Factorial of a factorial of a factorial

Number Theory Level 3

How many integer solutions are there for

$((x!)!)! = x?$

0 1 2 3 4

1 solution

Guilherme Niedu
Dec 16, 2016

Only $1$ and $2$ comply with this condition.

$0! = 1$ , so $((0!)!)! = 1 \neq 0$ , and for $x \geq 3$ , $((x!)!)!$ increases rapidly, being far greater than $x$

I think, the answer of this question is wrong. The correct answer would be $1$ .. Now, The given question is to find out integer solution for $x$ where $((x!)!)! =x$ .. If we look over the problem carefully we will notice that if $x=1$ then, $((1!)!)! =1=x$ .. But if $x=0$ then, $((0!)!)! =1≠x$ So, only one integer solution is possible.. But the official answer given there is 2, which can't be possible..

Prokash Shakkhar - 4 years, 6 months ago

What about $((2!)!)!=2$ ?

Nihar Mahajan - 4 years, 6 months ago

Oh! Thank you.. I haven't noticed it.. Just a slip of eye!

Prokash Shakkhar - 4 years, 6 months ago

Factorial of a factorial of a factorial

1 solution

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