For every positive even integer n, the function \(\mathrm{h(n)}\) is defined to be the product of all the even integers...

1. Translate the problem requirements: \(\mathrm{h(n)}\) means multiply all even numbers from 2 to n. So \(\mathrm{h(100)} = \mathrm{2 \times 4 \times 6 \times 8 \times \ldots \times 100}\). We need to find the smallest prime that divides \(\mathrm{h(100)} + \mathrm{1}\).
2. Recognize the key insight about h(100): Since \(\mathrm{h(100)}\) contains every even number up to 100 as factors, it's divisible by many small primes.
3. Apply the fundamental principle: If \(\mathrm{h(100)}\) is divisible by small primes like 2, 3, 5, etc., then \(\mathrm{h(100)} + \mathrm{1}\) cannot be divisible by these same primes.
4. Determine the range systematically: Check which small primes divide \(\mathrm{h(100)}\) to eliminate them as possibilities for the smallest prime factor of \(\mathrm{h(100)} + \mathrm{1}\).
5. Use elimination to narrow the answer: Since all small primes are ruled out, the smallest prime factor must be relatively large.

Execution of Strategic Approach

1. Translate the problem requirements

Let's first understand what we're working with in plain English. The function \(\mathrm{h(n)}\) takes a positive even integer and multiplies together all the even integers from 2 up to that number.

For example:

• \(\mathrm{h(4)} = \mathrm{2 \times 4} = \mathrm{8}\)
• \(\mathrm{h(6)} = \mathrm{2 \times 4 \times 6} = \mathrm{48}\)
• \(\mathrm{h(8)} = \mathrm{2 \times 4 \times 6 \times 8} = \mathrm{384}\)

So \(\mathrm{h(100)} = \mathrm{2 \times 4 \times 6 \times 8 \times 10 \times 12 \times \ldots \times 98 \times 100}\)

This is a massive number! We need to find the smallest prime number that divides \(\mathrm{h(100)} + \mathrm{1}\).

Process Skill: TRANSLATE - Converting the function definition into concrete understanding

2. Recognize the key insight about h(100)

Now let's think about what kinds of numbers divide \(\mathrm{h(100)}\). Since \(\mathrm{h(100)}\) is the product of all even numbers from 2 to 100, it contains many, many factors.

Let's see what primes divide \(\mathrm{h(100)}\):

• Does 2 divide \(\mathrm{h(100)}\)? Yes! Because 2 is one of our factors
• Does 3 divide \(\mathrm{h(100)}\)? Yes! Because \(\mathrm{6} = \mathrm{2 \times 3}\) is one of our factors
• Does 5 divide \(\mathrm{h(100)}\)? Yes! Because \(\mathrm{10} = \mathrm{2 \times 5}\) is one of our factors
• Does 7 divide \(\mathrm{h(100)}\)? Yes! Because \(\mathrm{14} = \mathrm{2 \times 7}\) is one of our factors
• Does 11 divide \(\mathrm{h(100)}\)? Yes! Because \(\mathrm{22} = \mathrm{2 \times 11}\) is one of our factors

In fact, for any prime \(\mathrm{p \leq 50}\), the number \(\mathrm{2p}\) will be \(\mathrm{\leq 100}\) and will be one of our even factors in \(\mathrm{h(100)}\).

3. Apply the fundamental principle

Here's the crucial insight: If a number is divisible by a prime p, then when you add 1 to that number, the result is NOT divisible by p.

Think about it with small examples:

• 6 is divisible by 2, but \(\mathrm{6 + 1} = \mathrm{7}\) is not divisible by 2
• 15 is divisible by 3, but \(\mathrm{15 + 1} = \mathrm{16}\) is not divisible by 3

Since \(\mathrm{h(100)}\) is divisible by 2, then \(\mathrm{h(100)} + \mathrm{1}\) is NOT divisible by 2.
Since \(\mathrm{h(100)}\) is divisible by 3, then \(\mathrm{h(100)} + \mathrm{1}\) is NOT divisible by 3.
Since \(\mathrm{h(100)}\) is divisible by 5, then \(\mathrm{h(100)} + \mathrm{1}\) is NOT divisible by 5.

And so on for all primes up to 50!

