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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 nn = n log n < n2

So n! always has fewer than n2 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.

Re: Arithmetic puzzle

Posted: Thu Aug 31, 2017 2:02 am UTC
by Schrödinger's Wolves
houlahop wrote: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?

I would guess that interpretation is lim(n->infinity) (n^2)/(d(n!)*pi(n))=1.
What is the function d in this case?

Re: Arithmetic puzzle

Posted: Sat Sep 02, 2017 10:06 am UTC
by Xias
I believe d(n) is just shorthand for "the number of digits in n."

I wonder about the fact that n^2 and pi(n) are not dependent on the base, but d(n) is.