1983 AHSME Problems/Problem 14

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

Then, we have $3^1001$ is congruent to $3$ (mod 10).

The number $7$ has a similar cycle, going: $1, 7, 9, 3, 1, ...$ . Following that, we have $7^1002$ is congruent to $9$ (mod 10).

$13^1003$ is congruent to $3^1003$ (mod 10) = $7$ .

$3\cdot 9\cdot 7$ is congruent to $\fbox9$ (mod 10).

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1983 AHSME Problems/Problem 14