Fatorial #2

If we solve for the variables $a,b,c,d$ in terms of $x$ , then we obtain:

$a = x+1, b =x+2, c=x+3, d=x+4$ .

Substituting these expressions into the latter factorial equation yields:

$x \cdot x! + a \cdot a! + b \cdot b! + c \cdot c! + d \cdot d! = 5040 - \log(100) \Rightarrow x \cdot x! + (x+1)(x+1)! + (x+2)(x+2)! + (x+3)(x+3)! + (x+4)(x+4)! = 7! - 2$ ;

or $[x + (x+1)^2 + (x+2)^2(x+1) + (x+3)^2(x+2)(x+1) + (x+4)^2(x+3)(x+2)(x+1)] \cdot x! = (7\cdot6\cdot5\cdot4\cdot3 - 1) \cdot 2!$ ;

or $\boxed{x = 2}.$

x . x! + a . a! + b . b! + c . c! + d . d! = 5040 - log 10²

x . x! + (x+1) . (x+1)! + (x+2) . (x+2)! + (x+3) . (x+3)! + (x+4) . (x+4)! = 5038

---

Deduction

If x = 0

0 . 0! + 1 . 1! + 2 . 2! + 3 . 3! +4 . 4! = 5038

0 . 1 + 1 . 1 + 2 . 2 + 3 . 6 + 4 . 24 = 5038

0 + 1 + 4 + 18 + 96 = 5038

119 = 5038 (False)

---

Deduction

If x = 1

1 . 1! + 2 . 2! + 3 . 3! + 4 . 4! + 5 . 5! = 5038

1 + 2 . 2 + 3 . 6 + 4 . 24 + 5 . 120 = 5038

1 + 4 + 18 + 96 + 600 = 5038

719 = 5038 (False)

---

Deduction

If x = 2

2 . 2! + 3 . 3! + 4 . 4! + 5 . 5! + 6 . 6! = 5038

2 . 2 + 3 . 6 + 4 . 24 + 5 . 120 + 6 . 720 = 5038

4 + 18 + 96 + 600 + 4320 = 5038

5038 = 5038 (True)