# Number of pairs from the first N natural numbers whose sum is divisible by K

Given the integer values of **N** and **K**. The task is to find the number of pairs from the set of natural numbers up to N{1, 2, 3……N-1, N} whose sum is divisible by K.

**Note :** 1 <= K <= N <= 10^6.

**Examples:**

Input :N = 10, K = 5

Output :9

Explanation :The possible pairs whose sum is divisible by 5 are (1, 4), (1, 9), (6, 4), (6, 9), (2, 3), (2, 8), (3, 7), (7, 8) and (5, 10). Hence the count is 9.

Input :N = 7, K = 3

Output :

Explanation :The possible pairs whose sum is divisible by 3 are (1, 2), (1, 5), (2, 4), (2, 7), (3, 6), (4, 5) and (5, 7). Hence the count is 7.

**Simple Approach:** A naive approach is to use a nested loop and check for all possible pairs and its divisibility by K. The time complexity of such an approach is O(N^2) which is not very efficient.

**Efficient Approach**: An efficient approach is to use basic Hashing technique.

**Firstly**, create array rem[K], where rem[i] contains the count of integers from 1 to N which gives the remainder i when divided by K. rem[i] can be calculated by the formula rem[i] = (N – i)/K + 1.

**Secondly**, the sum of two integers is divisible by K if:

- Both the integers are divisible by K. The count of which is calculated by rem[0]*(rem[0]-1)/2.
- Remainder of first integer is R and remainder of other number is K-R. The count of which is calculated by rem[R]*rem[K-R], where R varies from 1 to K/2.
- K is even and both the remainders are K/2. The count of which is calculated by rem[K/2]*(rem[K/2]-1)/2.

The sum of counts of all these cases gives the required count of pairs such that their sum is divisible by K.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the approach ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to find the number of pairs ` `// from the set of natural numbers up to ` `// N whose sum is divisible by K ` `int` `findPairCount(` `int` `N, ` `int` `K) ` `{ ` ` ` `int` `count = 0; ` ` ` ` ` `// Declaring a Hash to store count ` ` ` `int` `rem[K]; ` ` ` ` ` `rem[0] = N / K; ` ` ` ` ` `// Storing the count of integers with ` ` ` `// a specific remainder in Hash array ` ` ` `for` `(` `int` `i = 1; i < K; i++) ` ` ` `rem[i] = (N - i) / K + 1; ` ` ` ` ` `// Check if K is even ` ` ` `if` `(K % 2 == 0) { ` ` ` `// Count of pairs when both ` ` ` `// integers are divisible by K ` ` ` `count += (rem[0] * (rem[0] - 1)) / 2; ` ` ` ` ` `// Count of pairs when one remainder ` ` ` `// is R and other remainder is K - R ` ` ` `for` `(` `int` `i = 1; i < K / 2; i++) ` ` ` `count += rem[i] * rem[K - i]; ` ` ` ` ` `// Count of pairs when both the ` ` ` `// remainders are K / 2 ` ` ` `count += (rem[K / 2] * (rem[K / 2] - 1)) / 2; ` ` ` `} ` ` ` `else` `{ ` ` ` `// Count of pairs when both ` ` ` `// integers are divisible by K ` ` ` `count += (rem[0] * (rem[0] - 1)) / 2; ` ` ` ` ` `// Count of pairs when one remainder is R ` ` ` `// and other remainder is K - R ` ` ` `for` `(` `int` `i = 1; i <= K / 2; i++) ` ` ` `count += rem[i] * rem[K - i]; ` ` ` `} ` ` ` ` ` `return` `count; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `N = 10, K = 4; ` ` ` ` ` `// Print the count of pairs ` ` ` `cout << findPairCount(N, K); ` ` ` ` ` `return` `0; ` `} ` |

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## Java

`// Java implementation of the approach ` `class` `GfG ` `{ ` ` ` `// Function to find the number of pairs ` `// from the set of natural numbers up to ` `// N whose sum is divisible by K ` `static` `int` `findPairCount(` `int` `N, ` `int` `K) ` `{ ` ` ` `int` `count = ` `0` `; ` ` ` ` ` `// Declaring a Hash to store count ` ` ` `int` `rem[] = ` `new` `int` `[K]; ` ` ` ` ` `rem[` `0` `] = N / K; ` ` ` ` ` `// Storing the count of integers with ` ` ` `// a specific remainder in Hash array ` ` ` `for` `(` `int` `i = ` `1` `; i < K; i++) ` ` ` `rem[i] = (N - i) / K + ` `1` `; ` ` ` ` ` `// Check if K is even ` ` ` `if` `(K % ` `2` `== ` `0` `) ` ` ` `{ ` ` ` `// Count of pairs when both ` ` ` `// integers are divisible by K ` ` ` `count += (rem[` `0` `] * (rem[` `0` `] - ` `1` `)) / ` `2` `; ` ` ` ` ` `// Count of pairs when one remainder ` ` ` `// is R and other remainder is K - R ` ` ` `for` `(` `int` `i = ` `1` `; i < K / ` `2` `; i++) ` ` ` `count += rem[i] * rem[K - i]; ` ` ` ` ` `// Count of pairs when both the ` ` ` `// remainders are K / 2 ` ` ` `count += (rem[K / ` `2` `] * (rem[K / ` `2` `] - ` `1` `)) / ` `2` `; ` ` ` `} ` ` ` `else` ` ` `{ ` ` ` `// Count of pairs when both ` ` ` `// integers are divisible by K ` ` ` `count += (rem[` `0` `] * (rem[` `0` `] - ` `1` `)) / ` `2` `; ` ` ` ` ` `// Count of pairs when one remainder is R ` ` ` `// and other remainder is K - R ` ` ` `for` `(` `int` `i = ` `1` `; i <= K / ` `2` `; i++) ` ` ` `count += rem[i] * rem[K - i]; ` ` ` `} ` ` ` ` ` `return` `count; ` `} ` ` ` `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` ` ` `int` `N = ` `10` `, K = ` `4` `; ` ` ` ` ` `// Print the count of pairs ` ` ` `System.out.println(findPairCount(N, K)); ` `} ` `} ` ` ` `// This code is contributed by Prerna Saini ` |

