Problem Link: Contest Page | CodeChef

Author: @umar_07

Tester: @sharshach

Difficulty: Cakewalk

Prerequisite: None

# Explanation:

As it is easily noticeable, no matter what comes up in each dice, the sum will always be in the form 5*k (1<= k <= 11). Maximum possible sum in a single throw= 55, Minimum possible sum in a single throw= 5.

And each student throws both the dice M, so Maximum possible sum in M throws= 55*M, Minimum possible sum in M throws= 5 * M.

So for a sum to exist after M throws:

- sum%5 == 0
- 5 * M <= sum <= 55 * M

Setterâ€™s Code:

```
#include <bits/stdc++.h>
#define ll long long int
#define debug(x) cerr << #x << " is " << x << " ";
#define mod 1000000007
#define endl "\n"
#define fast_input ios_base::sync_with_stdio(false); cin.tie(NULL);
using namespace std;
int main()
{
ll t;
cin >> t;
while(t--)
{
ll n, m, c=0;
cin >> n >> m;
ll arr[n];
for(int i=0; i<n; i++)
{
cin >> arr[i];
if(arr[i]>=5*m && arr[i]<=55*m && arr[i]%5==0)
c++;
}
cout << c << endl;
}
return 0;
}
```