Question

The product of three consecutive positive integers is 120. Then the sum of the three positive numbers is:

• A. 16
• B. 24
• C. 15
• D. 12

Hint:

The product of three consecutive integers $(n-1)n(n+1) = n^3 - n$ is approximately equal to $n^3$ for larger numbers. To locate the middle number quickly, take the cube root of the target product: $$\sqrt[3]{120} \approx 4.93$$ Rounding to the nearest whole integer gives the middle term $n = 5$. Thus, the numbers are $4$, $5$, and $6$.

Answer 1

The Correct Option is C

Step 1: Set up.
Let integers be $n-1, n, n+1$. Product $= n(n^2-1) = 120$.
Step 2: Trial.
$n=5$: $4\times5\times6=120$ correct. Sum $=4+5+6$ \[ \boxed{15} \]