a) Let with n>1, and let p be a prime number. If p n!, prove that the exponent of p in the prime factorization of n! is [n/p] + [n/p^2] + [n/p^3] +......(Note that this sum is finite, since [n/p^m]=0 if p^m>n

my answer to part a)

this can also be said as the highest power of a prime number p contained in n !.

given p divides n !.

so, let k( n ! ) be the highest power of p contained in n ! .

then the multiples of p which are the divisors of n ! are p , 2p , 3p , -----,p

further , is the quotient when n is divided by p.

so, k ( n!) = +k( ! ) -------------(1)

replacing n by n / p we get

k( ! ) = +k( ! ) -----------(2)

again replacing n by n / p2 we get k( ! ) = + k( ! )-------(3)

continuing in the same by by replacement upto where becomes zero, and using (2) in (1) , (3) in (2) , ---------,(n) equation in (n-1) the equation, we get

k( n !) =

this process terminates at a finite stage while becomes for sufficiently large k .

thus, the highest power of p in n! is .

b) Use (part a) above to find the prime factorization of 20!

c) Find the number of zeros with each the decimal representaion of 100! terminates