Process Skill: INFER - Drawing the non-obvious conclusion about what cannot divide h(100) + 1

4. Determine the range systematically

Let's check all the primes up to 50:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47

For each prime p in this list, \(\mathrm{2p \leq 100}\), so \(\mathrm{2p}\) appears as a factor in \(\mathrm{h(100)}\). This means p divides \(\mathrm{h(100)}\), which means p does NOT divide \(\mathrm{h(100)} + \mathrm{1}\).

So the smallest prime factor of \(\mathrm{h(100)} + \mathrm{1}\) must be larger than 50.

The next prime after 47 is 53. But we need to check: does 53 divide \(\mathrm{h(100)}\)?

For 53 to divide \(\mathrm{h(100)}\), we'd need one of our even factors (2, 4, 6, ..., 100) to be divisible by 53.
Since \(\mathrm{53 \times 2} = \mathrm{106} > \mathrm{100}\), none of our factors from 2 to 100 are divisible by 53.

Therefore, 53 does not divide \(\mathrm{h(100)}\), which means 53 could potentially divide \(\mathrm{h(100)} + \mathrm{1}\).

5. Use elimination to narrow the answer

We've established that:

• All primes ≤ 47 cannot be factors of \(\mathrm{h(100)} + \mathrm{1}\)
• The smallest possible prime factor is ≥ 53

Looking at our answer choices:

• A. Between 2 and 10: No (we ruled out 2, 3, 5, 7)
• B. Between 10 and 20: No (we ruled out 11, 13, 17, 19)
• C. Between 20 and 30: No (we ruled out 23, 29)
• D. Between 30 and 40: No (we ruled out 31, 37)
• E. Greater than 40: Yes! The smallest prime factor must be ≥ 53

4. Final Answer

The smallest prime factor of \(\mathrm{h(100)} + \mathrm{1}\) is greater than 40.

The answer is E.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the function definition

Students may confuse \(\mathrm{h(n)}\) with factorial notation or misinterpret it as the product of ALL integers from 2 to n (instead of just even integers). For example, they might think \(\mathrm{h(6)} = \mathrm{2 \times 3 \times 4 \times 5 \times 6}\) instead of \(\mathrm{2 \times 4 \times 6}\). This fundamental misunderstanding would lead to completely incorrect analysis of what primes divide \(\mathrm{h(100)}\).

2. Missing the key insight about divisibility

Many students fail to recognize the crucial principle that if a prime p divides \(\mathrm{h(100)}\), then p cannot divide \(\mathrm{h(100)} + \mathrm{1}\). They might attempt to calculate \(\mathrm{h(100)}\) directly or miss that we need to think about what primes are ruled OUT as factors of \(\mathrm{h(100)} + \mathrm{1}\), rather than trying to find what primes are IN.

3. Incorrect identification of which primes divide h(100)

Students may not realize that every prime \(\mathrm{p \leq 50}\) divides \(\mathrm{h(100)}\) because \(\mathrm{2p \leq 100}\) appears as one of the even factors. They might think only primes that appear directly in the sequence (like 2) divide \(\mathrm{h(100)}\), missing that primes like 3, 5, 7, etc. also divide \(\mathrm{h(100)}\) through factors like 6, 10, 14, etc.

Errors while executing the approach

1. Incomplete enumeration of primes

Students may not systematically check all primes up to 50, potentially missing some primes in their analysis. For instance, they might check 2, 3, 5, 7 but forget to verify larger primes like 23, 29, 31, 37, 41, 43, 47, leading to an incorrect conclusion about where the smallest prime factor lies.

2. Boundary error at prime 50

Students may incorrectly include or exclude the boundary case around 50. They need to carefully verify that for prime \(\mathrm{p} = \mathrm{47}\), we have \(\mathrm{2p} = \mathrm{94 \leq 100}\) (so 47 divides \(\mathrm{h(100)}\)), but for prime \(\mathrm{p} = \mathrm{53}\), we have \(\mathrm{2p} = \mathrm{106} > \mathrm{100}\) (so 53 might not divide \(\mathrm{h(100)}\)). Confusion about this boundary leads to wrong range identification.

Errors while selecting the answer

1. Selecting based on incomplete analysis

After determining that primes ≤ 47 cannot divide \(\mathrm{h(100)} + \mathrm{1}\), students might hastily select answer choice D (Between 30 and 40) thinking that 47 falls in some nearby range, rather than recognizing that the smallest possible prime factor must be ≥ 53, which requires answer choice E (Greater than 40).