2024 AMC 12A Problem 6

To obtain a small positive sum, use two negative integers $-p,-q$ and one positive integer $r,$ where $pqr=60.$ Up to order, the positive factor triples of $60$ are $$\begin{gathered} (1,1,60),(1,2,30),\\ (1,3,20),(1,4,15),\\ (1,5,12),(1,6,10),\\ (2,2,15),(2,3,10),\\ (2,5,6),(3,4,5). \end{gathered}$$ A positive value of $r-p-q$ is smallest when the largest factor is chosen as $r.$ The positive values from the list are $58,27,16,10,6,3,11,5;$ the last two triples give negative values. Thus the least positive sum is $10-1-6=3.$ (Three positive integers have sum at least $3,$ and three negative integers cannot have positive product.) Thus, the correct answer is B.