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### Arithmetic puzzle

Posted: **Mon Jul 24, 2017 11:06 am UTC**

by **houlahop**

22!=1124000727777607680000

The number 22! is a 22 digit number

11240

00727

77760

76800

00

Find a number n such as n! is n^2 digit number

More generally find n such as n! is n^k digit number (k>2)

### Re: Arithmetic puzzle

Posted: **Mon Jul 24, 2017 11:57 am UTC**

by **jaap**

It is not possible.

For n>1 we have:

log n! < log n^{n} = n log n < n^{2}

So n! always has fewer than n^{2} digits (except for 1! and 0!).

### Re: Arithmetic puzzle

Posted: **Mon Jul 24, 2017 7:55 pm UTC**

by **houlahop**

Thank you

(n^2)/d(n!) is equal to pi(n) when n goes to infinity (where pi(n) is the counting function of primes)

Is there any interpretation of this "equality"?

My last post because I wanted to point out to this.

Good luck and good bye!

To the moderator :

No one asked me about clarification. Sir Gabriel answered without asking.

Conclusion : Either you did not read the post, either you are lying.

Where did you see any contempt from myself?

### Re: Arithmetic puzzle

Posted: **Thu Jul 27, 2017 1:34 am UTC**

by **SecondTalon**

Alright then. Bye.