Question

Find the number of zeros at the end of $1 \times 2 \times 3 \times 4 \times 5 \times 6 \times ..........$ upto $1000$ .

Answer 1

Hint: For this type of problem we use the factorial of a number. We write the terms that are multiplied while opening the factorial. Since each zero at the end is made up to a $5$ multiplied by $2$ , we count the number of terms that are divisible by $5$ . We use this method to get the number of zeros at the end. We write the problem as factorial $n! = n \times (n - 1) \times ...... \times 2 \times 1$ .

Complete step by step answer:
The given product is commonly known as the factorial of $1000$ , written $1000!$
The number of zeroes is determined by how many times $10 = 2 \times 5$ occurs in the prime factorization of $1000!$ .
There are plenty of factors of $2$ in it, so the number of zeroes is limited by the number of factors of $5$ in it.
These numbers have at least one factor $5$ .
Therefore $5,10,15,20,25,....,1000$ which is $\dfrac{{1000}}{5} = 200$ numbers.
These numbers have at least two factors: $5$ .
Therefore $25,50,75,100,....,1000$ which is $\dfrac{{1000}}{{25}} = 40$ numbers.
These numbers have at least three factors: $5$ .
Therefore $125,250,....,1000$ which is $\dfrac{{1000}}{{125}} = 8$ numbers
This number has four factors: $5$ .
Therefore $625$ which is a $1$ number.
So the total number of factors $5$ in $1000!$ is
Sum $ = 200 + 40 + 8 + 1$
$ = 249$ numbers
Hence there are $249$ zeros at the end of $1000!$ .i.e., $1 \times 2 \times 3 \times 4 \times 5 \times 6 \times .......... \times 1000$

Note:
Students might try to solve for the value of $1000!$ by multiplying all the values of the factorial given by $1000! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times .......... \times 1000$ . But since there are $1000$ numbers to be multiplied with each other, this will be a very long and complex calculation. Students are advised not to proceed in this manner. Also many students only write $20$ zeroes as they do not see the terms having ${5^2}$ in them.