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 1

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 $\boxed{\textbf{(E)}\ 9}$ .

Solution 2

By Euler's Totient Theorem, if $\gcd(a, 10) = 1$ , then $a^{\phi(10)} \equiv a^4 \equiv 1\ (\mathrm{mod}\ 10)$ . So $3^{1001} \cdot 7^{1002} \cdot 13^{1003} \equiv 3^{1001} \cdot 7^{1002} \cdot 3^{1003} \equiv 3^{2004} \cdot 7^{1002} \equiv 3^0 \cdot 7^2 \equiv 1 \cdot 9 \equiv \boxed{\textbf{(E)}\ 9}\ (\mathrm{mod}\ 10)$ .

-j314andrews

1983 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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1983 AHSME Problems/Problem 14

Contents

Problem

Solution 1

Solution 2

See Also