Call the integers $n-1$, $n$, and $n+1$. Their mean is $n$; their product is $(n-1)(n)(n+1)=120$ and their product divided by their mean is $(n-1)(n+1)=24$. Dividing the first equation by the second, we get $n=5$. The largest of the three is $n+1=\boxed{6}$.
Question:
The product of three consecutive integers is 120. That product divided by the mean of the three integers is 24. What is the largest of the three consecutive integers?
Answer:
Let's think step by step.

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6
The product of three consecutive integers is 120. That product divided by the mean of the three integers is 24. What is the largest of the three consecutive integers?
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