Introduction

In this article, we explore the process of finding the value of x such that x! 720. We will employ a structured approach to determine the value of x, using both traditional methods and advanced mathematical functions such as the Gamma function.

Factorial Function and Prime Factorization

One straightforward method to solve this problem is to use the factorial function and prime factorization. Let's begin by expanding the factorial function until we reach 720.

1! 1

2! 2

3! 6

4! 24

5! 120

6! 720

Thus, x 6.

We can also use prime factorization to verify this. The prime factors of 720 need to be reorganized to form an ascending list of natural numbers:

720 360 2 180 * 2 * 2 90 * 2 * 2 * 2 30 * 3 * 2 * 2 * 2 10 * 3 * 3 * 2 * 2 * 2 5 * 2 * 3 * 3 * 2 * 2 * 2 1 * 2 * 2 * 2 * 2 * 3 * 3 * 5 1 * 2 * 3 * 2 * 2 * 5 * 3 * 2 1 * 2 * 3 * 2 * 2 * 5 * 3 * 2 1 * 2 * 3 * 4 * 5 * 6 6!

Therefore, x 6.

Hit and Trial Method

Another practical approach involves a more direct hit and trial method. Since 720 has one trailing zero, we know that x must be at least 5 and less than 10. The prime factor 5 gives us a starting point. The next prime, 7, does not divide 720, narrowing our options to 5 and 6. Given that 720 is divisible by 9, which comes from the product of 2 and 3 (since 6 2 * 3), we can conclude that x 6.

Using the Gamma Function

To further validate our solution, we can use the Gamma function, which extends the factorial function to non-integer values. The Gamma function is defined as:

[ Gamma(n) (n-1)! ]

Setting ( n 7 ), we find:

[ Gamma(7) 6! ]

We can manually integrate the Gamma function to demonstrate:

Let's set up a reduction formula to simplify the process:

J_n  int x^n e^{-x} dx

Using integration by parts, we get:

Let ( u x^n ) and ( du n x^{n-1} dx ); ( dv e^{-x} dx ) and ( v -e^{-x} ), then:

[ J_n -e^{-x} x^n n int x^{n-1} e^{-x} dx ]

[ J_n -e^{-x} x^n n J_{n-1} ]

Some initial reductions:

[ J_0 -e^{-x} C ] [ J_1 -e^{-x} x - e^{-x} C ] [ J_2 -e^{-x} 2x - e^{-x} x^2 - e^{-x} x C ] [ J_3 -e^{-x} 6x^2 - e^{-x} 3x^3 - e^{-x} 3x C ] [ J_4 -e^{-x} 24x^3 - e^{-x} 12x^4 - e^{-x} 12x^2 C ] [ J_5 -e^{-x} 12^4 - e^{-x} 6^5 - e^{-x} 6^3 C ] [ J_6 -e^{-x} 72^5 - e^{-x} 36^6 - e^{-x} 36^4 C ]

Evaluating from 0 to infinity:

[ Gamma(7) J_6(0) - J_6(infty) 0 - (-720) 720 ]

Hence, we confirm that:

[ 6! Gamma(7) 720 ]

Conclusion: Through multiple methods, we have verified that the value of x such that x! 720 is indeed x 6. This exploration demonstrates the application of factorial functions, prime factorization, and the Gamma function in solving such problems.