I understand number of zeros means number of zeros at the end of #100!# i.e. trailing zeros.
If you dot know, #100! =100xx99xx98xx… xx2xx1#
How are the trailing zeros are formed. A trailing zero will be formed when a multiple of #5# is multiplied with a multiple of #2#. How many do we have in this long product?
First we should count the #5#’s - #5,10,15,20,25# and so on i.e. a total of #20#. However #25,50,75# and #100# have two #5#’s so for each of them, you count them twice, which makes for total #24#.
Now to count the number of #2#’s - #2,4,6,8,10# and so on i.e. a total of #50# multiples of #2#’s, #25# multiples of #4#’s (giving us #25# more #2#'s), #12# multiples of #8#’s (giving us #12# more #2#'s) and so on… i.e. far more than #24#
Now as each pair of #2# and #5# will give a trailing zero, but we have only #24# #5#’s and far more #2#'s,
we can only make #24# such pairs and
hence, the number of zeros in #100!# will be #24#.
