For each positive integer n, consider the highest common factor hn of the two numbers n! + 1 and (n + 1)!. For `n lt 100`, find the largest value of `h_n`

Answer

To solve the problem, we need to find the highest common factor \( h_n \) of the two numbers \( n! + 1 \) and \( (n + 1)! \) for each positive integer \( n \) less than 100. We will analyze the expressions step by step.

### Step 1: Understand the expressions

We have two expressions:
1. \( n! + 1 \)
2. \( (n + 1)! = (n + 1) \cdot n! \)

### Step 2: Find the highest common factor

We need to find \( h_n = \text{gcd}(n! + 1, (n + 1)!) \).

Using the property of gcd, we can express this as:
\[ h_n = \text{gcd}(n! + 1, (n + 1) \cdot n!) \]

### Step 3: Apply the Euclidean algorithm

Using the property of gcd, we can simplify:
\[ h_n = \text{gcd}(n! + 1, (n + 1) \cdot n!) = \text{gcd}(n! + 1, (n + 1) \cdot n! - (n + 1)(n! + 1)) \]
This simplifies to:
\[ = \text{gcd}(n! + 1, - (n + 1)) \]
Thus, we have:
\[ h_n = \text{gcd}(n! + 1, n + 1) \]

### Step 4: Analyze the gcd

Now, we need to analyze \( n! + 1 \) modulo \( n + 1 \):
- \( n! \) is divisible by all integers from \( 1 \) to \( n \), hence \( n! \equiv 0 \mod (n + 1) \).
- Therefore, \( n! + 1 \equiv 1 \mod (n + 1) \).

This means:
\[ \text{gcd}(n! + 1, n + 1) = \text{gcd}(1, n + 1) = 1 \]
for all \( n \) except when \( n + 1 \) is prime.

### Step 5: Special case for prime \( n + 1 \)

If \( n + 1 \) is a prime number \( p \), then:
\[ h_n = p \]
This occurs when \( n = p - 1 \).

### Step 6: Find the largest \( h_n \) for \( n < 100 \)

The largest prime number less than 101 is 97. Therefore, the largest value of \( h_n \) occurs when \( n = 96 \):
\[ h_{96} = 97 \]

### Conclusion

The largest value of \( h_n \) for \( n < 100 \) is:
\[ \boxed{97} \]