies 1 </math>
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<math> 3^5 \implies 3 </math>
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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
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<math> \boxed{{\textbf{(A)}\ 1}} </math>.
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==Solution 2==
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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}}
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{{MAA Notice}}

Latest revision as of 17:59, 15 April 2023

Problem

What is the units digit of $13^{2012}$?

$\textbf{(A)}\hspace{.05in}1\qquad\textbf{(B)}\hspace{.05in}3\qquad\textbf{(C)}\hspace{.05in}5\qquad\textbf{(D)}\hspace{.05in}7\qquad\textbf{(E)}\hspace{.05in}9$


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 $13^{2012}$, therefore, we can eliminate the tens digit of $13$, because the tens digit will not affect the final result. So our new expression is $3^{2012}$. Now we need to look for a pattern in the units digit.

$3^1 \implies 3$

$3^2 \implies 9$

$3^3 \implies 7$

$3^4 \implies 1$

$3^5 \implies 3$

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. $2011$ divided by $4$ leaves a remainder of $3$, so the answer is the units digit of $3^{3+1}$, or $3^4$. Thus, we find that the units digit of $13^{2012}$ is $\boxed{{\textbf{(A)}\ 1}}$.

Solution 2

Ignore the tens digit of $13$, we find a pattern in the units digit that $3^4 \implies 1$. We also find $2012$ can be divided by $4$ evenly, which is $2012/4=503$. So $3^{2012}$ = $(3^4)^{503}$. Because the units digit of $3^4 \implies 1$，so the units digit $1^{503} \implies 1$. Thus, the units digit of $13^{2012}$ is $\boxed{{\textbf{(A)}\ 1}}$. ---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

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