# Sum of largest divisor of numbers upto N not divisible by given prime number P

Given a number N and a prime number P, the task is to find the sum of largest divisors of each number in the range [1, N], which is not divisible by P.
Examples:

Input: N = 8, P = 2
Output: 22
Explanation: Numbers are in the range [1, 8].
Number                           Largest Divisor not divisible by P = 2
1                                                            1
2                                                            1
3                                                            3
4                                                            1
5                                                            5
6                                                            3
7                                                            7
8                                                            1
Sum of all divisors with given constraint = 22.

Input: N = 10, P = 5
Output: 43
Explanation: Numbers are in the range [1, 8].
Number                           Largest Divisor not divisible by P = 5
1                                                            1
2                                                            2
3                                                            3
4                                                            4
5                                                            1
6                                                            6
7                                                            7
8                                                            8
9                                                            9
10                                                          2
Sum of all divisors with given constraint = 43

Naive Approach: The naive idea is to find the divisors for each number in the range [1, N] and find the largest divisors which is not divisible by P and those numbers. Print the sum of all the those largest divisors.
Time Complexity: O(N3/2
Auxiliary Space: O(1)

Efficient Approach: The idea is to observe that the largest divisor of a number N not divisible by P would be N itself if N is not divisible by P. Else the required divisor will be the same as that of N/P. Below are the steps:

1. If N is not divisible by P, then the largest divisor will be N, add this to the final sum..
2. If N is divisible by P, the required divisor will be the same as that of N/P.
3. So, find the sum of all numbers which are not divisible by P and add to them divisors of those which are divisible by P separately.
4. The total sum would be N*(N + 1)/2. Subtract the sum of those which are divisible by P and add their corresponding value by recursively calling the function to find the sum for N/P.

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 largest ` `// divisors of numbers in range 1 to N ` `// not divisible by prime number P ` `int` `func(``int` `N, ``int` `P) ` `{ ` `    ``// Total sum upto N ` `    ``int` `sumUptoN = (N * (N + 1) / 2); ` `    ``int` `sumOfMultiplesOfP; ` ` `  `    ``// If no multiple of P exist up to N ` `    ``if` `(N < P) { ` `        ``return` `sumUptoN; ` `    ``} ` ` `  `    ``// If only P itself is in the range ` `    ``// from 1 to N ` `    ``else` `if` `((N / P) == 1) { ` `        ``return` `sumUptoN - P + 1; ` `    ``} ` ` `  `    ``// Sum of those that are divisible by P ` `    ``sumOfMultiplesOfP ` `        ``= ((N / P) * (2 * P + (N / P - 1) * P)) / 2; ` ` `  `    ``// Recursively function call to ` `    ``// find the sum for N/P ` `    ``return` `(sumUptoN ` `            ``+ func(N / P, P) ` `            ``- sumOfMultiplesOfP); ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``// Given N and P ` `    ``int` `N = 10, P = 5; ` ` `  `    ``// Function Call ` `    ``cout << func(N, P) << ``"\n"``; ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program for the above approach ` `class` `GFG{ ` ` `  `// Function to find the sum of largest ` `// divisors of numbers in range 1 to N ` `// not divisible by prime number P ` `static` `int` `func(``int` `N, ``int` `P) ` `{ ` `     `  `    ``// Total sum upto N ` `    ``int` `sumUptoN = (N * (N + ``1``) / ``2``); ` `    ``int` `sumOfMultiplesOfP; ` ` `  `    ``// If no multiple of P exist up to N ` `    ``if` `(N < P)  ` `    ``{ ` `        ``return` `sumUptoN; ` `    ``} ` ` `  `    ``// If only P itself is in the range ` `    ``// from 1 to N ` `    ``else` `if` `((N / P) == ``1``) ` `    ``{ ` `        ``return` `sumUptoN - P + ``1``; ` `    ``} ` ` `  `    ``// Sum of those that are divisible by P ` `    ``sumOfMultiplesOfP = ((N / P) * (``2` `* P +  ` `                         ``(N / P - ``1``) * P)) / ``2``; ` ` `  `    ``// Recursively function call to ` `    ``// find the sum for N/P ` `    ``return` `(sumUptoN + func(N / P, P) -  ` `            ``sumOfMultiplesOfP); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `     `  `    ``// Given N and P ` `    ``int` `N = ``10``, P = ``5``; ` ` `  `    ``// Function call ` `    ``System.out.println(func(N, P)); ` `} ` `} ` ` `  `// This code is contributed by Rajput-Ji `

## C#

 `// C# program for the above approach ` `using` `System; ` ` `  `class` `GFG{ ` ` `  `// Function to find the sum of largest ` `// divisors of numbers in range 1 to N ` `// not divisible by prime number P ` `static` `int` `func(``int` `N, ``int` `P) ` `{ ` `     `  `    ``// Total sum upto N ` `    ``int` `sumUptoN = (N * (N + 1) / 2); ` `    ``int` `sumOfMultiplesOfP; ` ` `  `    ``// If no multiple of P exist up to N ` `    ``if` `(N < P)  ` `    ``{ ` `        ``return` `sumUptoN; ` `    ``} ` ` `  `    ``// If only P itself is in the range ` `    ``// from 1 to N ` `    ``else` `if` `((N / P) == 1) ` `    ``{ ` `        ``return` `sumUptoN - P + 1; ` `    ``} ` ` `  `    ``// Sum of those that are divisible by P ` `    ``sumOfMultiplesOfP = ((N / P) * (2 * P +  ` `                         ``(N / P - 1) * P)) / 2; ` ` `  `    ``// Recursively function call to ` `    ``// find the sum for N/P ` `    ``return` `(sumUptoN + func(N / P, P) -  ` `            ``sumOfMultiplesOfP); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main(String[] args) ` `{ ` `     `  `    ``// Given N and P ` `    ``int` `N = 10, P = 5; ` ` `  `    ``// Function call ` `    ``Console.WriteLine(func(N, P)); ` `} ` `} ` ` `  `// This code is contributed by Amit Katiyar `

Output:

```43
```

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

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Improved By : Rajput-Ji, amit143katiyar