# Find the sum of digits of a number at even and odd places

Given a number N, the task is to find the sum of digits of a number at even and odd places.

Examples:

Input: N = 54873
Output:
Sum odd = 16
Sum even = 11

Input: N = 457892
Output:
Sum odd = 20
Sum even = 15

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

• First, calculate the reverse of the given number.
• To the reverse number we apply modulus operator and extract its last digit which is actually the first digit of a number so it is odd positioned digit.
• The next digit will be even positioned digit, and we can take the sum in alternating turns.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the reverse of a number ` `int` `reverse(``int` `n) ` `{ ` `    ``int` `rev = 0; ` `    ``while` `(n != 0) { ` `        ``rev = (rev * 10) + (n % 10); ` `        ``n /= 10; ` `    ``} ` `    ``return` `rev; ` `} ` ` `  `// Function to find the sum of the odd ` `// and even positioned digits in a number ` `void` `getSum(``int` `n) ` `{ ` `    ``n = reverse(n); ` `    ``int` `sumOdd = 0, sumEven = 0, c = 1; ` ` `  `    ``while` `(n != 0) { ` ` `  `        ``// If c is even number then it means ` `        ``// digit extracted is at even place ` `        ``if` `(c % 2 == 0) ` `            ``sumEven += n % 10; ` `        ``else` `            ``sumOdd += n % 10; ` `        ``n /= 10; ` `        ``c++; ` `    ``} ` ` `  `    ``cout << ``"Sum odd = "` `<< sumOdd << ``"\n"``; ` `    ``cout << ``"Sum even = "` `<< sumEven; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 457892; ` `    ``getSum(n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.*; ` ` `  `class` `GFG { ` ` `  `    ``// Function to return the reverse of a number ` `    ``static` `int` `reverse(``int` `n) ` `    ``{ ` `        ``int` `rev = ``0``; ` `        ``while` `(n != ``0``) { ` `            ``rev = (rev * ``10``) + (n % ``10``); ` `            ``n /= ``10``; ` `        ``} ` `        ``return` `rev; ` `    ``} ` ` `  `    ``// Function to find the sum of the odd ` `    ``// and even positioned digits in a number ` `    ``static` `void` `getSum(``int` `n) ` `    ``{ ` `        ``n = reverse(n); ` `        ``int` `sumOdd = ``0``, sumEven = ``0``, c = ``1``; ` ` `  `        ``while` `(n != ``0``) { ` ` `  `            ``// If c is even number then it means ` `            ``// digit extracted is at even place ` `            ``if` `(c % ``2` `== ``0``) ` `                ``sumEven += n % ``10``; ` `            ``else` `                ``sumOdd += n % ``10``; ` `            ``n /= ``10``; ` `            ``c++; ` `        ``} ` ` `  `        ``System.out.println(``"Sum odd = "` `+ sumOdd); ` `        ``System.out.println(``"Sum even = "` `+ sumEven); ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``int` `n = ``457892``; ` `        ``getSum(n); ` `    ``} ` `} ` ` `  `// This code is contributed by ` `// Surendra_Gangwar `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function to return the  ` `# reverse of a number ` `def` `reverse(n): ` `    ``rev ``=` `0` `    ``while` `(n !``=` `0``): ` `        ``rev ``=` `(rev ``*` `10``) ``+` `(n ``%` `10``) ` `        ``n ``/``/``=` `10` `    ``return` `rev ` ` `  `# Function to find the sum of the odd ` `# and even positioned digits in a number ` `def` `getSum(n): ` ` `  `    ``n ``=` `reverse(n) ` `    ``sumOdd ``=` `0` `    ``sumEven ``=` `0` `    ``c ``=` `1` ` `  `    ``while` `(n !``=` `0``): ` ` `  `        ``# If c is even number then it means ` `        ``# digit extracted is at even place ` `        ``if` `(c ``%` `2` `=``=` `0``): ` `            ``sumEven ``+``=` `n ``%` `10` `        ``else``: ` `            ``sumOdd ``+``=` `n ``%` `10` `        ``n ``/``/``=` `10` `        ``c ``+``=` `1` ` `  `    ``print``(``"Sum odd ="``, sumOdd) ` `    ``print``(``"Sum even ="``, sumEven) ` ` `  `# Driver code ` `n ``=` `457892` `getSum(n) ` ` `  `# This code is contributed  ` `# by mohit kumar `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG { ` ` `  `    ``// Function to return the reverse of a number ` `    ``static` `int` `reverse(``int` `n) ` `    ``{ ` `        ``int` `rev = 0; ` `        ``while` `(n != 0) { ` `            ``rev = (rev * 10) + (n % 10); ` `            ``n /= 10; ` `        ``} ` `        ``return` `rev; ` `    ``} ` ` `  `    ``// Function to find the sum of the odd ` `    ``// and even positioned digits in a number ` `    ``static` `void` `getSum(``int` `n) ` `    ``{ ` `        ``n = reverse(n); ` `        ``int` `sumOdd = 0, sumEven = 0, c = 1; ` ` `  `        ``while` `(n != 0) { ` ` `  `            ``// If c is even number then it means ` `            ``// digit extracted is at even place ` `            ``if` `(c % 2 == 0) ` `                ``sumEven += n % 10; ` `            ``else` `                ``sumOdd += n % 10; ` `            ``n /= 10; ` `            ``c++; ` `        ``} ` ` `  `        ``Console.WriteLine(``"Sum odd = "` `+ sumOdd); ` `        ``Console.WriteLine(``"Sum even = "` `+ sumEven); ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `n = 457892; ` `        ``getSum(n); ` `    ``} ` `} ` ` `  `// This code is contributed by ` `// Akanksha Rai `

## PHP

 ` `

Output:

```Sum odd = 20
Sum even = 15
```

Another approach: The problem can be solved without reversing the number. We can extract all the digits from the number one by one from the end. If the original number was odd then the last digit must be odd positioned else it will be even positioned. After processing a digit, we can invert the state from odd to even and vice versa.

Below is the implementation of the above approach:

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find the sum of the odd ` `// and even positioned digits in a number ` `void` `getSum(``int` `n) ` `{ ` ` `  `    ``// If n is odd then the last digit ` `    ``// will be odd positioned ` `    ``bool` `isOdd = (n % 2 == 1) ? ``true` `: ``false``; ` ` `  `    ``// To store the respective sums ` `    ``int` `sumOdd = 0, sumEven = 0; ` ` `  `    ``// While there are digits left process ` `    ``while` `(n != 0) { ` ` `  `        ``// If current digit is odd positioned ` `        ``if` `(isOdd) ` `            ``sumOdd += n % 10; ` ` `  `        ``// Even positioned digit ` `        ``else` `            ``sumEven += n % 10; ` ` `  `        ``// Invert state ` `        ``isOdd = !isOdd; ` ` `  `        ``// Remove last digit ` `        ``n /= 10; ` `    ``} ` ` `  `    ``cout << ``"Sum odd = "` `<< sumOdd << ``"\n"``; ` `    ``cout << ``"Sum even = "` `<< sumEven; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 457892; ` `    ``getSum(n); ` ` `  `    ``return` `0; ` `} `

Output:

```Sum odd = 20
Sum even = 15
```

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