How much zeros has the number $1000!$ at the end?

Question

I know that it depends of the factors of five and two.

But the number is too long to figure how much factos of five and two there are.

Any hints?

Answer 1

There are always twos a-plenty. The exponent of prime $p$ occuring in $n!$ is well-known to be $$ \lfloor n/p\rfloor +\lfloor n/p^2\rfloor + \lfloor n/p^3\rfloor +\ldots$$ hence for $n=1000$ and $p=5$ we find $$ \lfloor 1000/5\rfloor +\lfloor 1000/25\rfloor + \lfloor 1000/125\rfloor +\lfloor 1000/625\rfloor + \ldots= 200+40+8+1+0+\ldots=249$$ (Just to check, for $p=2$ we get $$500+250+125+62+31+15+7+3+1\gg249 $$ so really more than enough)

Answer 2

yes it depends on $2$ and $5$. Note that there are plenty of even numbers. Also note that $25\times 4 = 100$ which gives two zeros. Also note that there $125\times 8 = 1000$ gives three zeroes and $5^4 \times 2^4 = 10^4$. Each power of $5$ add one extra zero. So, count the multiple of $5$ and it's power less than $1000$.

Answer 3

the number of factor 2's between 1-1000 is more than 5's.so u must count the number of 5's that exist between 1-1000.can u continue?

Answer 4

If a number ends with $n$ zeros than it is divisible by $10^n$, that is $2^n5^n$.
A factorial clearly has more $2$s than $5$s in its factorization so you only need to count how many $5$s are there in the factorization of $1000!$
There are $\lfloor\frac{1000}{5}\rfloor=200$ numbers below $1000$ that can be divided by $5$, but you also have to consider that there are $\lfloor\frac{1000}{5^2}\rfloor=40$ numbers divisible by $25$, $\lfloor\frac{1000}{5^3}\rfloor=8$ and $\lfloor\frac{1000}{5^4}\rfloor=1$ number divisible by $625$, for a total of $249$ zeros.

Answer 5

Two questions:

Can you work out which of $2$ and $5$ will be the critical one for your question? A little thought should make this obvious.

You say that "the number is too long" - but actually the number of factors can be computed quite quickly even by hand. Can you see how many factors $521$ there might be, or factors $257$ in the product? How about factors $31$? Can you apply this to your choice of $2$ or $5$?