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

(Video Solution 4)
(Solution 2)
 
(14 intermediate revisions by 7 users not shown)
Line 25: Line 25:
  
 
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>
We can do casework on <math>a</math>:
+
We apply casework to <math>a</math>:
  
 
If <math>a=1</math>, then there are <math>3</math> cases:
 
If <math>a=1</math>, then there are <math>3</math> cases:
Line 43: Line 43:
 
~MathFun1000
 
~MathFun1000
  
==Video Solution 1==
+
==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
 
https://www.youtube.com/watch?v=Ij9pAy6tQSg&t=142
  
 
~Interstigation
 
~Interstigation
  
==Video Solution 2==
+
==Video Solution 4==
 
https://youtu.be/LHnC_Wz6fOU
 
https://youtu.be/LHnC_Wz6fOU
  
 
~savannahsolver
 
~savannahsolver
  
==Video Solution 3==
+
==Video Solution 5==
 
https://youtu.be/Q0R6dnIO95Y?t=98
 
https://youtu.be/Q0R6dnIO95Y?t=98
  
 
~STEMbreezy
 
~STEMbreezy
  
==Video Solution 4 (CREATIVE THINKING!!!)==
+
==Video Solution 6==
https://youtu.be/5-6zj2mBBSA
+
https://www.youtube.com/watch?v=KkZ95iNlFyc
  
~Education, the Study of Everything
+
~harungurcan
  
 
==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}}

Latest revision as of 00:00, 18 January 2024

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 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

See Also

2022 AMC 8 (ProblemsAnswer KeyResources)
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