Solving the Unsolvable: Factorial Equation Challenge

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Solving the Unsolvable: Factorial Equation Challenge

Join us on an exhilarating mathematical journey as we take on the unsolvable – the Factorial Equation Challenge! 🧮💡 We'll explore complex mathematical equations involving factorials, apply clever strategies, and work through some mind-bending examples. Whether you're a math enthusiast or looking to level up your algebra skills, this video is sure to spark your curiosity and test your problem-solving abilities. Can you solve the unsolvable? Let's find out together! 🔍✨

You will learn:
1. Factorial Equations
2. Olympiad Exam
3. Mathematics
4. Problem Solving
5. Value of X
6. Exam Preparation
7. Math Tutorial
8. Step-by-Step Guide
9. Math Olympiad
10. Algebraic identities
11. Math Olympiad Preparation
12. Factorial notation
13. Recursive formula

8 Key moments of this video:
0:00 Introduction
0:32 Factorials
0:58 Recursive formula
2:23 Algebraic identities
4:50 Substitution
6:30 Algebraic expansion
8:16 Logical conclusions
10:08 Verification

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#algebra #math #factorials #mathematics #infyGyan

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(x + 2)! + x^3 = x^5

Thus, x => -2 and is an integer

If x = -2, then 0! + (-8) != (-32)
If x = -1, then 1! + (-1) != (-1)
If x = 0, then 2! + 0 != 0
If x = 1, then 3! + 1 != 1
If x = 2, then 4! + 8 = 32 so x = 2 is a solution.

Now suppose x => 3.

(x + 2)! = x^5 - x^3
(x + 2)! = x^3 (x^2 - 1)
(x + 2)! = x^3 (x - 1)(x + 1)
(x + 2)(x + 1)(x)(x - 1)(x - 2)(x - 3)! = x^3 (x - 1)(x + 1)
(x + 2)(x - 2)(x - 3)! = x^2
(x^2 - 4)(x - 3)! = x^2
(x - 3)! = x^2 / (x^2 - 4)
(x - 3)! = 1 + 4 / (x^2 - 4)
(x - 3)! - 1 = 4 / (x^2 - 4)

Since x => 3, then
x^2 => 9
x^2 - 4 => 5 > 4
x^2 - 4 > 4
4 / (x^2 - 4) < 1

The LHS is an integer and the RHS is between 0 and 1, a contradiction.

Thus, x = 2 is the only solution

Packerfan
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Real men would solve this with the gamma function

ShadowGamingof
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thank you so much sir, I love you man

MAARSHN
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how does this video only have 21 likes this is beautiful

mahadkhan