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

Latest revision as of 01:48, 19 November 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

Video Solution 7 by Dr. David

https://youtu.be/EbLGPhGVz6E

@@ Line 3: / Line 3: @@
 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?
-<math>\textbf{(A)} ~0\qquad\textbf{(B)} ~1\qquad\textbf{(C)} ~2\qquad\textbf{(D)} ~3\qquad\textbf{(E)} ~4</math>
+<math>\textbf{(A) } 0 \qquad \textbf{(B) } 1\qquad\textbf{(C) } 2\qquad\textbf{(D) } 3\qquad\textbf{(E) } 4</math>
-==Solution==
+==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
+==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 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==
 {{AMC8 box|year=2022|num-b=2|num-a=4}}
 {{MAA Notice}}

2022 AMC 8 (Problems • Answer Key • Resources)
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