Arithmetic puzzle

A forum for good logic/math puzzles.

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houlahop
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Joined: Thu Jul 13, 2017 9:57 am UTC

Arithmetic puzzle

Postby houlahop » Mon Jul 24, 2017 11:06 am UTC

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)

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jaap
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Re: Arithmetic puzzle

Postby jaap » Mon Jul 24, 2017 11:57 am UTC

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!).

houlahop
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Joined: Thu Jul 13, 2017 9:57 am UTC

Re: Arithmetic puzzle

Postby houlahop » Mon Jul 24, 2017 7:55 pm UTC

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.
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SecondTalon
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Re: Arithmetic puzzle

Postby SecondTalon » Thu Jul 27, 2017 1:34 am UTC

Alright then. Bye.
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