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

Revision as of 22:41, 8 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?

$\textbf{(A) } 0 \qquad \textbf{(B) } 1\qquad\textbf{(C) } 2\qquad\textbf{(D) } 3\qquad\textbf{(E) } 4$

4 ways

Solution 2

The positive divisors of $100$ are $1,2,4,5,10,20,25,50,100.$ 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 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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 5: / Line 5: @@
 <math>\textbf{(A) } 0 \qquad \textbf{(B) } 1\qquad\textbf{(C) } 2\qquad\textbf{(D) } 3\qquad\textbf{(E) } 4</math>
-==Solution 1==
+ways
-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).
-* 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}$ ways.
-~MRENTHUSIASM
 ==Solution 2==

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