Zeros are obtained if there is any zero at the end of any multiplicand and if 5 or multiple of 5 are multiplied by any even number. i.e. (5)n(2)m has n zeros if $n Now we obtain the index of 5 as follows: Index =[
1000
5
]+[
1000
52
]+[
1000
53
]+[
1000
54
] =200+40+8+1=249 Certainly, n will be less than m. ∴ Number of zeros =249