Number of trailing zeroes in $10^{n}!$

Okay, this is something I've noticed. The number of trailing zeroes in

$1!$ is $0$ .

$10!$ is $2$ .

$100!$ is $24$ .

$1000!$ is $249$ .

$10000!$ is $2499$ .

$100000!$ is $24999$ .

Is there, like, a generalisation or a proof or something for this? And is this result, or observation, useful? Comment any of your thoughts you believe add to this.

#NumberTheory #Factorial

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered 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"

Math	Appears 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

Well, I can give you this formula ${ 10 }^{ n }!\Rightarrow \sum _{ k=1 }^{ \infty }{ \left\lfloor \frac { { 10 }^{ n } }{ 5^{ k } } \right\rfloor }$

And the pattern only continues until $n$ is $5$ , beyond that, there will still be a long chain of $9$ 's but it is not exactly as what you put here. For example, for $n=200$ , number of trailing $0$ 's is $24999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999959$

Julian Poon - 6 years, 4 months ago

Omkar Kulkarni - 6 years, 4 months ago

I can't spot a pattern in how close it is to what you conjectured (it always seems to be quite close though) but I found a reason why the number of trailing zeroes will tend to be very close to 10^n/4.

Clearly, the number of trailing zeroes is simply the exponent of the 5's in the prime factorization of (10^n)!. If we go by from 1 to 10^n in the expansion of (10^n)! (1234...10^n) and write down the exponent of the greatest power of 5 that divides each number, we notice that 4 in every 25 numbers in the series has a 1, 4 in every 125 has a two, and so on. This allows us to make the following approximation for the exponent of the 5's:

4/25 * 10^n +2 * 4/125 * 10^n... = 4 * 10^n * [1/25+2/125+3/625...]

We can evaluate [1/25+2/125+3/625...] pretty easily:

5S= [1/5+1/25...] + [1/25+1/125...]+...

5S=1/4 + 1/20 + 1/100...

S=1/16

Which shows that we'll get about 10^n/4 trailing zeroes. As n becomes larger, the first few terms will become more accurate (percentage wise) because the error is within a certain range. I'll give an example:

   Let's say I'm looking at the 4/25 * k term (where k! is the number you are trying to find the trailing zeroes for). If n is small, say 35, this term is 5.6 (when we wanted 6). If n is large, say 20,035, we get 3205.6 (when we wanted 3206). As you can see, adding 25's makes no difference to the error, but adding 1-24 does. This means that the error, percentage wise, becomes less as n becomes large! The same concept can be applied to the 4/125 * k  term, and the accuracy for each term goes up.

So for a large n, the result of the approximation becomes more accurate. Your observation was mostly true :)

Dylan Pentland - 6 years, 4 months ago

Perhaps we should notice that they are to do with 5. From 10/ 5 = 2, 1000/ 5 = 200, 10000/ 5 = 2000 and 100000/ 5 = 20000, the rough idea is there.

Lu Chee Ket - 6 years, 4 months ago

Math	Appears 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}`	$\boxed{123}$

Number of trailing zeroes in \(10^{n}!\)

Comments