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 "2022 AMC 8 Problems/Problem 3"

(Created page with "==Problem== When three positive integers <math>a</math>, <math>b</math>, and <math>c</math> are multiplied together, their product is <math>100</math>. Suppose <math>a < b <...")
 
Line 6: Line 6:
  
 
==Solution==
 
==Solution==
−
IN PROGRESS
+
 
 +
The positive divisors of <math>100</math> are <cmath>1,2,4,5,10,20,25,50,100.</cmath>
 +
It is clear that <math>c\geq10,</math> so we apply casework to <math>c:</math>
 +
 
 +
* If <math>c=10,</math> then <math>(a,b,c)=(2,5,10).</math>
 +
 
 +
* If <math>c=20,</math> then <math>(a,b,c)=(1,5,20).</math>
 +
 
 +
* If <math>c=25,</math> then <math>(a,b,c)=(1,4,25).</math>
 +
 
 +
* If <math>c=50,</math> then <math>(a,b,c)=(1,2,50).</math>
 +
 
 +
Together, the numbers <math>a,b,</math> and <math>c</math> can be chosen in <math>\boxed{\textbf{(E)} ~4}</math> ways.
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM

Revision as of 11:01, 28 January 2022

Problem

When three positive integers $a$, $b$, and $c$ are multiplied together, their product is $100$. Suppose $a < b < c$. In how many ways can the numbers be chosen?

$\textbf{(A)} ~0\qquad\textbf{(B)} ~1\qquad\textbf{(C)} ~2\qquad\textbf{(D)} ~3\qquad\textbf{(E)} ~4$

Solution

The positive divisors of $100$ are \[1,2,4,5,10,20,25,50,100.\] It is clear that $c\geq10,$ so we apply casework to $c:$

  • If $c=10,$ then $(a,b,c)=(2,5,10).$
  • If $c=20,$ then $(a,b,c)=(1,5,20).$
  • If $c=25,$ then $(a,b,c)=(1,4,25).$
  • If $c=50,$ then $(a,b,c)=(1,2,50).$

Together, the numbers $a,b,$ and $c$ can be chosen in $\boxed{\textbf{(E)} ~4}$ ways.

~MRENTHUSIASM

See Also

2022 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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
All AJHSME/AMC 8 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=2022_AMC_8_Problems/Problem_3&oldid=170424"