# Cool fact about number of digits in n!

• B
• Kyuubi
Kyuubi
This may have already been found by many people but I discovered the pattern on my own out of curiosity with some coding.

There are only 4 natural numbers whose factorial contains the same number of digits as the number itself. That is to say n = digits_in(n!).

The trivial case is obviously just 1. 1! has only one digit. The other three are surprising, because they are consecutive.

They are 22, 23, and 24.

Unfortunately for our pal 25, it has 26 digits, and after 25, the number of digits begins increasing with an average >1 and the digits can't keep up with their mere increment of +1.

Just thought that was cool and wanted to share it.

AndreasC, sophiecentaur, jedishrfu and 1 other person
I wonder how it would work if you played with different number bases like 2, 3, 5, 7, 11, 13... or powers of 2 ie 2,4,8,16...

AndreasC, SammyS and Kyuubi
jedishrfu said:
I wonder how it would work if you played with different number bases like 2, 3, 5, 7, 11, 13... or powers of 2 ie 2,4,8,16
I tried with powers as well, but I did it in the form n^n. I just felt like taking powers of 2 would be a bit arbitrary. I want the entire expression to only depend on n. Anyways looking for numbers that satisfy n = digits_in(n^n), we get 1 (as expected), but we also simply get 8 and 9. 8^8 = 16777216 and 9^9 = 387420489. No other numbers satisfy this rule. Granted, I only tried up to 4,000, but I'm confident that won't happen :)

My ideal goal would be to find some expression (ideally a simple one too) that increases the number of digits with an average of +1 increment such that you have a bunch of scattered answers rather than cluttered up consecutive answers.

Kyuubi said:
My ideal goal would be to find some expression (ideally a simple one too) that increases the number of digits with an average of +1 increment
Do you know what a logarithm is?

Tom.G

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