Question

If 720 is the product of the consecutive integers beginning with 2 and ending with n, what is the value of n-1?

• A. 5
• B. 6
• C. 8
• D. 11
• E. 23

Hint:

Recognizing common factorial values can be very helpful. \(720\) is \(6!\) (6 factorial), which is \(1 \times 2 \times 3 \times 4 \times 5 \times 6\). Since the product in the problem starts from 2, it is \(2 \times 3 \times 4 \times 5 \times 6\), which is also 720. This immediately tells you that the last number, \(n\), must be 6.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The problem describes a product of consecutive integers starting from 2. This is similar to a factorial product. We need to find the last integer in this sequence, \(n\), and then calculate \(n-1\).
Step 2: Key Formula or Approach:
We are given the equation \(2 \times 3 \times 4 \times \dots \times n = 720\). The most direct approach is to start multiplying the integers from 2 upwards and see when the product reaches 720.
Step 3: Detailed Explanation:
Let's calculate the product of consecutive integers starting with 2:
- Start with 2.
- Multiply by the next integer, 3: \(2 \times 3 = 6\).
- Multiply the result by the next integer, 4: \(6 \times 4 = 24\).
- Multiply the result by the next integer, 5: \(24 \times 5 = 120\).
- Multiply the result by the next integer, 6: \(120 \times 6 = 720\).
We have reached the target product, 720. The sequence of integers is 2, 3, 4, 5, 6. The last integer in this sequence is \(n\).
Therefore, \(n=6\).
The question asks for the value of \(n-1\).
\[ n - 1 = 6 - 1 = 5 \] Step 4: Final Answer:
The value of \(n\) is 6, so the value of \(n-1\) is 5.