2012 AMC 8 Problems/Problem 12

Revision as of 21:30, 20 August 2021 by Larryflora (talk | contribs) (Solution 2)

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 in the units digit that $3^4 \implies 1$. We also find $2012$ can be divided by $4$ evenly, meaning $2012/4=503$. So $3$ power of 2012 = $3^4$ power of $503$. Because the units digit of $(3^4)$ power of 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

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