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For any integer n greater than 1, \(\mathrm{n!}\) denotes the product of all the integers from 1 to n, inclusive. How many prime numbers are there between \(\mathrm{6! + 2}\) and \(\mathrm{6! + 6}\), inclusive?
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:
Process Skill: TRANSLATE - Converting the bracket notation and range requirements into clear mathematical tasks
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:
So we need to check if any of these numbers are prime: 722, 723, 724, 725, 726
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:
Process Skill: INFER - Recognizing that factorial + k shares factors with k when k divides the factorial
Based on our reasoning above, let's verify that each number is composite (not prime):
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.
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.
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.
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).
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.