We Need To Go Deeper - 5

( ( ( ( ( x ! ) ! ) ! ) ! ) ! Infinitely many factorials = x \large \underbrace{(\ldots((((x!)!)!)! \ldots)!}_{\text{Infinitely many factorials}} =x

Find the sum of all integer solution(s) that satisfy the equation above.

Notation :

! ! denotes the factorial notation. For example, 10 ! = 1 × 2 × 3 × × 10 10! = 1\times2\times3\times\cdots\times10 .

1 3 2 4

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Colin Carmody
Mar 17, 2016

1 and 2 work because 1!=1 and 2!=2. So 1+2=3.

Isn't this problem a bit overrated?

Edit: It is not overrated now

Arulx Z - 5 years, 2 months ago

Log in to reply

Hm yeah I guess so . As The only solutions to x!=x is clear by view only.

Aditya Narayan Sharma - 5 years, 2 months ago

Now it isn't :P

Mehul Arora - 5 years, 2 months ago

I think -1 and -2 work too.

donkey kong - 5 years, 2 months ago

Log in to reply

Factorials are not defined for 1 , 2 -1, -2 Infact they are not defined for any negative integer.

Mehul Arora - 5 years, 2 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...