Waste less time on Facebook — follow Brilliant.

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.

Note by Omkar Kulkarni
1 year, 9 months ago

No vote yet
1 vote


Sort by:

Top Newest

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 · 1 year, 9 months ago

Log in to reply

@Julian Poon Oh Omkar Kulkarni · 1 year, 9 months ago

Log in to reply

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...


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 · 1 year, 9 months ago

Log in to reply

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 · 1 year, 9 months ago

Log in to reply


Problem Loading...

Note Loading...

Set Loading...