### PROBLEM LINK:

**Author:** Chandan Boruah

**Tester:** Chandan Boruah

**Editorialist:** Chandan Boruah

### DIFFICULTY:

EASY

### PREREQUISITES:

Maths

### PROBLEM:

You need to print sum of all the numbers that are divisible by 10 and are less than and equal to given number N.

### QUICK EXPLANATION:

Divide N by 10, then use formula (x*(x+1)*10)/2 to find the sum.

### EXPLANATION:

```
#include<iostream>
using namespace std;
int main()
{
int t;
cin>>t;
while(t--)
{
long long n,sum=0;
cin>>n;
long long x=n/10;
sum=x*(x+1)*5;
cout<<sum<<endl;
}
}
```

c# code:

```
using System;
using System.Collections.Generic;
class some
{
public static void Main()
{
int n=Int32.Parse(Console.ReadLine());
for(int t=0;t<n;t++)
{
long nn=Int32.Parse(Console.ReadLine());
if(nn<10)Console.WriteLine(0);
else
{
nn/=10;
Console.WriteLine(nn*(nn+1)*5);
}
}
}
}
```

lets say you have to print sum of numbers less than equal to number 20, divide 20 by 10 gives you 2.

This is a pattern. Everytime you divide it by 10 you get count of numbers less than or equal to N that are divisible by 10. Now, x*(x+1)/2 is the sum of first x positive integers. Since, there are 2 such numbers and we have 10+20=30, we divided by 10 so we have 1+2=3. We multiply that result by 10 to get 30. This, solution comes naturally to mind, because our psychology is to think step by step.

### AUTHOR’S AND TESTER’S SOLUTIONS:

Author’s solution can be found above (c# code).

Tester’s solution can be found above (c# code).