2022 AMC 8 Problems/Problem 3

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

The positive divisors of $100$ are $1,2,4,5,10,20,25,50,100.$ It is clear that $c\geq10,$ 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

2022 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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All AJHSME/AMC 8 Problems and Solutions

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2022 AMC 8 Problems/Problem 3

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