Brazil | A nice factorial equation | math Olympiad

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[js]

The last part of the proof was not really necessary, because when you get the equation:

x[(x - 1)! - 1] = 3

It suffices to note that 3 is prime, so x can only be 1 or 3, and the former is impossible.

[bharatbhandari]

I observe while solving the questions and where u apply a formula, you right THEREFORE sign to write formulae, instead of SINCE sign.

[skyLoon]

I really dont understand, since at the end u are still trying the number 1 by 1, why not try it earlier. U giving me a hope thats a algebra way to solve it....end

[LilCletus]

If you were going to trial and error it half way through the solution, why not just do it at the beginning?

[sidgar]

I considered the first instance where m^3 was close to m!. That excluded 1-4. Then I found that 5^3-5 = 120, which is also 5!. Of course, that was just a small number which was easy to guess. For a generalized case which could have been too large to calculate easily, it's best to solve this type of problem algebraically.

[Pietruszka]

It's easier to check the first few numbers because m! grows much faster than m^3 for big numbers

[HyperFocusMarshmallow]

3:12 —
x((x-1)!-1)=3.
Since we’re looking for integer solutions here, it’s enligt to note that the equation is in the form a*b=3.
Either we must have (a, b)=(1, 3) or (a, b)=(3, 1). (Negative solutions are ruled out by the domain of the original equality.)
3((3-1)!-1)=3(2!-1)=3(2-1)=3*1=3,
So x = 3 works.

which is not =3.

So at that point we plug those results back into the original equation, and can consider ourselves done with the problem.

——
Dividing by x just introduces fractions which are more complicated. The only result from the further analysis is that x could only be 1 or 3 anyway so there is really no new information gained by doing that step.

[Prypak]

m³-m = (m+1)m(m-1)
m! = m(m-1)(m-2)...
Simplify :
(m+1)=(m-2), thats unsolvable to I take the next factor,
m+1 = (m-2)(m-3) then you can solve it with the quadratic formula but you really don't need to, the answer is m=5
If the answer wasn't an integer I would have taken another factor from m! But I figured this strategy would work since this video seems like not-too complicated maths

[katiehall]

why does x have to be equal to or under 3

[HenryBriskin]

Yes factorial of a numbers e.g 5 is 5x4x3x2x1

[HyperFocusMarshmallow]

Another attempt.
Find m such that m! = m^3-m

Alright:
Factor the RHS to get:
m! = (m-1)m(m+1).

Shift index m=n+1:
(n+1)!=n(n+1)(n+2).
Well let’s go one more step actually:
n=k+1
(k+2)!=(k+1)(k+2)(k+3)

If we multiply both sides by k!, we get
k!(k+2)!=(k+3)!

k!=k+3

Recall that m is non negative and thus k is at least 2.
k! Has a bunch of factors of 2 in it. It is clearly even.
Hence k must be odd. So it is at least 3.

The lowest number with s prime factors is 2^s.
All factorials k! where k>3 has at least 4 prime factors 2*3*2*2… .
So either the number is 3 or:
k+3>=2^4=16
In other words k>=13.
But that’s just ridiculous since (13)! >> 10^4.

There was ever only one real candidate all along. k=3.

3!=3+3=6 checks out.

So if any m works, it must be m=5. 5!=120, 5^3=125, 5=5 (the last one is deep math for sure). I’ll leave subtraction as an exercise for the reader. It checks out.

In summary: m=5.
Which we could have guessed in about 12 seconds of mental arithmetic.

[Математиканапять]

Ой как много букв) оценка в строчку и ответ..

[-WarMapping-]

You did guessing at 4:21 so why not do it for the original problem?

[herbertklumpp]

One easily Sees m=5 is a solution.N! Is stronger ingreasing than linear function is m=5 the only solution

[Iihgtff]

X cant be 2. As factorial notation cant be in later decomal places

[HyperFocusMarshmallow]

Find m: m!=m^3-m.
(m! grows much faster than m^3 so any candidate must be quite small. A quick guess or some trial and error makes it quite obvious that m=5. But we can still simplify it at bit to make it really trivial.)

RHS=m(m^2-1)=(m-1)m(m+1).
Dividing both sides by (m-1)m we have (for m>1):
(m-2)!=m+1.
Shift the index n=m-2:
n!=n+3.

Now if m! grows faster than m^3, well then of course n! out grows n even more quickly.

There are so many interesting simplifying arguments here even though the answer is obviously n=3.

Factorials are either very small or have many factors.

The number of prime factors of n! grows faster than n since (spoiler alert) there exists numbers that aren’t prime numbers.
But all numbers k<8 have <3 prime factors. (<3<3<3)
The only factorials n! with n>1 that have less than three prime factors are as 2! And 3!.
We could immediately have seen that 2+3, would fail to have both 2 and 3 as factors which would disqualify it as a factorial number immediately. We simply have too many arguments for why n=3.

As was obvious from the beginning n=3 => m=5.

120 = 125 - 5.

[juandepuntagiamichelli]

En el minuto 2:31, ya se resuelve, X solo puede valer 3.

[justincase]

How is it that you can say that
m! = m(m-1)(m-2)!
How does that make sense?

[jacobcombs]

One look and I knew it was 5 because of all the 5! Memes everyone should know 5! is 120 and anyone who has made it past the 3rd grade should know that 5^3 is 125, and the difference between them is conveniently 5.

[shirimewows]

Can someone explain where the upper bound x <=3 comes from?