Difference between revisions of "2012 AMC 8 Problems/Problem 12"

m (Reverted edits by Bahatopo (talk) to last revision by Pi is 3.14)
(Tag: Rollback)
Line 7: Line 7:
 
https://youtu.be/7an5wU9Q5hk?t=1186
 
https://youtu.be/7an5wU9Q5hk?t=1186
  
==Solution==
+
==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.
  
Line 22: Line 22:
 
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  
 
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==
 +
We find a pattern that <math> 3^4 \implies 1 </math>. <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)^503</math> 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 21:27, 20 August 2021

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

https://youtu.be/7an5wU9Q5hk?t=1186

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

We find a pattern that $3^4 \implies 1$. $2012$ can be divided by $4$ evenly, meaning $2012/4=503$. So it gives us the units digit of $(3^4)^503$ is the same as $(3^4)$. Thus the answer is $\boxed{{\textbf{(A)}\ 1}}$. ---LarryFlora


See Also

2012 AMC 8 (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS