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Difference between revisions of "2022 AMC 8 Problems/Problem 3"
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In total, there are <math>3+1=\boxed{\textbf{(E) } 4}</math> ways to choose distinct positive integer values of <math>a,b,c</math>.
 
In total, there are <math>3+1=\boxed{\textbf{(E) } 4}</math> ways to choose distinct positive integer values of <math>a,b,c</math>.
 +
 +
~MathFun1000
  
 
==Video Solution==
 
==Video Solution==

Revision as of 22:12, 16 January 2023

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 1

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

Solution 2

The positive divisors of $100$ are $1,2,4,5,10,20,25,50,$ and $100$. We can do casework on $a$:

If $a=1$, then there are $3$ cases:

  • $b=2,c=50$
  • $b=4,c=25$
  • $b=5,c=20$

If $a=2$, then there is only $1$ case:

  • $b=5,c=10$

In total, there are $3+1=\boxed{\textbf{(E) } 4}$ ways to choose distinct positive integer values of $a,b,c$.

~MathFun1000

Video Solution

https://www.youtube.com/watch?v=Ij9pAy6tQSg&t=142 ~Interstigation

Video Solution 2

https://youtu.be/LHnC_Wz6fOU

~savannahsolver

Video Solution

https://youtu.be/Q0R6dnIO95Y?t=98

~STEMbreezy

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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