Brazil | A Nice Factorial Equation | Math Olympiad | Find m

Brazil | A Nice Factorial Equation | Math Olympiad | Find m
#olympiadmathematics #olympiad #exponential #algebra #mathematics #mathtricks #squareroot
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[@subamnaut]

I would rather go for hit and trial in the first place within 15 sec after reading the question

[@sebastiangudino9377]

Factorial grows faster than cubic polynomials. So the answer needs to be before the points where x³ - x goes below x! (Quite early) That means the answer is not very large, we can probably brute-force it! Lets write a few cubes 1³ = 1 2³ = 6 3³ = 9 4³ = 64 5³ = 125 And a few factorial 2! = 1*2 3! = 2*3 4! =6*4 5! = 20*5 6! = 120*6 Ok, so 5! (120) Is really close to 5³. The intersection point is near. And we can see that 4 doesn't work 20 ≠ 64 - 4. So it HAS to be either 5 or 6 (Else we overshoot the intersection of the two graphs) 5! = 120 5³ = 125 5³ - 5 = 120 Ok, so it WAS 5. m = 5 satisfies m! = m³ - m Quick and easy educated brute-force to the rescue

[@DCWornock]

Without viewing the video, I solved it in my head by guessing M =3 then 4, and finally 5 which is answer. Similar problems with very large M would require many more guesses. I don't know how to solve with algebra.

[@Libertarian1208]

It can be done simpler. Until 1:46 it is ok. But then (m-2)! >= (m-2)(m-3) = m^2 - 5m +6 = m(m-5) + 6 >= m +6 if m>= 6. Thus m can't be greater than 5. It is enough to check m=1,2,3,4,5 to see that the only solution is 5

[@superzerdax]

x [(x-1)!-1] = 3 But 3 is prime, so the only numbers wich makes this product correct is 3 . 1 = 3, wich leads to x=3. And then, m=5. Is this solution correct?

[@segi450]

Way overengeneered. m! = m * (m-1) * (m-2)! m^3 - m = m * (m^2 - 1) = m * (m - 1) * (m +1) We can tell right away that m is not 0 or 1, so we can divide by m*(m-1) and are left with (m-2)! = m+1. From here it's evident that m can't be a large number, must be larger than 3, and that m+1 can't be a prime number such as 5 or 7. So only m+1=6 -> m=5 is really an option.

[@brunonicolasgonzalez1103]

Saludos de Argentina!

[@MeesterG]

Only 1 minute in and just a primary school teacher. When you write n! = n (n-1)(n-2)(n-3)! That wouldn't work if n=3 right? Because then 3 = 3 * 2 * 1 * 0! Oh wait, should I read the last part as 0 factorial, which is 1? Than is does work. But how about n = 2 2! = 2 * 1 * 0 * -1! Is that correct? I am learning about minus factorials now. However, there is a 0 there, so it should add up to 2! = 0 Or where am I wrong.

[@MeesterG]

And it seems like you can intuitively feel that m can't be a big number, so is brute forcing the solution allowed in an Math Olympiad? Then you would try the first few numbers, only then you don't prove that that would be the only solution.

[@Maurycy5]

@@MeesterG To be fair if brute forcing isn't allowed (as long as you provde a proof that no other answers exist, which is easy here because the factorial majorizes any polynomial), then the olympiad isn't worth much. Brute forcing is increasingly often how real math is done now anyway, even if cool ideas are still required along the way.

[@Maurycy5]

@@MeesterG You are also correct to note that the method is technically incorrect for small m, but (-1)! is not 0, it's undefined. Best case scenario you can extend the factorial function to the complex plane, known as the Gamma function, at which point it remains undefined for the negative integers.