# 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 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

~Math-X

## Video Solution 2 (CREATIVE THINKING!!!)

~Education, the Study of Everything

~Interstigation

~savannahsolver

~STEMbreezy

## Video Solution 6

~harungurcan

 2022 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 2 Followed byProblem 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
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