2012 AMC 8 Problems/Problem 12

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

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.

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