Write a program to find sum of middle 2 or 3 digits of an entered number in python

For example take 2345, how we add middle numbers 3 and 4 and gives output ..the sum of middle numbers is 7.

xavier This code works for 4 digit number. If you want code that works for any digit, you can tell me. I will give you. #include int main() { int a,sum; printf("Enter a number: "); scanf("%d",&a); sum=(a/10)%10+(a/100)%10; printf("Answer: %d",sum); return 0; }

The easiest way is to convert the number into an array of characters, take the substring and then find the sum.

i have tried but no use can you provide me with complete code

If you are sure it is a four digit number then #include /* Function to get sum of digits */ int getSum(int n) { int sum; sum=0; /* Single line that calculates sum */ sum=sum+(n/100)%10+(n%100)/10; return sum; } // Driver code int main() { int n = 3547; printf(" %d ", getSum(n)); return 0; }

Can you suugest me with code

/* Function to get sum of digits */ int getSum(int n) { int sum; /* Single line that calculates sum */ for (sum = 0; n > 0; sum += n % 10, n /= 10) ; return sum; } // Driver code int main() { int n = 3547; printf(" %d ", getSum(n)); return 0; }

Use while loop to get number of digits

i dont know to use while loop that is why i request you to provide code

Am new here who can teach me python directly

Given a number, find the sum of its digits.

Examples : 

Input: n = 687
Output: 21

Input: n = 12
Output: 3

Follow the below steps to solve the problem:

• Get the number
• Declare a variable to store the sum and set it to 0
• Repeat the next two steps till the number is not 0
• Get the rightmost digit of the number with help of the remainder ‘%’ operator by dividing it by 10 and adding it to the sum.
• Divide the number by 10 with help of ‘/’ operator to remove the rightmost digit.
• Print or return the sum

Below is the implementation of the above approach:

C++

#include

using namespace std;

class gfg {

public:

    int getSum(int n)

    {

        int sum = 0;

        while (n != 0) {

            sum = sum + n % 10;

            n = n / 10;

        }

        return sum;

    }

};

int main()

{

    gfg g;

    int n = 687;

    cout << g.getSum(n);

    return 0;

}

C

#include

int getSum(int n)

{

    int sum = 0;

    while (n != 0) {

        sum = sum + n % 10;

        n = n / 10;

    }

    return sum;

}

int main()

{

    int n = 687;

    printf(" %d ", getSum(n));

    return 0;

}

Java

import java.io.*;

class GFG {

    static int getSum(int n)

    {

        int sum = 0;

        while (n != 0) {

            sum = sum + n % 10;

            n = n / 10;

        }

        return sum;

    }

    public static void main(String[] args)

    {

        int n = 687;

        System.out.println(getSum(n));

    }

}

Python3

def getSum(n):

    sum = 0

    while (n != 0):

        sum = sum + int(n % 10)

        n = int(n/10)

    return sum

if __name__ == "__main__":

    n = 687

    print(getSum(n))

C#

using System;

class GFG {

    static int getSum(int n)

    {

        int sum = 0;

        while (n != 0) {

            sum = sum + n % 10;

            n = n / 10;

        }

        return sum;

    }

    public static void Main()

    {

        int n = 687;

        Console.Write(getSum(n));

    }

}

PHP

function getsum($n)

{

    $sum = 0;

    while ($n != 0)

    {

        $sum = $sum + $n % 10;

        $n = $n/10;

    }

    return $sum;

}

$n = 687;

$res = getsum($n);

echo("$res");

?>

Javascript

Time Complexity: O(log N)
Auxiliary Space: O(1)

How to compute in a single line?

The below function has three lines instead of one line, but it calculates the sum in one line using for loop. It can be made one-line function if we pass the pointer to the sum. 

