Problem: The ratio of the factorial of a number x to the square of the factorial of another number which when increased by 50% gives the required number is 1.25. Find the number x.

Solution:

Let's break down the problem into simpler steps.

Step 1: Understand the problem statement
- We are given a ratio of factorials and we need to find the value of x.
- Another number, which is 50% greater than x, when factorial is taken and squared, gives the required number in the ratio.
- The ratio of factorials is 1.25.

Step 2: Write the equation
- Let's assume the other number is y and the required number is z.
- According to the problem statement, we have:
- x! / (y!^2) = 1.25
- y * 1.5! = z

Step 3: Simplify the equation
- Let's simplify the second equation:
- y * 1.5! = z
- y * 1.5 * 0.5! = z
- y * 1.5 * 0.5 = z
- y = z / 1.125

- Substituting this value of y in the first equation:
- x! / ((z/1.125)!^2) = 1.25

Step 4: Solve for x
- Let's simplify the denominator:
- ((z/1.125)!)^2 = ((z/1.125) * (z/1.125 - 1) * (z/1.125 - 2) * ... * (1.125)) ^ 2
- ((z/1.125)!)^2 = (z^2 * (z/1.125 - 1)^2 * (z/1.125 - 2)^2 * ... * (1.125)^2) / (1.125^2)

- Substituting this value in the first equation:
- x! / (z^2 * (z/1.125 - 1)^2 * (z/1.125 - 2)^2 * ... * (1.125)^2 / (1.125^2)) = 1.25
- x! * 1.125^2 = 1.25 * z^2 * (z/1.125 - 1)^2 * (z/1.125 - 2)^2 * ... * (1.125)^2

- Taking the logarithm of both sides:
- log(x!) + 2 * log(1.125) = log(1.25) + 2 * log(z) + 2 * log(z/1.125 - 1) + 2 * log(z/1.125 - 2) + ... + 2 * log(1.125)

- Using Stirling's approximation, we can approximate log(x!) as:
- log(x!) = x * log(x) - x + O(log(x))

- Substituting this value in the equation:
- x * log(x) - x + O(log(x))