It means 100! is divisible by 3

^{48}

How do you find it?

Let p be a prime number and n be a positive integer. Then find the last integer in the sequence 1,2,…,n which is divisible by p.

Express this integer as [n/p]p.

[n/p] denotes the greatest integer less than or equal to n/p

In case of 3 (p) and 100 (n); [n/p] is 33 and n/p is 33 and 1/3.

Let E

_{p}(n) denote the exponent of the prime p in the positive integer n. Then,

E

_{p}(n!) = E

_{p}(1.2.3…(n-1).n)

This will be equal to E

_{p}(p.2p.3p…[n/p]p)

= [n/p]+ E

_{p}(1.2.3...[n/p])

This process continues and we get the answer

E

_{p}(n!) = [n/p] + [n/p²]+…+[n/p

^{s}]

Where s is the largest positive integer such that p

^{s}≤n≤p

^{s+1}

Hence applying the formula to find exponent of prime 3 in 100!

E

_{3}(100!) = [100/3] + [100/3²] + [100/3³] + [100/3

^{4}]

= 33+11+3+1 = 48

Note: remember the meaning of notation [100/3] or [n/p]

Chapter: Permutations