1961 AHSME Problems/Problem 28

Problem 28

If $2137^{753}$ is multiplied out, the units' digit in the final product is:

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

Solution

$7^1$ has a unit digit of $7$ . $7^2$ has a unit digit of $9$ . $7^3$ has a unit digit of $3$ . $7^4$ has a unit digit of $1$ . $7^5$ has a unit digit of $7$ .

Notice that the unit digit eventually cycles to itself when the exponent is increased by $4$ . It also does not matter what the other digits are in the base because the units digit is found by multiplying by only the units digit. Since $753$ leaves a remainder of $1$ after being divided by $4$ , the units digit of $2137^{753}$ is $7$ , which is answer choice $\boxed{\textbf{(D)}}$ .

SOLUTION 2

$Lemma$ ( $Fermat's$ $Theorem$ ): If $p$ is a prime and $a$ is an integer prime to $p$ then we have $a^p-1 \equiv 1\ (\textrm{mod}\ p).

Let's define$ (Error compiling LaTeX. Unknown error_msg)U( $x$ ) $as units digit funtion of$ x $. We can clearly observe that,$ U( $7^1$ ) $=$ 7$$ (Error compiling LaTeX. Unknown error_msg).

    .
    .$$ (Error compiling LaTeX. Unknown error_msg)U(=

and we can see by Fermat's Theorem that this cycle repeats with the cyclicity of $4$ . Now $753$ = $4k + 1$ \Rightarrow $U($ 7^753 $)$ = $7$ .

1961 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 27	Followed by Problem 29
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1961 AHSME Problems/Problem 28

Contents

Problem 28

Solution

SOLUTION 2

See Also