15 × 14 × 13 × … × 3 × 2 × 1 = 3 m × n 12. 15 × 14 × 13 × … × 3 × 2 × 1 = 3 m × n Where m and n are positive integers, then what is the maximum value of m?

The problem asks for the maximum value of 'm' in the equation 15! = 3^m * n, where 15! represents 15 × 14 × 13 × … × 3 × 2 × 1, and 'n' is a positive integer not divisible by 3. This means 'm' is the highest power of 3 that divides 15!.

To find the highest power of a prime number 'p' that divides n!, we use Legendre's formula: m = floor(n/p) + floor(n/p^2) + floor(n/p^3) + ...

In this case, n = 15 and p = 3.

1. Calculate the first term: floor(15/3) = floor(5) = 5. (This counts numbers like 3, 6, 9, 12, 15, each contributing at least one factor of 3).

2. Calculate the second term: floor(15/3^2) = floor(15/9) = floor(1.66...) = 1. (This counts numbers like 9, which contribute an additional factor of 3 because 9 = 3*3. The first term already counted one 3 from 9, this term counts the second 3 from 9).

3. Calculate the third term: floor(15/3^3) = floor(15/27) = floor(0.55...) = 0. (We stop here as subsequent terms will also be 0).

Now, sum the terms: m = 5 + 1 = 6.

Alternatively, we can directly count the factors of 3 in the numbers from 1 to 15:

• Multiples of 3: 3, 6, 9, 12, 15.
• Each of these contributes at least one factor of 3: (3=3^1), (6=23^1), (9=3^2), (12=43^1), (15=5*3^1).
• Count the factors of 3:
  • From 3: one 3
  • From 6: one 3
  • From 9: two 3s (since 9 = 3 * 3)
  • From 12: one 3
  • From 15: one 3
• Total number of factors of 3 = 1 + 1 + 2 + 1 + 1 = 6.

Thus, the maximum value of m is 6.

The final answer is B