1983 AHSME Problems/Problem 14

Problem

The units digit of $3^{1001} 7^{1002} 13^{1003}$ is

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

Solution

First, we notice that $3^0$ is congruent to $1 \ \text{(mod 10)}$, $3^1$ is $3 \ \text{(mod 10)}$, $3^2$ is $9 \ \text{(mod 10)}$, $3^3$ is $7 \ \text{(mod 10)}$, $3^4$ is $1 \ \text{(mod 10)}$, and so on. This turns out to be a cycle repeating every $4$ terms, so $3^{1001}$ is congruent to $3 \ \text{(mod 10)}$.

The number $7$ has a similar cycle, going $1, 7, 9, 3, 1, ...$ Hence we have that $7^{1002}$ is congruent to $9 \ \text{(mod 10)}$. Finally, $13^{1003}$ is congruent to $3^{1003} \equiv 7 \ \text{(mod 10)}$. Thus the required units digit is $3\cdot 9\cdot 7 \equiv 9 \ \text{(mod 10)}$, so the answer is $\boxed{\textbf{(E)}\ 9}$.

See Also

1983 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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