1*2*3*4*5*6.....upto 1000 Find the number of zeroes at the end?

1 Answer
Apr 24, 2016

#249#

Explanation:

This product is commonly known as the factorial of #1000#, written #1000!#

The number of zeros is determined by how many times #10=2xx5# occurs in the prime factorisation of #1000!#.

There are plenty of factors of #2# in it, so the number of zeros is limited by the number of factors of #5# in it.

These numbers have at least one factor #5#:

#5, 10, 15, 20, 25,..., 1000# which is #1000/5 = 200# numbers.

These numbers have at least two factors #5#:

#25, 50, 75, 100,..., 1000# which is #1000/25 = 40# numbers.

These numbers have at least three factors #5#:

#125, 250, 375, 500,..., 1000# which is #1000/125 = 8# numbers

This number has four factors #5#:

#625# which is #1# number.

So the total number of factors #5# in #1000!# is:

#200+40+8+1 = 249#

Hence there are #249# zeros at the end of #1000!#