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Sum of product of all integers upto N with their count of divisors | Function to find the sum of the product of all the integers and their positive divisors up to N.
Iterate for every number between 1 and N.
Find the first multiple of i between 1 and N.
Find the last multiple of i between 1 and N.
Find the total count of multiple of in [ 1 , N ].
Compute the contribution of i using the formula.
Add the contribution of i to the answer.
Return the result.
Given N.
Function call"
Solution in Python:
assistant
#Here is code to sum of product of all integers upto N with their count of divisors and the other steps required.
 def sumOfFactors ( N ) :
	ans = 0
	for i in range ( 1 , N + 1 ) :
		first = i
		last = ( N // i ) * i
		factors = ( last - first ) // i + 1
		totalContribution = ( ( ( factors * ( factors + 1 ) ) // 2 ) * i )
		ans += totalContribution
	return ans
N = 3
print ( sumOfFactors ( N ) )