There is only one prime
p such that
p! is
p digits long.
Find it and prove its uniqueness.
If it's true, then the answer is 23.
We can use x=Floor[10^n/n!] as a rough proxy for candidates. x rises to 2755 at n=9, n=10, then falls again.
If N is 25 or more, x=0. 22!, 23! and 24! return x in single figures, and all of those have their digit length equal to their value; but only 23 is prime.
Edited on April 28, 2016, 9:20 am
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Posted by broll
on 2016-04-28 08:49:47 |
Possible approach
