# Find the sum of the number of divisors

Given three integers A, B, C, the task is to find
ΣAi=1 ΣBj=1ΣCk=1 d(i.j.k), where d(x) is the number of divisors of x. Answer can be very large, So, print answer modulo 109+7.

Examples:

```Input: A = 2, B = 2, c = 2
Output: 20
Explanation: d(1.1.1) = d(1) = 1;
d(1·1·2) = d(2) = 2;
d(1·2·1) = d(2) = 2;
d(1·2·2) = d(4) = 3;
d(2·1·1) = d(2) = 2;
d(2·1·2) = d(4) = 3;
d(2·2·1) = d(4) = 3;
d(2·2·2) = d(8) = 4.

Input: A = 5, B = 6, C = 7
Output: 1520
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

Below is the implementation of the above approach:

## C++

 `#include ` `using` `namespace` `std; ` ` `  `#define N 100005 ` `#define mod 1000000007 ` ` `  `// To store the number of divisors ` `int` `cnt[N]; ` ` `  `// Function to find the number of divisors ` `// of all numbers in  the range 1 to n ` `void` `Divisors() ` `{ ` `    ``memset``(cnt, 0, ``sizeof` `cnt); ` ` `  `    ``// For every number 1 to n ` `    ``for` `(``int` `i = 1; i < N; i++) { ` ` `  `        ``// Increase divisors count for every number ` `        ``for` `(``int` `j = 1; j * i < N; j++) ` `            ``cnt[i * j]++; ` `    ``} ` `} ` ` `  `// Function to find the sum of divisors ` `int` `Sumofdivisors(``int` `A, ``int` `B, ``int` `C) ` `{ ` `    ``// To store sum ` `    ``int` `sum = 0; ` ` `  `    ``Divisors(); ` ` `  `    ``for` `(``int` `i = 1; i <= A; i++) { ` `        ``for` `(``int` `j = 1; j <= B; j++) { ` `            ``for` `(``int` `k = 1; k <= C; k++) { ` `                ``int` `x = i * j * k; ` ` `  `                ``// Count the diviosrs ` `                ``sum += cnt[x]; ` `                ``if` `(sum >= mod) ` `                    ``sum -= mod; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``return` `sum; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` ` `  `    ``int` `A = 5, B = 6, C = 7; ` ` `  `    ``// Function call ` `    ``cout << Sumofdivisors(A, B, C); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java code for above given approach ` `class` `GFG  ` `{ ` ` `  `    ``static` `int` `N = ``100005``; ` `    ``static` `int` `mod = ``1000000007``; ` ` `  `    ``// To store the number of divisors  ` `    ``static` `int` `cnt[] = ``new` `int``[N]; ` ` `  `    ``// Function to find the number of divisors  ` `    ``// of all numbers in the range 1 to n  ` `    ``static` `void` `Divisors()  ` `    ``{ ` ` `  `        ``// For every number 1 to n  ` `        ``for` `(``int` `i = ``1``; i < N; i++)  ` `        ``{ ` ` `  `            ``// Increase divisors count for every number  ` `            ``for` `(``int` `j = ``1``; j * i < N; j++)  ` `            ``{ ` `                ``cnt[i * j]++; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// Function to find the sum of divisors  ` `    ``static` `int` `Sumofdivisors(``int` `A, ``int` `B, ``int` `C)  ` `    ``{ ` `        ``// To store sum  ` `        ``int` `sum = ``0``; ` ` `  `        ``Divisors(); ` ` `  `        ``for` `(``int` `i = ``1``; i <= A; i++)  ` `        ``{ ` `            ``for` `(``int` `j = ``1``; j <= B; j++)  ` `            ``{ ` `                ``for` `(``int` `k = ``1``; k <= C; k++)  ` `                ``{ ` `                    ``int` `x = i * j * k; ` ` `  `                    ``// Count the diviosrs  ` `                    ``sum += cnt[x]; ` `                    ``if` `(sum >= mod)  ` `                    ``{ ` `                        ``sum -= mod; ` `                    ``} ` `                ``} ` `            ``} ` `        ``} ` ` `  `        ``return` `sum; ` `    ``} ` ` `  `    ``// Driver code  ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `A = ``5``, B = ``6``, C = ``7``; ` ` `  `        ``// Function call  ` `        ``System.out.println(Sumofdivisors(A, B, C)); ` `    ``} ` `} ` ` `  `/* This code contributed by PrinciRaj1992 */`

