Waste less time on Facebook — follow Brilliant.
×

Finding number of digits in \(n!\)

Hi!

I have encountered a number of questions asking for the number of digits in \(n!\). If you know varied methods of solving the same, please do share here.

Terms and conditions:

  • No use of Computer Science is allowed.

Note by Swapnil Das
2 years, 1 month ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

The first two terms of Stirling's approximation to n! are

\(\ln (n!) = n \ln (n) - n\)

So the number of digits in \(n!\) can be estimated as \(\lfloor n \ln(10) (\ln (n) - 1) \rfloor + 1\)

Thomas Jones - 2 years, 1 month ago

Log in to reply

One of the ways : Stirling's Formula.

Swapnil Das - 2 years, 1 month ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...