Below is the implementation of the above approach:

C++

#include

using namespace std;

class gfg {

public:

    int getSum(int n)

    {

        int sum;

        for (sum = 0; n > 0; sum += n % 10, n /= 10)

            ;

        return sum;

    }

};

int main()

{

    gfg g;

    int n = 687;

    cout << g.getSum(n);

    return 0;

}

C

#include

int getSum(int n)

{

    int sum;

    for (sum = 0; n > 0; sum += n % 10, n /= 10)

        ;

    return sum;

}

int main()

{

    int n = 687;

    printf(" %d ", getSum(n));

    return 0;

}

Java

import java.io.*;

class GFG {

    static int getSum(int n)

    {

        int sum;

        for (sum = 0; n > 0; sum += n % 10, n /= 10)

            ;

        return sum;

    }

    public static void main(String[] args)

    {

        int n = 687;

        System.out.println(getSum(n));

    }

}

Python3

def getSum(n):

    sum = 0

    while(n > 0):

        sum += int(n % 10)

        n = int(n/10)

    return sum

if __name__ == "__main__":

    n = 687

    print(getSum(n))

C#

using System;

class GFG {

    static int getSum(int n)

    {

        int sum;

        for (sum = 0; n > 0; sum += n % 10, n /= 10)

            ;

        return sum;

    }

    public static void Main()

    {

        int n = 687;

        Console.Write(getSum(n));

    }

}

PHP

function getsum($n)

{

    for ($sum = 0; $n > 0; $sum += $n % 10,

                                  $n /= 10);

    return $sum;

}

$n = 687;

echo(getsum($n));

?>

Javascript

Time Complexity: O(log N)
Auxiliary Space: O(1)

Sum of the digits of a given number using recursion:

Follow the below steps to solve the problem:

• Get the number
• Get the remainder and pass the next remaining digits
• Get the rightmost digit of the number with help of the remainder ‘%’ operator by dividing it by 10 and adding it to the sum.
• Divide the number by 10 with help of the ‘/’ operator to remove the rightmost digit.
• Check the base case with n = 0
• Print or return the sum

Below is the implementation of the above approach:

C++

#include

using namespace std;

class gfg {

public:

    int sumDigits(int no)

    {

        if (no == 0) {

            return 0;

        }

        return (no % 10) + sumDigits(no / 10);

    }

};

int main(void)

{

    gfg g;

    cout << g.sumDigits(687);

    return 0;

}

C

#include

int sumDigits(int no)

{

    if (no == 0) {

        return 0;

    }

    return (no % 10) + sumDigits(no / 10);

}

int main()

{

    printf("%d", sumDigits(687));

    return 0;

}

Java

import java.io.*;

class GFG {

    static int sumDigits(int no)

    {

        if (no == 0) {

            return 0;

        }

        return (no % 10) + sumDigits(no / 10);

    }

    public static void main(String[] args)

    {

        System.out.println(sumDigits(687));

    }

}

Python3

def sumDigits(no):

    return 0 if no == 0 else int(no % 10) + sumDigits(int(no/10))

if __name__ == "__main__":

    print(sumDigits(687))

C#

using System;

class GFG {

    static int sumDigits(int no)

    {

        return no == 0 ? 0 : no % 10 + sumDigits(no / 10);

    }

    public static void Main()

    {

        Console.Write(sumDigits(687));

    }

}

PHP

function sumDigits($no)

{

return $no == 0 ? 0 : $no % 10 +

                      sumDigits($no / 10) ;

}

echo sumDigits(687);

?>

Javascript

Time Complexity: O(log N)
Auxiliary Space: O(log N)

Sum of the digits of a given number with input as string:

When the number of digits of that number exceeds 1019 , we can’t take that number as an integer since the range of long long int doesn’t satisfy the given number. So take input as a string, run a loop from start to the length of the string and increase the sum with that character(in this case it is numeric)

Follow the below steps to solve the problem:

• Declare a variable sum equal to zero
• Run a loop from zero to the length of the input string
  • Add the value of each character into the sum, by converting the character into it’s integer value
• Return sum

Below is the implementation of the above approach:

Sum of digits is 15