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

Concept: Consecutive positive integers are numbers that follow each other in order without gaps, each being exactly 1 greater than the preceding number. If the middle integer is designated as $n$, then the three integers can be written symmetric to $n$ as $(n - 1)$, $n$, and $(n + 1)$, which simplifies algebraic expansions.

Step-by-step Explanation:

Step 1: Establishing the algebraic equation.
Let the three consecutive positive integers be defined systematically as: $$x_1 = n - 1, \quad x_2 = n, \quad x_3 = n + 1 \quad (\text{where } n \in \mathbb{Z}^+ \text{ and } n \gt 1)$$ We are given that the product of these three elements equals 120: $$(n - 1) \cdot n \cdot (n + 1) = 120$$ Using the algebraic identity for the difference of squares, $(n - 1)(n + 1) = n^2 - 1$, we can rewrite this product expression as: $$n(n^2 - 1) = 120 \quad \Rightarrow \quad n^3 - n - 120 = 0$$

Step 2: Analysis using Prime Factorization.
While solving a cubic polynomial directly is one path, analyzing the prime factorization of 120 provides a faster breakthrough since $n$ must be an integer. Let us decompose 120 into its prime components: $$120 = 2 \times 2 \times 2 \times 3 \times 5 = 2^3 \times 3 \times 5$$ We must cluster these factors into three distinct groups that form consecutive integers.

• Grouping the first two factors together gives: $2 \times 2 = 4$

• The prime factor 5 remains unchanged: $5$

• Grouping the remaining factors together gives: $2 \times 3 = 6$
This rearrangement reveals the consecutive integers: $4$, $5$, and $6$.

Step 3: Verification of the found numbers.
Let us verify if their product matches our original algebraic condition: $$\text{Product} = 4 \times 5 \times 6 = 20 \times 6 = 120$$ This satisfies the constraint exactly, meaning $n - 1 = 4$, $n = 5$, and $n + 1 = 6$.

Step 4: Evaluating the requested summation value.
The question asks for the explicit sum of these three consecutive positive numbers: $$\text{Sum} = (n - 1) + n + (n + 1) = 3n$$ $$\text{Sum} = 4 + 5 + 6 = 15$$