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: Understand that \(\mathrm{n!}\) means factorial (product of integers 1 to n), so \(\mathrm{6! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 = 720}\). We need to find how many prime numbers exist between \(\mathrm{6! + 2 = 722}\) and \(\mathrm{6! + 6 = 726}\), inclusive.
2. Recognize the factorial divisibility pattern: Since \(\mathrm{6!}\) contains factors 2, 3, 4, 5, and 6, adding these same numbers to \(\mathrm{6!}\) creates numbers that are divisible by those factors.
3. Test each number for primality using the divisibility insight: Check if \(\mathrm{6! + 2}\), \(\mathrm{6! + 3}\), \(\mathrm{6! + 4}\), \(\mathrm{6! + 5}\), and \(\mathrm{6! + 6}\) have obvious factors greater than 1.
4. Count the prime numbers: Determine how many numbers in our range have no factors other than 1 and themselves.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what this problem is asking. The symbol \(\mathrm{n!}\) represents a factorial, which means we multiply all integers from 1 up to n together.

So \(\mathrm{6!}\) means: \(\mathrm{1 \times 2 \times 3 \times 4 \times 5 \times 6 = 720}\)

Now we need to find how many prime numbers exist between \(\mathrm{6! + 2}\) and \(\mathrm{6! + 6}\), inclusive. Let's calculate these boundary values:

• \(\mathrm{6! + 2 = 720 + 2 = 722}\)
• \(\mathrm{6! + 6 = 720 + 6 = 726}\)

So we're looking for prime numbers in the range: 722, 723, 724, 725, 726 (inclusive).

Remember: A prime number is a number greater than 1 that has no positive divisors other than 1 and itself.

Process Skill: TRANSLATE - Converting the factorial notation and range into concrete numbers we can work with

2. Recognize the factorial divisibility pattern

Here's the key insight that makes this problem much simpler than checking each number individually for primality.

Since \(\mathrm{6! = 1 \times 2 \times 3 \times 4 \times 5 \times 6}\), we know that \(\mathrm{6!}\) contains all the factors 2, 3, 4, 5, and 6 within it.

This means:

• \(\mathrm{6!}\) is divisible by 2
• \(\mathrm{6!}\) is divisible by 3
• \(\mathrm{6!}\) is divisible by 4
• \(\mathrm{6!}\) is divisible by 5
• \(\mathrm{6!}\) is divisible by 6

Now, when we add these same numbers to \(\mathrm{6!}\), something interesting happens:

• \(\mathrm{6! + 2 = 720 + 2 = (2 \times 360) + 2 = 2 \times (360 + 1) = 2 \times 361}\)
• \(\mathrm{6! + 3 = 720 + 3 = (3 \times 240) + 3 = 3 \times (240 + 1) = 3 \times 241}\)
• \(\mathrm{6! + 4 = 720 + 4 = (4 \times 180) + 4 = 4 \times (180 + 1) = 4 \times 181}\)

Do you see the pattern? Each of these numbers has an obvious factor!

Process Skill: INFER - Recognizing the non-obvious connection between factorial structure and divisibility

3. Test each number for primality using the divisibility insight

Let's apply our insight to check each number in our range:

For 722 = \(\mathrm{6! + 2}\):
Since \(\mathrm{6!}\) is divisible by 2, we can write \(\mathrm{6! = 2k}\) for some integer \(\mathrm{k = 360}\).
So \(\mathrm{6! + 2 = 2k + 2 = 2(k + 1) = 2(361)}\)
This shows 722 is divisible by 2, so it's not prime.

For 723 = \(\mathrm{6! + 3}\):
Since \(\mathrm{6!}\) is divisible by 3, we can write \(\mathrm{6! = 3j}\) for some integer \(\mathrm{j = 240}\).
So \(\mathrm{6! + 3 = 3j + 3 = 3(j + 1) = 3(241)}\)
This shows 723 is divisible by 3, so it's not prime.

For 724 = \(\mathrm{6! + 4}\):
Since \(\mathrm{6!}\) is divisible by 4, we can write \(\mathrm{6! = 4m}\) for some integer \(\mathrm{m = 180}\).
So \(\mathrm{6! + 4 = 4m + 4 = 4(m + 1) = 4(181)}\)
This shows 724 is divisible by 4, so it's not prime.

For 725 = \(\mathrm{6! + 5}\):
Since \(\mathrm{6!}\) is divisible by 5, we can write \(\mathrm{6! = 5n}\) for some integer \(\mathrm{n = 144}\).
So \(\mathrm{6! + 5 = 5n + 5 = 5(n + 1) = 5(145)}\)
This shows 725 is divisible by 5, so it's not prime.

For 726 = \(\mathrm{6! + 6}\):
Since \(\mathrm{6!}\) is divisible by 6, we can write \(\mathrm{6! = 6p}\) for some integer \(\mathrm{p = 120}\).
So \(\mathrm{6! + 6 = 6p + 6 = 6(p + 1) = 6(121)}\)
This shows 726 is divisible by 6, so it's not prime.

4. Count the prime numbers

After checking all five numbers in our range (722, 723, 724, 725, 726), we found that:

• 722 is divisible by 2 (not prime)
• 723 is divisible by 3 (not prime)
• 724 is divisible by 4 (not prime)
• 725 is divisible by 5 (not prime)
• 726 is divisible by 6 (not prime)

Therefore, there are zero prime numbers in the given range.

Final Answer

The number of prime numbers between \(\mathrm{6! + 2}\) and \(\mathrm{6! + 6}\), inclusive, is zero.

The answer is (A) None.

Common Faltering Points

Errors while devising the approach

• Misinterpreting factorial notation: Students might not recognize that \(\mathrm{n!}\) represents n! (factorial), or they might confuse it with other mathematical notations like absolute value or floor function. This would lead them to calculate completely wrong boundary values.
• Missing the key insight about factorial divisibility: Students might plan to check each number (722, 723, 724, 725, 726) individually for primality using traditional methods like trial division, missing the elegant pattern that \(\mathrm{n! + k}\) is divisible by k when \(\mathrm{k \leq n}\).
• Misunderstanding the range 'inclusive': Students might incorrectly interpret the range as exclusive, checking numbers between 723 and 725 only, rather than including both endpoints 722 and 726.

Errors while executing the approach

• Arithmetic errors in factorial calculation: Students might miscalculate \(\mathrm{6! = 720}\), getting values like 700 or 750, which would shift all subsequent boundary calculations and lead to checking the wrong range of numbers.
• Incorrect application of the divisibility pattern: Students might recognize that \(\mathrm{6!}\) contains factors 2, 3, 4, 5, 6 but fail to correctly apply this to show that \(\mathrm{6! + k}\) is divisible by k, potentially making algebraic errors in the factorization step.
• Incomplete verification: Students might correctly identify that some numbers in the range are composite but fail to check all five numbers (722, 723, 724, 725, 726), potentially missing one and incorrectly concluding there is one prime number.

Errors while selecting the answer

• Confusing composite count with prime count: Students might correctly identify all five numbers as composite but then select the wrong answer choice, perhaps choosing (E) Four instead of (A) None due to misreading what the question asks for.