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.
Answers
Answer:
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Explanation:
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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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