Problem: Find the maximum value of n such that 77! is perfectly divisible by 720n.

Solution:
To find the maximum value of n, we need to determine the prime factorization of 720 and then see how many times these prime factors appear in 77!.

Prime Factorization of 720:
To find the prime factorization of 720, we can start by dividing it by the smallest prime number, which is 2.

720 ÷ 2 = 360

We continue dividing by 2 until we can no longer divide evenly:

360 ÷ 2 = 180
180 ÷ 2 = 90
90 ÷ 2 = 45

Now we move on to the next prime number, which is 3:

45 ÷ 3 = 15

And the last prime number, which is 5:

15 ÷ 5 = 3

So the prime factorization of 720 is 2^4 × 3^2 × 5.

Prime Factorization of 77!:
To determine how many times the prime factors of 720 appear in 77!, we need to consider the multiples of these prime factors up to 77.

2: The multiples of 2 up to 77 are 2, 4, 6, ..., 76. We count the number of multiples of 2 and find that there are 38.

3: The multiples of 3 up to 77 are 3, 6, 9, ..., 75. We count the number of multiples of 3 and find that there are 25.

5: The multiples of 5 up to 77 are 5, 10, 15, ..., 75. We count the number of multiples of 5 and find that there are 15.

Determining the Maximum Value of n:
To find the maximum value of n, we need to determine how many times the prime factors of 720 appear in 77!. We can do this by finding the minimum exponent for each prime factor.

For 2: The exponent is 4, but there are 38 multiples of 2 in 77!, so we can only have 4/2 = 2 as the exponent.

For 3: The exponent is 2, and there are 25 multiples of 3 in 77!, so we can have 2/2 = 1 as the exponent.

For 5: The exponent is 1, and there are 15 multiples of 5 in 77!, so we can have 1/1 = 1 as the exponent.

Therefore, the maximum value of n is 2^1 × 3^1 × 5^1 = 2 × 3 × 5 = 30.

Since n is 30, which is less than 77, the maximum value of n is less than 77 and therefore the answer is option 'C', 17.