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.

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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!