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## Python3

`# Python3 implementation of the approach ` ` ` `# Function to find the number of pairs ` `# from the set of natural numbers up to ` `# N whose sum is divisible by K ` `def` `findPairCount(N, K) : ` ` ` `count ` `=` `0` `; ` ` ` ` ` `# Declaring a Hash to store count ` ` ` `rem ` `=` `[` `0` `] ` `*` `K; ` ` ` ` ` `rem[` `0` `] ` `=` `N ` `/` `/` `K; ` ` ` ` ` `# Storing the count of integers with ` ` ` `# a specific remainder in Hash array ` ` ` `for` `i ` `in` `range` `(` `1` `, K) : ` ` ` `rem[i] ` `=` `(N ` `-` `i) ` `/` `/` `K ` `+` `1` `; ` ` ` ` ` `# Check if K is even ` ` ` `if` `(K ` `%` `2` `=` `=` `0` `) : ` ` ` ` ` `# Count of pairs when both ` ` ` `# integers are divisible by K ` ` ` `count ` `+` `=` `(rem[` `0` `] ` `*` `(rem[` `0` `] ` `-` `1` `)) ` `/` `/` `2` `; ` ` ` ` ` `# Count of pairs when one remainder ` ` ` `# is R and other remainder is K - R ` ` ` `for` `i ` `in` `range` `(` `1` `, K ` `/` `/` `2` `) : ` ` ` `count ` `+` `=` `rem[i] ` `*` `rem[K ` `-` `i]; ` ` ` ` ` `# Count of pairs when both the ` ` ` `# remainders are K / 2 ` ` ` `count ` `+` `=` `(rem[K ` `/` `/` `2` `] ` `*` `(rem[K ` `/` `/` `2` `] ` `-` `1` `)) ` `/` `/` `2` `; ` ` ` ` ` `else` `: ` ` ` ` ` `# Count of pairs when both ` ` ` `# integers are divisible by K ` ` ` `count ` `+` `=` `(rem[` `0` `] ` `*` `(rem[` `0` `] ` `-` `1` `)) ` `/` `/` `2` `; ` ` ` ` ` `# Count of pairs when one remainder is R ` ` ` `# and other remainder is K - R ` ` ` `for` `i ` `in` `rage(` `1` `, K` `/` `/` `2` `+` `1` `) : ` ` ` `count ` `+` `=` `rem[i] ` `*` `rem[K ` `-` `i]; ` ` ` ` ` `return` `count; ` ` ` `# Driver code ` `if` `__name__ ` `=` `=` `"__main__"` `: ` ` ` ` ` `N ` `=` `10` `; K ` `=` `4` `; ` ` ` ` ` `# Print the count of pairs ` ` ` `print` `(findPairCount(N, K)); ` ` ` `# This code is contributed by Ryuga ` |

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## C#

`// C# implementation of the approach ` `class` `GfG ` `{ ` ` ` `// Function to find the number of pairs ` `// from the set of natural numbers up to ` `// N whose sum is divisible by K ` `static` `int` `findPairCount(` `int` `N, ` `int` `K) ` `{ ` ` ` `int` `count = 0; ` ` ` ` ` `// Declaring a Hash to store count ` ` ` `int` `[] rem = ` `new` `int` `[K]; ` ` ` ` ` `rem[0] = N / K; ` ` ` ` ` `// Storing the count of integers with ` ` ` `// a specific remainder in Hash array ` ` ` `for` `(` `int` `i = 1; i < K; i++) ` ` ` `rem[i] = (N - i) / K + 1; ` ` ` ` ` `// Check if K is even ` ` ` `if` `(K % 2 == 0) ` ` ` `{ ` ` ` `// Count of pairs when both ` ` ` `// integers are divisible by K ` ` ` `count += (rem[0] * (rem[0] - 1)) / 2; ` ` ` ` ` `// Count of pairs when one remainder ` ` ` `// is R and other remainder is K - R ` ` ` `for` `(` `int` `i = 1; i < K / 2; i++) ` ` ` `count += rem[i] * rem[K - i]; ` ` ` ` ` `// Count of pairs when both the ` ` ` `// remainders are K / 2 ` ` ` `count += (rem[K / 2] * (rem[K / 2] - 1)) / 2; ` ` ` `} ` ` ` `else` ` ` `{ ` ` ` `// Count of pairs when both ` ` ` `// integers are divisible by K ` ` ` `count += (rem[0] * (rem[0] - 1)) / 2; ` ` ` ` ` `// Count of pairs when one remainder is R ` ` ` `// and other remainder is K - R ` ` ` `for` `(` `int` `i = 1; i <= K / 2; i++) ` ` ` `count += rem[i] * rem[K - i]; ` ` ` `} ` ` ` ` ` `return` `count; ` `} ` ` ` `// Driver code ` `static` `void` `Main() ` `{ ` ` ` `int` `N = 10, K = 4; ` ` ` ` ` `// Print the count of pairs ` ` ` `System.Console.WriteLine(findPairCount(N, K)); ` `} ` `} ` ` ` `// This code is contributed by mits ` |

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## PHP

**Output:**

10

**Time Complexity**: O(K).

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