**Divisible Sum Pairs**

### Problem Statement :

Given an array of integers and a positive integer k, determine the number of (i, j) pairs where i<j and ar[i] + ar[j] is divisible by k. Example ar = [1, 2, 3, 4, 5, 6] k=5 Three pairs meet the criteria: [1, 4], [2, 3] and [4, 6]. Function Description Complete the divisibleSumPairs function in the editor below. divisibleSumPairs has the following parameter(s): int n: the length of array int ar[n]: an array of integers int k: the integer divisor Returns - int: the number of pairs Input Format The first line contains 2 space-separated integers, n and k. The second line contains n space-separated integers, each a value of arr[i]. Constraints 2 <= n <= 100 1 <= k <=100 1 <= arr[i] <= 100

### Solution :

` ````
Solution in C :
python3 :
#!/bin/python3
import sys
n,k = input().strip().split(' ')
n,k = [int(n),int(k)]
a = [int(a_temp) for a_temp in input().strip().split(' ')]
result = 0
for i in range(n-1):
for j in range(i+1, n):
if not ((a[i] + a[j]) % k):
result += 1
print(result)
Java :
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static void main(String[] args) {
int cnt=0;
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int k = in.nextInt();
int a[] = new int[n];
for(int a_i=0; a_i < n; a_i++){
a[a_i] = in.nextInt();
}
for(int a_i=0; a_i < n-1; a_i++){
for(int a_j=a_i+1; a_j< n; a_j++){
if( (a[a_i]+a[a_j])%k==0)
cnt++;
}
}
System.out.println(cnt);
}
}
C++ :
#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <string>
#include <bitset>
#include <cstdio>
#include <limits>
#include <vector>
#include <climits>
#include <cstring>
#include <cstdlib>
#include <fstream>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <unordered_map>
using namespace std;
int main(){
int n;
int k;
int count = 0;
cin >> n >> k;
vector<int> a(n);
for(int a_i = 0;a_i < n;a_i++){
cin >> a[a_i];
}
for(int i =0 ; i < n -1 ; i++){
for(int j=i+1 ; j < n ; j++){
if( (a[i]+a[j])%k ==0){
count++;
}
}
}
cout << count;
return 0;
}
C :
#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>
int main(){
int n;
int k;
int ans = 0, i, j, t;
scanf("%d %d",&n,&k);
int *a = malloc(sizeof(int) * n);
for(int a_i = 0; a_i < n; a_i++){
scanf("%d",&a[a_i]);
}
for (i=0;i<n;i++)
{
for(j=i+1;j<n;j++)
{
if ((a[i]+a[j])%k==0)
ans++;
}
}
printf("%d\n",ans);
return 0;
}
```

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