## Addition of digits

// A Simple C++ program to compute sum of digits in numbers from 1 to n

#include<bits/stdc++.h>

using namespace std;

int sumOfDigits(int );

// Returns sum of all digits in numbers from 1 to n

int sumOfDigitsFrom1ToN(int n)

{

int result = 0; // initialize result

// One by one compute sum of digits in every number from

// 1 to n

for (int x = 1; x <= n; x++)

result += sumOfDigits(x);

return result;

}

// A utility function to compute sum of digits in a

// given number x

int sumOfDigits(int x)

{

int sum = 0;

while (x != 0)

{

sum += x %10;

x   = x /10;

}

return sum;

}

// Driver Program

int main()

{

int n = 328;

cout << “Sum of digits in numbers from 1 to ” << n << ” is “

<< sumOfDigitsFrom1ToN(n);

return 0;

}

## Addition of even digits

// C++ implementation to find sum of

// first n even numbers

#include <bits/stdc++.h>

using namespace std;

// function to find sum of

// first n even numbers

int evenSum(int n)

{

int curr = 2, sum = 0;

// sum of first n even numbers

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

sum += curr;

// next even number

curr += 2;

}

// required sum

return sum;

}

// Driver program to test above

int main()

{

int n = 20;

cout << “Sum of first ” << n

<< ” Even numbers is: ” << evenSum(n);

return 0;

}

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### 3 .  Write a program to Get mark 1 to mark 5 and create a method which has argument pass mark if mark 1 to mark 5 are greater than pass mark ,return true ,else return false

```#include<stdio.h>
void main()
{
int m1,m2,m3,total;
float per;
clrscr();
printf("Enter 3 Nos.");
scanf("%D%D%D",&m1,&m2,&m3);
total=m1+m2+m3;
per=total*100/300;
if(per>=60&&per<=100)
printf("You are 1st :");
else if(per>=50&&per<=60)
printf("You are 2nd");
else if(per>=40&&per<=50)
printf("You are 3rd");
else
printf("You are Fail");
getch();
}```

### REFERENCES :

https://www.geeksforgeeks.org/count-sum-of-digits-in-numbers-from-1-to-n/

https://www.geeksforgeeks.org/sum-first-n-even-numbers/