# Is there any way to find the number of zeroes at the end of $n!$ for $n\ge5$?

Generalised Question : Is there any way to find the number of zeroes at the end of $$n!$$ for $$n\ge5$$ ?$$\\$$
I recently came across a question in which it was asked to find the number of zeroes at the end of $$1000!$$ , I tried doing that and I noticed that every $$5k!$$ where $$k \in \mathbb N$$ had one greater $$0$$ than it's predecessor i.e. ($$4! = 24$$, $$5! = 120$$) ; ($$9! = 362880$$, $$10! = 362800$$), hence by induction I came to prove that $$1000!$$ should have $$200$$ zeroes but the answer was $$249$$. $$\\$$
Where did I go wrong ? $$\\$$
Any help would be appreciated. Thanks !!

Yes, there is. It is called de-Polignac's formula, which I suggest you look up. It is given by $$p=\sum_{i=0}^\infty \lfloor \frac {n}{5^i} \rfloor$$. Note that after $$n<5^m$$, for some $$m$$, all subsequent terms would be zero and no further calculation would be required.

You can also find out the power of $$5$$ in the prime factorization of $$n!$$, which would be the number of $$0s$$ required.

• Welcome to Mathematics Stack Exchange. It is also known as Legendre's formula May 11, 2021 at 22:02

You have only counted the $$5$$ of numbers like $$25,125, 250, 1000$$ etc. once each. But these have more than one $$5$$ in their factorisations. The correct method is as follows:

Count one $$5$$ from each of these numbers once: $$5,10,15,20,25,30,.....1000$$ We get $$\left\lfloor\frac{1000}{5}\right\rfloor=200$$ of these $$5$$'s.

Now count another $$5$$ from each multiple of $$25$$ (note that we have already counted one $$5$$ from each of these!) $$25,50,75,100,.....1000$$ We get $$\left\lfloor\frac{1000}{25}\right\rfloor=40$$ of these $$5$$'s.

Now count another $$5$$ from each multiple of $$125$$ (we have already counted two $$5$$s from each) $$125,250,375,.....1000$$ We get $$\left\lfloor\frac{1000}{125}\right\rfloor=8$$ of these $$5$$'s.

Finally, count another $$5$$ from each multiple of $$625$$ (we have already counted three $$5$$s from them) $$625$$ We get $$\left\lfloor\frac{1000}{625}\right\rfloor=1$$ of these $$5$$'s.

Adding all of these up, the maximum power of $$5$$ in $$1000!$$ is $$200+40+8+1=249$$. Obviously, the maximum power of $$2$$ is much more, so we won't bother about it. The maximum power of $$10$$ is therefore equal to the maximum power of $$5$$, that is $$249$$.

So $$1000!$$ has $$249$$ zeroes at the end.

There is a technique that works due to Legendre's formula .

• Convert $$n$$ to base $$5$$.
• Subtract the sum from $$n$$
• Divide by $$4$$
Converting from base $$10$$ to base $$5$$ is one of the easier conversions, especially starting with the least significant digit. Since $$1000_{10}$$ = $$13000_{5}$$, $$z(1000!) = \frac{1000 - (1 + 3 + 0 + 0 + 0)}{4} = 249$$