Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1961 AHSME Problems/Problem 28"

(→‎SOLUTION 2)
(→‎SOLUTION 2)
Line 35: Line 35:
 
and we can see by Fermat's Theorem that this cycle repeats with the cyclicity of <math>4</math>.  
 
and we can see by Fermat's Theorem that this cycle repeats with the cyclicity of <math>4</math>.  
  
−
Now <math>753</math> = <math>4k</math> + <math>1</math>  \Rightarrow  
+
Now <math>753</math> = <math>4k</math> + <math>1</math>  \Rightarrow <math>U(</math>7<math></math>^753<math>)</math> = <math>7</math>.
−
 
 
−
<math>U(</math>7<math></math>^753<math>)</math> = <math>7</math>.
 
 
   
 
   
 
<math> ~GEOMETRY-WIZARD </math>
 
<math> ~GEOMETRY-WIZARD </math>

Revision as of 08:01, 31 December 2023

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 $U$($x$) 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 $753$ = $4k$ + $1$ \Rightarrow $U($7$$ (Error compiling LaTeX. Unknown error_msg)^753$)$ = $7$.

$~GEOMETRY-WIZARD$

See Also

1961 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 27 		Followed by
Problem 29
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1961_AHSME_Problems/Problem_28&oldid=209179"