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Difference between revisions of "2022 AMC 8 Problems/Problem 3"

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The positive divisors of <math>100</math> are <cmath>1,2,4,5,10,20,25,50,100.</cmath>
 
The positive divisors of <math>100</math> are <cmath>1,2,4,5,10,20,25,50,100.</cmath>
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It is clear that <math>c\geq10,</math> so we apply casework to <math>c:</math>
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It is clear that <math>10\leq c\leq50,</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=10,</math> then <math>(a,b,c)=(2,5,10).</math>

Revision as of 00:52, 2 February 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 $10\leq c\leq50,$ 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

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