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 $\boxed{\textbf{(D)}}$ .

Alternate Solution

$Lemma$ ( $Fermat's$ $Theorem$ ): 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 $U$ ( $x$ ) as units digit funtion of $x$ .

We can clearly observe that,

 ()= 
  .      .
  .      .
  .      .
 (1$$ (Error compiling LaTeX. Unknown error_msg)

and we can see by Fermat's Theorem that this cycle repeats with the cyclicity of $4$ . Now $753$ = $4k$ + $1$ $=>$ $U($ 7 $^{753}$ ) $=$ 7 $.$ ~GEOMETRY-WIZARD $

1961 AHSC (Problems • Answer Key • Resources)
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1961 AHSME Problems/Problem 28

Contents

Problem 28

Solution

Alternate Solution

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