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

Revision as of 01:12, 20 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 $1,2,4,5,10,20,25,50,100.$ 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

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

Video Solution 2

https://youtu.be/LHnC_Wz6fOU

Video Solution

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

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 43: / Line 43: @@
 ~MathFun1000
-==Video Solution==
+==Video Solution 1==
 https://www.youtube.com/watch?v=Ij9pAy6tQSg&t=142

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