If 5a is a factor of n!, and the greatest integer value of a is 6, what is the largest possible value

of b such that 7b

is a factor of the largest possible value of n!?

To solve this problem, let's break it down into parts.

### Problem Statement Recap

- Given that \( 5^a \) is a factor of \( n! \) and the greatest integer value of \( a \) is 6, we need to determine the largest possible value of \( b \) such that \( 7^b \) is a factor of \( n! \).

### Step 1: Understanding \( 5^a \) as a Factor of \( n! \)

The problem tells us that \( 5^a \) is a factor of \( n! \), and the largest possible value of \( a \) is 6. This means that \( a \) represents the sum of all powers of 5 that divide \( n! \).

Mathematically, we calculate the number of factors of 5 in \( n! \) as follows:

\[
a = \left\lfloor \frac{n}{5} \right\rfloor + \left\lfloor \frac{n}{25} \right\rfloor + \left\lfloor \frac{n}{125} \right\rfloor + \cdots
\]

We want \( a = 6 \), so let's find the smallest \( n \) such that this sum equals 6.

### Step 2: Calculate \( n \) such that \( a = 6 \)

Start with small values of \( n \):

For \( n = 25 \):

\[
a = \left\lfloor \frac{25}{5} \right\rfloor + \left\lfloor \frac{25}{25} \right\rfloor = 5 + 1 = 6
\]

So, \( n = 25 \) gives \( a = 6 \). Therefore, \( n = 25 \) is the smallest \( n \) that satisfies the condition.

### Step 3: Determine the Largest \( b \) for \( 7^b \) to Be a Factor of \( 25! \)

Now, we need to calculate how many factors of 7 are in \( 25! \). The number of factors of 7 is calculated similarly:

\[
b = \left\lfloor \frac{25}{7} \right\rfloor + \left\lfloor \frac{25}{49} \right\rfloor
\]

Calculating each term:

\[
\left\lfloor \frac{25}{7} \right\rfloor = 3
\]

\[
\left\lfloor \frac{25}{49} \right\rfloor = 0
\]

Adding these together:

\[
b = 3 + 0 = 3
\]

### Conclusion

The largest possible value of \( b \) is 3.

---

If you have any more questions or need additional details, feel free to ask! Here are five related questions to explore this topic further:

1. How do you find the largest power of a prime number dividing \( n! \)?
2. What is the general formula for finding factors of \( p^k \) in \( n! \), where \( p \) is a prime?
3. Can you extend this approach to non-prime factors?
4. How does this method change for larger values of \( n \)?
5. What happens if \( p \) is a composite number rather than a prime?

**Tip:** Always break down factorial problems by analyzing the powers of primes dividing the factorial!

Explore how to determine the largest prime factor in factorials using Legendre's formula and prime factorization. This problem focuses on calculating powers of primes in factorials, with detailed steps and examples. Suitable for advanced high school students.

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