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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 <...")
 
(→‎Video Solution (A Clever Explanation You’ll Get Instantly))
 
(43 intermediate revisions by 17 users not shown)
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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==
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==Solution 1==
−
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>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=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
 +
 +
==Solution 2==
 +
 +
The positive divisors of <math>100</math> are <cmath>1,2,4,5,10,20,25,50,100.</cmath>
 +
We apply casework to <math>a</math>:
 +
 +
If <math>a=1</math>, then there are <math>3</math> cases:
 +
 +
* <math>b=2,c=50</math>
 +
 +
* <math>b=4,c=25</math>
 +
 +
* <math>b=5,c=20</math>
 +
 +
If <math>a=2</math>, then there is only <math>1</math> case:
 +
 +
* <math>b=5,c=10</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 (A Creative Way To Think)==
 +
https://youtu.be/tYWp6fcUAik?si=V8hv_zOn_zYOi9E5&t=135
 +
~hsnacademy
 +
 +
==Video Solution 1 by Math-X (First understand the problem!!!)==
 +
https://youtu.be/oUEa7AjMF2A?si=tkBYOey2NioTPPPq&t=221
 +
 +
~Math-X
 +
 +
==Video Solution 2 (CREATIVE THINKING!!!)==
 +
https://youtu.be/5-6zj2mBBSA
 +
 +
~Education, the Study of Everything
 +
 +
==Video Solution 3==
 +
https://www.youtube.com/watch?v=Ij9pAy6tQSg&t=142
 +
 +
~Interstigation
 +
 +
==Video Solution 4==
 +
https://youtu.be/LHnC_Wz6fOU
 +
 +
~savannahsolver
 +
 +
==Video Solution 5==
 +
https://youtu.be/Q0R6dnIO95Y?t=98
 +
 +
~STEMbreezy
 +
 +
==Video Solution 6==
 +
https://www.youtube.com/watch?v=KkZ95iNlFyc
 +
 +
~harungurcan
 +
 +
==Video Solution 7 by Dr. David==
 +
 +
https://youtu.be/EbLGPhGVz6E
  
 
==See Also==  
 
==See Also==  
 
{{AMC8 box|year=2022|num-b=2|num-a=4}}
 
{{AMC8 box|year=2022|num-b=2|num-a=4}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
 +
[[Category:Introductory Number Theory Problems]]

Latest revision as of 01:16, 3 November 2025

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 \[1,2,4,5,10,20,25,50,100.\] We apply casework to $a$:

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 (A Creative Way To Think)

https://youtu.be/tYWp6fcUAik?si=V8hv_zOn_zYOi9E5&t=135 ~hsnacademy

Video Solution 1 by Math-X (First understand the problem!!!)

https://youtu.be/oUEa7AjMF2A?si=tkBYOey2NioTPPPq&t=221

~Math-X

Video Solution 2 (CREATIVE THINKING!!!)

https://youtu.be/5-6zj2mBBSA

~Education, the Study of Everything

Video Solution 3

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

~Interstigation

Video Solution 4

https://youtu.be/LHnC_Wz6fOU

~savannahsolver

Video Solution 5

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

~STEMbreezy

Video Solution 6

https://www.youtube.com/watch?v=KkZ95iNlFyc

~harungurcan

Video Solution 7 by Dr. David

https://youtu.be/EbLGPhGVz6E

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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