A mysterious factorial equation.

We solve a nice equation over the natural numbers involving exponentiation and factorials. Our approach employs Wilson's theorem and other modular reduction tricks.

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[HarmonicEpsilonDelta]

10:47 I think we used the hypothesis when we multiplied 2*(n/2), because for that to be true, we need both 2 and (n/2) in the product, and for that, we need n/2 to be different than 2, since they come from the factorial, and therefore n>4, which implies n\geq 5.

[NO_ON-jb]

Wilson + inequality + bionomial + basic congruence = amazing problem

[factsverse]

It appeared on the SMO (Singapore Mathematical Olympiad), Open Round 2 on the year 2008, as problem number 1. Hope it helps!

[roboto]

I was trying that exact problem, it was in an LTE article. Thanks

[Kettwiesel]

I wasn't aware of Wilson's Theorem, but it's not too hard to see that n+1 has to be a prime in this problem, this is the easy part of Wilson's Theorem. If n+1 has a factor, which is smaller than it, then it appears in n! as well as (n+1)^k, so the two can not be just 1 apart (they are not coprime)

[TwilightBrawl]

3:25 Not a big mistake but the left hand side is even and the right hand side is odd.

[mihamihailovic]

If it isn't too much of a hastle can you please put the video link in the description for theorems you used and proved in another video (like wilsons theorem here)

[a_llama]

10:47 im guessing n>5 because you assume that there exists an n/2 >= 3, meaning n>=6

[mcwulf]

Clever solution. Especially noticing n/2 and 2 are factors of n! when n even.

[mathissupereasy]

I would love to see you working on inequality problems.

[stevenwilson]

6:53 why is n + 1 a prime? Does this cover cases where n + 1 is not prime?

[prunodagen]

You don"t need Wilson Theorem.

(n+1)^k = n! + 1

n! + 1 is odd for n > 1, then n + 1 must be odd, then n must be even. (we don't need n + 1 prime, just the fact n is even to get k = 0[n])

[ffggddss]

First thing I thought when I saw the thumbnail: "Hey, that looks like Wilson's Theorem! . . Oh wait, not quite." Wilson's Theorem says that
p | [(p–1)! + 1], iff p is prime. Here we would make n = p–1 and this becomes
(n+1) | (n! + 1), iff (n+1) is prime; vs our problem:

(n+1)ᵏ = n! + 1

which is a stronger condition on (n! + 1) than mere divisibility by (n+1).

k & n are restricted to the +ve integers, so start with k=1:
N+1 – 1 – n = n!
Works for n = 1 and 2; no others.
Solutions: (n, k) = (1, 1), (2, 1)

k=2:
n! + 1 will have to be a square. I think the only cases of that are n = 4, 5, and 7. n = 4 works for this equation; 5 and 7 do not:
4! + 1 = 25 = 5²
5! + 1 = 121 = 11²
7! + 1 = 5041 = 71²
So:
(4+1)² – 1 = 4! . . 24 = 24
Solution: (n, k) = (4, 2)

I suspect there are no others, but I don't see a way to prove that. Let's see what Michael has...

... Very nicely done! So my suspicion was borne out. Thanks to Michael for the proof.

Fred

[angelvalera]

I love those videos.
Greetings from Spain :D

[chhabisarkar]

A nice little problem from INMO 2014, problem 2
Prove that [n] + [n/2] + [n/3] ....+ [n/n] + [√n] is even, where [n] is the floor of n
Plz see it . Thank you

[joaopedrobmenezes]

Vp(x!) >= 2*Vp(x) for all non prime x such that x> 4, then you can see that if n>4 and it is not prime( which is the case for n >4) k would have to be at least n, but (n + 1)^n - 1 > n! for n > 4. This is just an alternative solution:)

[WriteRightMathNation]

Nice problem, but at 3:28 (someone else may have already pointed this out...), you mixed up "left-hand" and "right-hand"... 😊

[hans-juergenbrasch]

n=1, 2 imply k=1. Now for n>2 we must have n+1=p is a prime, otherwise n+1 would have a proper divisor 1 < d =< (n+1)/2 < n which yields a contradiction modulo d:
0=n!=(n+1)^k-1=-1 and hence we can rewrite p^k-1=(p-1)! with p>=5 or
1+p+...+p^(k-1)=(p-2)! Note that gcd(1+p+...+p^(k-1), p-1)=gcd(k, p-1)
Since (p-1)/2 < p-2 we must have
k = t (p-1)/2 for some t. Since p^(p-1)-1=
= (1+(p-1))^(p-1)-1>(p-1)^(p-1)
we must have t=1, i.e.
p^((p-1)/2)-1=(p-1)!
For p=5 we get a solution corresponding to n=4, k=2 and for p>5 we get p^((p-1)/2)-1=
=prod_(i=1)^((p-1)/2) [i(p-i)]>

>p^((p-1)/2), a contradiction. Note, that we have split the product into two factors, one for i=1, 2 given by (p-1)2 (p-2) and the other for i=3, ..., (p-1)/2

[gregsavitt]

Just got out of Calc 3 thinking that I had gotten pretty good at math. Then I watched this and my brain melted.

[TechToppers]

Let s(n) denote the sum of the digits of a positive integer n in base 10. If s(m) = 20 and s(33m) = 120,
what is the value of s(3m)?

Can someone please help me with this?
It came in India. PRMO 2019 25th August Paper