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,

 $U$($7^1$)= $7$
  .      .
  .      .
  .      .
 $U$($7^4)=$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 $

See Also

1961 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 27
Followed by
Problem 29
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