## Python3

 `# Python3 code for above given approach ` `N ``=` `100005` `mod ``=` `1000000007` ` `  `# To store the number of divisors  ` `cnt ``=` `[``0``] ``*` `N;  ` ` `  `# Function to find the number of divisors  ` `# of all numbers in the range 1 to n  ` `def` `Divisors() :  ` ` `  `    ``# For every number 1 to n  ` `    ``for` `i ``in` `range``(``1``, N) : ` ` `  `        ``# Increase divisors count ` `        ``# for every number  ` `        ``for` `j ``in` `range``(``1``, N ``/``/` `i) :  ` `            ``cnt[i ``*` `j] ``+``=` `1``;  ` ` `  `# Function to find the sum of divisors  ` `def` `Sumofdivisors(A, B, C) :  ` `     `  `    ``# To store sum  ` `    ``sum` `=` `0``;  ` ` `  `    ``Divisors();  ` ` `  `    ``for` `i ``in` `range``(``1``,A ``+` `1``) : ` `        ``for` `j ``in` `range``(``1``, B ``+` `1``) :  ` `            ``for` `k ``in` `range``(``1``, C ``+` `1``) : ` `                ``x ``=` `i ``*` `j ``*` `k; ` `                 `  `                ``# Count the diviosrs ` `                ``sum` `+``=` `cnt[x]; ` `                ``if` `(``sum` `>``=` `mod) : ` `                    ``sum` `-``=` `mod;  ` ` `  `    ``return` `sum``;  ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `:  ` ` `  `    ``A ``=` `5``; B ``=` `6``; C ``=` `7``;  ` ` `  `    ``# Function call  ` `    ``print``(Sumofdivisors(A, B, C));  ` ` `  `# This code is contributed by Ryuga `

## C#

 `// C# code for above given approach ` `using` `System; ` `     `  `class` `GFG  ` `{ ` ` `  `    ``static` `int` `N = 100005; ` `    ``static` `int` `mod = 1000000007; ` ` `  `    ``// To store the number of divisors  ` `    ``static` `int` `[]cnt = ``new` `int``[N]; ` ` `  `    ``// Function to find the number of divisors  ` `    ``// of all numbers in the range 1 to n  ` `    ``static` `void` `Divisors()  ` `    ``{ ` ` `  `        ``// For every number 1 to n  ` `        ``for` `(``int` `i = 1; i < N; i++)  ` `        ``{ ` ` `  `            ``// Increase divisors count for every number  ` `            ``for` `(``int` `j = 1; j * i < N; j++)  ` `            ``{ ` `                ``cnt[i * j]++; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// Function to find the sum of divisors  ` `    ``static` `int` `Sumofdivisors(``int` `A, ``int` `B, ``int` `C)  ` `    ``{ ` `        ``// To store sum  ` `        ``int` `sum = 0; ` ` `  `        ``Divisors(); ` ` `  `        ``for` `(``int` `i = 1; i <= A; i++)  ` `        ``{ ` `            ``for` `(``int` `j = 1; j <= B; j++)  ` `            ``{ ` `                ``for` `(``int` `k = 1; k <= C; k++)  ` `                ``{ ` `                    ``int` `x = i * j * k; ` ` `  `                    ``// Count the diviosrs  ` `                    ``sum += cnt[x]; ` `                    ``if` `(sum >= mod)  ` `                    ``{ ` `                        ``sum -= mod; ` `                    ``} ` `                ``} ` `            ``} ` `        ``} ` ` `  `        ``return` `sum; ` `    ``} ` ` `  `    ``// Driver code  ` `    ``public` `static` `void` `Main(String[] args) ` `    ``{ ` `        ``int` `A = 5, B = 6, C = 7; ` ` `  `        ``// Function call  ` `        ``Console.WriteLine(Sumofdivisors(A, B, C)); ` `    ``} ` `} ` ` `  `// This code contributed by Rajput-Ji `

Output:

```1520
```

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