# 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 the 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 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` `import` `java.io.*;` `public` `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`

## Python3

 `# Python3 program for the` `# above approach`   `# Function to find the sum ` `# of largest divisors of ` `# numbers in range 1 to N` `# not divisible by prime number P` `def` `func(N, P):` `  `  `    ``# Total sum upto N` `    ``sumUptoN ``=` `(N ``*` `(N ``+` `1``) ``/` `2``);` `    ``sumOfMultiplesOfP ``=` `0``;`   `    ``# 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` `    ``elif` `((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` `if` `__name__ ``=``=` `'__main__'``:` `  `  `    ``# Given N and P` `    ``N ``=` `10``;` `    ``P ``=` `5``;`   `    ``# Function call` `    ``print``(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`

## Javascript

 ``

Output:

`43`

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

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