To determine the largest possible integer $n$ such that $n!$ (n factorial) is divisible by 210, we need to factor 210 and ensure that $n!$ contains all of its prime factors.

Step 1: Prime factorization of 210

$210 = 2 \times 3 \times 5 \times 7$ Thus, for $n!$ to be divisible by 210, $n!$ must include at least one factor of 2, 3, 5, and 7.

Step 2: Find the smallest $n$ such that $n!$ includes all these factors

We need to determine the smallest $n$ such that $n!$ contains the numbers 2, 3, 5, and 7. Let’s check the factorials of consecutive integers:

• $4! = 4 \times 3 \times 2 \times 1 = 24$
  • Prime factors: $2^3 \times 3$. Does not contain 5 or 7.
• $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$
  • Prime factors: $2^3 \times 3 \times 5$. Does not contain 7.
• $6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$
  • Prime factors: $2^4 \times 3^2 \times 5$. Does not contain 7.
• $7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040$
  • Prime factors: $2^4 \times 3^2 \times 5 \times 7$. Contains 2, 3, 5, and 7.

Step 3: Conclusion

The smallest $n$ such that $n!$ is divisible by 210 is $n = 7$.

Thus, the largest possible integer $n$ such that $n!$ is divisible by 210 is: $\boxed{7}$

Would you like more details on factorials or prime factorizations?

Here are 5 related questions to consider:

1. What is the prime factorization of a factorial number?
2. How is the divisibility of factorial numbers determined?
3. What are the properties of prime numbers within factorial sequences?
4. How can you calculate the number of trailing zeros in a factorial?
5. What is the smallest $n$ such that $n!$ is divisible by a larger number, like 420?

Tip: When working with divisibility in factorials, remember that factorials grow rapidly and often contain many factors of smaller primes.