# Sum of product of all integers upto N with their count of divisors

Given a positive integer N, the task is to find the sum of the product of all the integers in the range [1, N] with their count of divisors

Examples:

Input: N = 3
Output: 11
Explanation:
Number of positive divisors of 1 is 1( i.e. 1 itself). Therefore, f(1) = 1.
Number of positive divisors of 2 is 2( i.e. 1, 2). Therefore, f(2) = 2.
Number of positive divisors of 3 is 2( i.e. 1, 3). Therefore, f(3) = 2.
So the answer is 1*f(1) + 2*f(2) + 3*f(3) = 1*1 + 2*2 + 3*2 = 11.

Input: N = 4
Output: 23
Explanation:
Here f(1) = 1, f(2) = 2, f(3) = 2 and f(4) = 3. So, the answer is 1*1 + 2*2 + 3*2 + 4*3 = 23.

Naive Approach: The naive approach is to traverse from 1 to N and find the sum of all the numbers with their count of divisors.

Time Complexity: O(N*sqrt(N))
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, the idea is to consider the total contribution each number makes to the answer. Below are the steps:

1. For any number X in the range [1, N], X contributes K to the sum for each K from 1 to N such that K is a multiple of X.
2. Observe that the list of these K is of form i, 2i, 3i, …, Fi where F is the number of multiples of i between 1 and N.
3. Hence, to find the sum of these numbers generated above is given by

i*(1+2+3 + … F) = i*(F*(F+1))/2.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ` `using` `namespace` `std;`   `// Function to find the sum of the` `// product of all the integers and` `// their positive divisors up to N` `long` `long` `sumOfFactors(``int` `N)` `{` `    ``long` `long` `ans = 0;`   `    ``// Iterate for every number` `    ``// between 1 and N` `    ``for` `(``int` `i = 1; i <= N; i++) {`   `        ``// Find the first multiple` `        ``// of i between 1 and N` `        ``long` `long` `frst = i;`   `        ``// Find the last multiple` `        ``// of i between 1 and N` `        ``long` `long` `last = (N / i) * i;`   `        ``// Find the total count of` `        ``// multiple of in [1, N]` `        ``long` `long` `factors` `            ``= (last - frst) / i + 1;`   `        ``// Compute the contribution of i` `        ``// using the formula` `        ``long` `long` `totalContribution` `            ``= (((factors) * (factors + 1)) / 2) * i;`   `        ``// Add the contribution` `        ``// of i to the answer` `        ``ans += totalContribution;` `    ``}`   `    ``// Return the result` `    ``return` `ans;` `}`   `// Driver code` `int` `main()` `{` `    ``// Given N` `    ``int` `N = 3;`   `    ``// function call` `    ``cout << sumOfFactors(N);` `}`

## Java

 `// Java program for the above approach` `class` `GFG{` `  `  `// Function to find the sum of the` `// product of all the integers and` `// their positive divisors up to N` `static` `int` `sumOfFactors(``int` `N)` `{` `    ``int` `ans = ``0``;` ` `  `    ``// Iterate for every number` `    ``// between 1 and N` `    ``for` `(``int` `i = ``1``; i <= N; i++) ` `    ``{` ` `  `        ``// Find the first multiple` `        ``// of i between 1 and N` `        ``int` `frst = i;` ` `  `        ``// Find the last multiple` `        ``// of i between 1 and N` `        ``int` `last = (N / i) * i;` ` `  `        ``// Find the total count of` `        ``// multiple of in [1, N]` `        ``int` `factors = (last - frst) / i + ``1``;` ` `  `        ``// Compute the contribution of i` `        ``// using the formula` `        ``int` `totalContribution = (((factors) * ` `                                  ``(factors + ``1``)) / ``2``) * i;` ` `  `        ``// Add the contribution` `        ``// of i to the answer` `        ``ans += totalContribution;` `    ``}` ` `  `    ``// Return the result` `    ``return` `ans;` `}` ` `  `// Driver code` `public` `static` `void` `main(String[] args)` `{` `    ``// Given N` `    ``int` `N = ``3``;` ` `  `    ``// function call` `    ``System.out.println(sumOfFactors(N));` `}` `}`   `// This code is contributed by Ritik Bansal`

## Python3

 `# Python3 implementation of` `# the above approach`   `# Function to find the sum of the` `# product of all the integers and` `# their positive divisors up to N` `def` `sumOfFactors(N):`   `    ``ans ``=` `0`   `    ``# Iterate for every number` `    ``# between 1 and N` `    ``for` `i ``in` `range``(``1``, N ``+` `1``):`   `        ``# Find the first multiple` `        ``# of i between 1 and N` `        ``frst ``=` `i`   `        ``# Find the last multiple` `        ``# of i between 1 and N` `        ``last ``=` `(N ``/``/` `i) ``*` `i`   `        ``# Find the total count of` `        ``# multiple of in [1, N]` `        ``factors ``=` `(last ``-` `frst) ``/``/` `i ``+` `1`   `        ``# Compute the contribution of i` `        ``# using the formula` `        ``totalContribution ``=` `(((factors ``*` `                              ``(factors ``+` `1``)) ``/``/` `2``) ``*` `i)`   `        ``# Add the contribution` `        ``# of i to the answer` `        ``ans ``+``=` `totalContribution`   `    ``# Return the result` `    ``return` `ans`   `# Driver Code`   `# Given N` `N ``=` `3`   `# Function call ` `print``(sumOfFactors(N))`   `# This code is contributed by Shivam Singh`

## C#

 `// C# program for the above approach` `using` `System;` `class` `GFG{` `  `  `// Function to find the sum of the` `// product of all the integers and` `// their positive divisors up to N` `static` `int` `sumOfFactors(``int` `N)` `{` `    ``int` `ans = 0;` ` `  `    ``// Iterate for every number` `    ``// between 1 and N` `    ``for` `(``int` `i = 1; i <= N; i++) ` `    ``{` ` `  `        ``// Find the first multiple` `        ``// of i between 1 and N` `        ``int` `frst = i;` ` `  `        ``// Find the last multiple` `        ``// of i between 1 and N` `        ``int` `last = (N / i) * i;` ` `  `        ``// Find the total count of` `        ``// multiple of in [1, N]` `        ``int` `factors = (last - frst) / i + 1;` ` `  `        ``// Compute the contribution of i` `        ``// using the formula` `        ``int` `totalContribution = (((factors) * ` `                                  ``(factors + 1)) / 2) * i;` ` `  `        ``// Add the contribution` `        ``// of i to the answer` `        ``ans += totalContribution;` `    ``}` ` `  `    ``// Return the result` `    ``return` `ans;` `}` ` `  `// Driver code` `public` `static` `void` `Main(String[] args)` `{` `    ``// Given N` `    ``int` `N = 3;` ` `  `    ``// function call` `    ``Console.WriteLine(sumOfFactors(N));` `}` `}`   `// This code is contributed by gauravrajput1`

Output:

```11

```

Time Complexity: O(N)
Auxiliary Space: O(1)

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