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Difference between revisions of "2022 AMC 8 Problems/Problem 3"
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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 < c</math>. In how many ways can the numbers be chosen?
 
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 < c</math>. In how many ways can the numbers be chosen?
  
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<math>\textbf{(A)} ~0\qquad\textbf{(B)} ~1\qquad\textbf{(C)} ~2\qquad\textbf{(D)} ~3\qquad\textbf{(E)} ~4</math>
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<math>\textbf{(A) } 0 \qquad \textbf{(B) } 1\qquad\textbf{(C) } 2\qquad\textbf{(D) } 3\qquad\textbf{(E) } 4</math>
  
 
==Solution==
 
==Solution==
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* If <math>c=50,</math> then <math>(a,b,c)=(1,2,50).</math>
 
* If <math>c=50,</math> then <math>(a,b,c)=(1,2,50).</math>
  
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Together, the numbers <math>a,b,</math> and <math>c</math> can be chosen in <math>\boxed{\textbf{(E)} ~4}</math> ways.
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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 10:14, 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

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