Question

Given a number N, let P be the set of all the prime divisors of N. Now, let S be another set of all the divisors of N (including 1 and N). Let's define the term Prime Score of a prime divisor as the number of elements in S which have x as a prime divisor. Output the product of prime scores of all the elements in P modulo 1000003.

Answer 1

Answer:

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Explanation:

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Answer 2

Answer:

def primeFactors(n):  #calculating prime factor of number

   p=[]

   for i in range(2, n+ 1):

       if(n% i == 0):

           isprime = 1

       for j in range(2, (i //2 + 1)):

           if(i % j == 0):

               isprime = 0

               break

       if (isprime == 1):

           p.append(i)

   return p

           

def factors(x):     #calculating  factors of number

   s=[]

   for i in range(1, x + 1):

       if x % i == 0:

           s.append(i)

   return s

n = int(input())

a=factors(n)

b=primeFactors(n)

r=[]

for i in range(0,len(b)):

   c=0

   for j in range(0,len(a)):

       if(a[j]%b[i] == 0):    

           c=c+1

   r.append(c)

m=1

for k in range (0,len(r)):

   m=m*r[k]

print(m%1000003)

Explanation:

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