A Very Nice Factorial Equation | Diophantine Equation | Math Olympiad Preparation

A Very Nice Factorial Equation | Diophantine Equation | Math Olympiad Preparation

Welcome to infyGyan! In today's video, we dive into a fascinating factorial equation that's perfect for Math Olympiad preparation. This challenging diophantine problem will test your algebraic skills and help you improve your problem-solving techniques. Whether you're preparing for a math competition or just love tackling tough math problems, this video is for you!

🔍 In this video, you will learn:

How to approach and solve a complex factorial equation
Key concepts and strategies for tackling Diophantine equations
Step-by-step solution to enhance your understanding

📚 Don't forget to:

Like the video if you find it helpful
Subscribe to infyGyan for more challenging math problems and solutions
Share your thoughts and solutions in the comments below!
Stay tuned for more exciting math content and keep sharpening those math skills.
Happy solving!

Additional Resources:

Time-stamps:

00:00 Introduction
00:22 Recursive factorial formula
02:10 Algebraic manipulations
04:44 Substitution
06:23 Solving factorial equation
08:25 Integer solution
09:00 Verification

#matholympiad #factorialequation #mathchallenge #algebra #matholympiadpreparation #mathcompetition #problemsolving #diophantineequations #math #tutorial #integersolutions

Thank You for Watching Videos !!

Do not forget to subscribe our channel infyGyan.

[florianbuerzle]

I also used substitution. Let u = x - 4, so the equation will be u! = u + 3. As 3 must be divisible by u, only 1, 2 or 3 are possible solutions. So we get u = 3 and x = 3 + 4 = 7.

[Shobhamaths]

Let a=(x-4), a! =a+3, a=3 then, x=7👍

[RashmiRay-cy]

Let x-4=t Then, t!=t+3 > t=3 > x=7.

[narendrabahali]

x-4 =u
Then problem is reduced to
u[(u-1)! -1] =3
So u =1 or u=3
u can't be 1 as 2nd term is -ve
u =3
Then x = 7

[mohamedbrahim]

Solved in 10’’ : 4, 5, 6 do not work, 7 works inequality is obvious starting from 8. X=7.

[RealQinnMalloryu]

(x ➖ 4)^2=( x^2 ➖ 16)= 14 2^7 (x ➖ 7x+2). (x ➖ 1)^2= (x^2 ➖ 1)= x^1 (x ➖ 1x+1) .