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Count of numbers up to N having at least one prime factor common with N | Function to calculate the value of Euler 's totient function.
Initialize result with N.
Find all prime factors of N and subtract their multiples.
Check if p is a prime factor.
If found to be true , then update N and result.
If N has a prime factor greater than sqrt ( N ) , then there can be at - most one such prime factor.
Function to count all the numbers in the range [ 1 , N ] having common factor with N other than 1.
Stores the resultant count.
Print the count.
Driver Code"
Solution in Python:
assistant
#Here is code to count of numbers up to N having at least one prime factor common with N and the other steps required.
 def phi ( N ) :
	result = N
	for p in range ( 2 , int ( pow ( N , 1 / 2 ) ) + 1 ) :
		if ( N % p == 0 ) :
			while ( N % p == 0 ) :
				N = N / p
			result -= result // p
	if ( N > 1 ) :
		result -= result // N
	return result
def countNumbers ( N ) :
	count = N - phi ( N )
	print ( count )
N = 5
countNumbers ( N )