This is my solution, with JavaScript:

function divisibleSumPairs(n, k, ar) {
    // Write your code here
    let count = 0;
    while(ar.length>1){
        let i = ar.shift();
        for(let j=0; j<ar.length;j++){    
        let sum = i + ar[j];
        if(sum%k===0) count++;
        }
    }
    return count;
}

def divisibleSumPairs(n, k, ar): freq = [0] * k count = 0

for num in ar:
    mod = num % k
    complement = (k - num) % k
    count += freq[complement]
    freq[mod] += 1

return count

My solution in python:

def divisibleSumPairs(n, k, arr):

return sum([1 for i in range(n) for j in range (i + 1 , n) if i < j and (arr[i] + arr[j]) % k == 0])

JAVA Solution

public static int divisibleSumPairs(int n, int k, List ar) {

    int count = 0;

    for(int i = 0; i<n; i++){

        for(int j = i+1; j<n; j++){

            if((ar.get(i)+ ar.get(j)) % k == 0){
                count++;
            }
        }
    }

    return count;

}

C# var remainderCounts = new int[k]; var count = 0;

foreach (var num in ar) { var remainder = num % k; var complement = (k - remainder) % k; count += remainderCounts[complement]; remainderCounts[remainder]++; }

return count;