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

Problem

What is the units digit of $13^{2012}$? $\textbf{(A)}\hspace{.05in}1\qquad\textbf{(B)}\hspace{.05in}3\qquad\textbf{(C)}\hspace{.05in}5\qquad\textbf{(D)}\hspace{.05in}7\qquad\textbf{(E)}\hspace{.05in}9$

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

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