For any integer n greater than 1, n! denotes the product of all the integers from 1 to n, inclusive....

1. Translate the problem requirements: First, calculate \(\mathrm{6!}\) (factorial of 6), then find the range between \(\mathrm{6! + 2}\) and \(\mathrm{6! + 6}\) inclusive, and determine how many numbers in this range are prime
2. Calculate the factorial and establish the range: Compute \(\mathrm{6! = 720}\), so we need to check numbers from 722 to 726 inclusive
3. Apply factorial divisibility reasoning: Since \(\mathrm{6!}\) contains factors 2, 3, 4, 5, and 6, any number of the form \(\mathrm{6! + k}\) will inherit these factors when k itself shares common factors
4. Test each number for primality using factor analysis: Check if 722, 723, 724, 725, and 726 can be factored using the insight that they're all close to a large factorial

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what the problem is asking. The bracket notation [n means factorial - the product of all integers from 1 to n. So \(\mathrm{[6]}\) means \(\mathrm{6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1}\).

We need to:

• Calculate \(\mathrm{[6]}\) (which is \(\mathrm{6!}\))
• Find the range from \(\mathrm{[6] + 2}\) to \(\mathrm{[6] + 6}\) (inclusive)
• Count how many prime numbers exist in that range

Process Skill: TRANSLATE - Converting the bracket notation and range requirements into clear mathematical tasks

2. Calculate the factorial and establish the range

Let's calculate \(\mathrm{[6]}\) step by step:

\(\mathrm{[6] = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720}\)

Now we can establish our range:

• Starting point: \(\mathrm{[6] + 2 = 720 + 2 = 722}\)
• Ending point: \(\mathrm{[6] + 6 = 720 + 6 = 726}\)

So we need to check if any of these numbers are prime: 722, 723, 724, 725, 726

3. Apply factorial divisibility reasoning

Here's the key insight: Since \(\mathrm{[6] = 720}\) contains all the factors 1, 2, 3, 4, 5, and 6, we can use this to our advantage.

Think about it this way: 720 already has factors of 2, 3, 4, 5, and 6 built into it. When we add small numbers to 720, we often create numbers that can be divided by these same factors.

Let's see what happens:

• \(\mathrm{722 = 720 + 2 = (multiple\ of\ 2) + 2 = 2 \times (360 + 1) = 2 \times 361}\), so 722 is divisible by 2
• \(\mathrm{723 = 720 + 3 = (multiple\ of\ 3) + 3 = 3 \times (240 + 1) = 3 \times 241}\), so 723 is divisible by 3
• \(\mathrm{724 = 720 + 4 = (multiple\ of\ 4) + 4 = 4 \times (180 + 1) = 4 \times 181}\), so 724 is divisible by 4
• \(\mathrm{725 = 720 + 5 = (multiple\ of\ 5) + 5 = 5 \times (144 + 1) = 5 \times 145}\), so 725 is divisible by 5
• \(\mathrm{726 = 720 + 6 = (multiple\ of\ 6) + 6 = 6 \times (120 + 1) = 6 \times 121}\), so 726 is divisible by 6

Process Skill: INFER - Recognizing that factorial + k shares factors with k when k divides the factorial

4. Test each number for primality using factor analysis

Based on our reasoning above, let's verify that each number is composite (not prime):

• \(\mathrm{722 = 2 \times 361}\) (composite - has factor 2)
• \(\mathrm{723 = 3 \times 241}\) (composite - has factor 3)
• \(\mathrm{724 = 4 \times 181}\) (composite - has factor 4, and also 2)
• \(\mathrm{725 = 5 \times 145}\) (composite - has factor 5)
• \(\mathrm{726 = 6 \times 121}\) (composite - has factor 6, and also 2 and 3)

Since every number in our range can be factored (none are prime), there are zero prime numbers between \(\mathrm{[6] + 2}\) and \(\mathrm{[6] + 6}\), inclusive.

Final Answer

There are no prime numbers between \(\mathrm{[6] + 2}\) and \(\mathrm{[6] + 6}\), inclusive.

The answer is (A) None.

This makes intuitive sense: when you add small numbers to a large factorial, you often create composite numbers because the factorial already contains many factors that can combine with the numbers you're adding.

Common Faltering Points

Errors while devising the approach

Faltering Point 1: Misunderstanding the bracket notation
Students may confuse the bracket notation [n with other mathematical symbols or think it represents something other than factorial. They might interpret \(\mathrm{[6]}\) as 6 squared, or as some kind of absolute value, rather than recognizing it as \(\mathrm{6! = 720}\).

Faltering Point 2: Missing the key factorial divisibility insight
Many students will approach this as a brute-force primality testing problem, planning to check each number (722, 723, 724, 725, 726) individually for prime factors. They miss the elegant insight that \(\mathrm{n! + k}\) is always divisible by k when \(\mathrm{k \leq n}\), which makes the solution much more efficient.

Faltering Point 3: Misinterpreting the range boundaries
Students might incorrectly interpret "between \(\mathrm{[6] + 2}\) and \(\mathrm{[6] + 6}\), inclusive" and either exclude one of the endpoints or include additional numbers outside the specified range, leading them to check the wrong set of numbers for primality.

Errors while executing the approach

Faltering Point 1: Arithmetic errors in factorial calculation
When calculating \(\mathrm{[6] = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1}\), students may make multiplication errors and arrive at an incorrect value instead of 720, which would throw off the entire range of numbers to be tested.

Faltering Point 2: Incorrect application of the divisibility rule
Even if students recognize the factorial divisibility pattern, they might apply it incorrectly. For example, they might incorrectly conclude that \(\mathrm{720 + 2}\) is divisible by 2 without properly understanding why (that both 720 and 2 are divisible by 2, so their sum is also divisible by 2).

Errors while selecting the answer

Faltering Point 1: Miscounting or misremembering findings
After correctly determining that all numbers in the range are composite, students might lose track of their work and incorrectly think they found one or more prime numbers, leading them to select (B), (C), (D), or (E) instead of (A) None.