The number of digits in n! (Part 1)

When I was in school, perhaps my favorite pet project was trying to find a formula for the number of digits in n!. For starters:

  • 0! = 1: 1 digit
  • 1! = 1: 1 digit
  • 2! = 2: 1 digit
  • 3! = 6: 1 digit
  • 4! = 24: 2 digits
  • 5! = 120: 3 digits
  • 6! =720: 3 digits
  • 7! = 5040: 4 digits
  • 8! = 40,320: 5 digits

I owned what was then a top-of-the-line scientific calculator (with approximately the same computational capability as a modern TI-30), and I distinctly remember making a graph like the following on graph paper. The above calculations contribute the points (0,1), (1,1), (2,1), (3,1), (4,2), (5,3), (6,3), (7,4), and (8,5).

factorialdigitsI had to stop (or, more accurately, I thought I had to stop) at 69! because my calculator couldn’t handle numbers larger than 10^{100}.

I stared at this graph for weeks, if not months, trying to figure out an equation that would fit these points. And I never could figure it out.

And, to this day, I’m somewhat annoyed at my adolescent self that I wasn’t able to figure out this puzzle for myself… since I had all the tools in my possession needed to solve the puzzle, though I didn’t know how to use the tools.

In this series of posts, I’ll answer this question with the clever application of some concepts from calculus and precalculus.

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