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Count number of integers less than or equal to N which has exactly 9 divisors | Function to count numbers having exactly 9 divisors.
Sieve array , initially prime [ i ] = i.
use sieve concept to store the first prime factor of every number.
mark all factors of i.
check for all numbers if they can be expressed in form p * q.
p prime factor.
q prime factor.
if both prime factors are different if p * q <= n and q !=.
Check if it can be expressed as p ^ 8.
Driver Code"
Solution in Python:
 What is a code to count number of integers less than or equal to N which has exactly 9 divisors and the other steps required?
assistant
#Here is code to count number of integers less than or equal to N which has exactly 9 divisors and the other steps required.
 def countNumbers ( n ) :
	c = 0
	limit = int ( n ** ( 0.5 ) )
	prime = [ i for i in range ( limit + 1 ) ]
	i = 2
	while i * i <= limit :
		if prime [ i ] == i :
			for j in range ( i * i , limit + 1 , i ) :
				if prime [ j ] == j :
					prime [ j ] = i
		i += 1
	for i in range ( 2 , limit + 1 ) :
		p = prime [ i ]
		q = prime [ i // prime [ i ] ]
		if p * q == i and q != 1 and p != q :
			c += 1
		elif prime [ i ] == i :
			if i ** 8 <= n :
				c += 1
	return c
if __name__ == " _ _ main _ _ " :
	n = 1000
	print ( countNumbers ( n ) )