Promo: Number of Trailing Zeros in 100! Video Lecture | Sets and Functions - JEE

what is the number of trailing zeros in
100 factorial pause this video and try
to figure it out
let's see how we can calculate the
number of trailing zeros in the
factorial of a number the answer to this
problem is 24 that is there are 24 zeros
trailing in 100 factorial
a trailing zero is formed when we
multiply a multiple of five with a
multiple of two so each pair of two and
five will cause a trailing zero now all
we have to do is count the number of
pairs of five sent to s in the
multiplication this will give us the
total number of trailing zeros
let's count the number of fives first
five occurs in five 10 15 20 25 and so
on 100 this makes a total of twenty
fives the numbers 25 50 75 and 100
consists of two fives each so we have to
count them twice
this makes the grand total 24 we can
write the formula to calculate the total
number of five says 100 divided by five
plus 100 divided by v square plus 100
divided by five cube and so on the
brackets implies greatest integer
function that is we consider the integer
part of each term in this sum so this
equals 20 plus 4 plus 0 which equals 24
let's count the number of two's to
occurs in two four six eight ten and so
on
this makes a total of fifty twos the
multiples of four consists of two twos
so we have to count them again they are
25 in number the multiples of eight
consists of three twos so we have to
count them once more they are twelve in
number the number of twos equals 100
divided by two plus 100 divided by 2
square plus 100 divided by 2 cube and so
on the brackets implies greatest integer
function this equals 50 plus 25 plus 12
plus 6 plus 3 plus 1 plus 0 which equals
97
so the number of five sequels 24 and the
number of two's equals 97 this implies
there are only 24 pairs of 2s and 5s
since each pair cause one trailing 0 the
total number of trailing zeros in 100
factorial equals 24 in fact to calculate
the number of trailing zeros in a
factorial we can only count the number
of 5s because it equals the number of
pairs of 5 sent to s in the
multiplication