x!+y!+z!=w!, A Factorial Equation

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[RexxSchneider]

The factorial of a non-negative integer is >= 1, so w must be strictly greater than x, y, z. So (w-1) >= x, y, z
w! = w.(w-1)! So for all w > 3, we can see that w! > 3.(w-1)!
Therefore for w > 3, we have w! > x! + y! + z! and it follows that w <= 3
Also 0! + 0! + 0! = 3 > 2! so 2! < x! + y! + z! and therefore w! > 2! and so w > 2
The only remaining value for w is 3 and 3! = 2! + 2! +2!, giving the sole solution.

[MrLidless]

Substitute the first five minutes with this:
3*max(x, y, z)!>=w!, so w<=3. That’s all you needed.

[JohnSmith-nxzj]

Since x, y, z are symmetrical in this equation, wlog assume x <= y <= z

Now since a non-negative factorial is always >= 1 we can say:

1. 2+z! <= w! and
2. 3z! >= w!

Now assume z is >= 3. To satisfy inequality 1, w must be >= z+1

So w! >= (z+1)! = (z+1)z! >= 4z!

But this contradicts 2.

So z <= 2.

By checking cases:

2, 2, 2 gives 6 which does equal 3!
0, 0, 0 gives 3 which does not equal a factorial.
All other possible values of x, y, z will give a value between 3 and 5 which does not equal a factorial. So 2, 2, 2 is the only solution.

[MathTutor]

Very interesting problem. You also extend this for any finite number of variables. See whether you can also do this type of calculation for products as well. Thank you.

[WahranRai]

Recall : 0! = 1! = 1 ; 2!=1*2; 3! = 1*2*3 =6 Then min (a! + b! + c! ) = 3
the first factorial w ! > 3 and a! + b! + c! = w! is w! = 1*2*3 = 6 with a = b = c =2 (unique solution )
w! = 3! = 2! + 2! +2! is also the maximum of a! + b! + c! satisfying a! + b! + c! = w!
Let us show that w > 3 are not possible
The maximum possible value of a, b, c is w-1 (due the constraint a! + b! + c! = w! ) then max (a! + b! + c! ) = 3*(w-1)!
For w > 3 : 3* (w-1)! <= w*(w-1)! = w! and it is clear that only one possibility (equality) with w= 3

[pavelkotsev]

Very nice start of the year! :) Thank you SyberMath!

[BorisRaifler]

We can generalize this for N > 2 variables: X_1! + X_2! + ... + X_N! = Y! => N*(N-1)! = N! => all X_K = (N-1)! and Y=N!
For N=2 : Y=2! and each X_K = {0, 1}

[udic]

7:55 there is no need to check w=1.
Since 1=<u=<w-1 => 2=<w

[titassamanta]

We observe that x=y=z=2 is a solution. Now we argue that no more solutions exist. For integral values of x, y and z less than 2, there are no solutions. Now let us consider 2<x<=y<=z. Then we can say z!<x!+y!+z!<3z!. Thus z!<w!<(z+1)! whence z<w<z+1 implying that thre exists an integer between two consecutive integers, which is certainly false. Thus our assumption that there are solutions other than x=y=z=2 was wrong. So, we have established that the only solution is x=y=z=2.

[jovisha]

I have a different method.
x!+y!+z!=w! and let x<y<z could be also equal and because of symmetry could be any order but x, y, z<w
then exist p, q, s>=1 and p.q.s from Z+ where y!=p x! and z!=pq x! and w!=pqs x! then
x!+p x!+pq x!=pqs x! divide by x! gives 1+p+pq=pqs then 1+p+pq-pqs=0 then 1+p=pqs-pq then 1+p=pq(s-1)
then s-1=(1+p)/pq or s=1+1/pq + 1/q, as s from Z+ this is possible when p=1 and q=2 from that s=2, or p=1 and q=1 gives s=3. The first pair gives x!+x!+2x!=w!, 4x!=w! then x=3 and w=4 but as p=1 then y!=x!=6 and q=2 then z!=2x!=12 and that is not possible so we reject pair one.
If p=q=1 and s=3 then
w!=pqs x! then w!=3x! and as x!<w! the only solution is x=2, from p=1 and q=1 and y!=p x! and z!=pq x!
we have y!=x! and z!=x! and x=y=z=2 and w!=pqs x!= 3 x! then w!=6 and w=3

[cube]

Your videos have the power to distract me to study. 😂

[ardaehi]

I feel like i'm just getting better at math just by watching this channel

[-basicmaths]

x=y=z=2, w=3
w is greater than x, y, z and x, y, z are factor of w! Multiplied by 3 both sides of the equation and solving the equation We get the values of
x, y, z, w

[sonaruo]

2!+2!+2!=3!
an obvious solution looking the problem

[JohnRandomness]

w = 4 can't work: 4! = 4*3! > 3*3! ≥ three factorials of at most three added together. (All factorials are at least 1 so none of x, y, and z can be 4.)
w = 1 or 2 can't work: three factorials add to at least three.
Therefore w = 3 is the only solutions: 3! = 3*2!. x, y, and z are two.

[Youtuber]

We suppose x<y<z and x>2 then no solution

[davideiannotta]

You sound exactly like Mr. Garrison from South Park

[matthewnoriega]

This is easy, just divide out the exclamation mark

[pranavamali]

That was really insane i couldn't solve it

[devansh]

Hey Brother would you like to cover some problems from JEE Advanced any year some questions from 2016 and 2021(Calculus) were very good.