Sum of Even & Odd Digits Problem Statement

Create a program that accepts an integer N and outputs two sums: the sum of all even digits and the sum of all odd digits found in the integer.

Explanation:

Digits refer to the numbers themselves, not their positional values. For example, if N is "13245", the even digits are 2 & 4, and the odd digits are 1, 3 & 5.

Input:

Integer N

Output:

Sum_of_Even_Digits Sum_of_Odd_Digits

First print the sum of even digits, followed by the sum of odd digits, separated by a space.

Example:

Input:

13245

Output:

6 9

Explanation:

In the input 13245, the even digits are 2 and 4, summing to 6. The odd digits are 1, 3, and 5, summing to 9.

Constraints:

0

Question

Sum of Even & Odd Digits Problem Statement

Create a program that accepts an integer N and outputs two sums: the sum of all even digits and the sum of all odd digits found in the integer.

Explanation:

Digits refer to the numbers themselves, not their positional values. For example, if N is "13245", the even digits are 2 & 4, and the odd digits are 1, 3 & 5.

Input:

Integer N

Output:

Sum_of_Even_Digits Sum_of_Odd_Digits

First print the sum of even digits, followed by the sum of odd digits, separated by a space.

Example:

Input:

Output:

6 9

Explanation:

In the input 13245, the even digits are 2 and 4, summing to 6. The odd digits are 1, 3, and 5, summing to 9.

Constraints:

0 <= N <= 10^8

Accepted Answer

Create a program to find the sum of even and odd digits in an integer.

Iterate through each digit of the integer and check if it is even or odd
Keep track of the sum of even digits and odd digits separately
Print the sums of even and odd digits at the end

Accepted Answer

k=[1,3,2,4,5]
even = []
odd = []
for i in k:
    #print(i)
    if i % 2 == 0:
        even.append(i)
    else:
        odd.append(i)

print("odd: ", odd)
print("even: ", even)

Accepted Answer

import java.util.Scanner;

public class SumEvenAndOddDigits {

public static void main(String[] args) {

Scanner scanner = new Scanner(System.in);

System.out.print("Enter an integer N: ");

int N = scanner.nextInt();

int evenSum = 0;

int oddSum = 0;

while (N != 0) {

int digit = N % 10;

if (digit % 2 == 0) {

evenSum += digit;

} else {

oddSum += digit;

}

N /= 10;

}

System.out.println("Sum of even digits: " + evenSum);

System.out.println("Sum of odd digits: " + oddSum);

}

Accepted Answer

a = int(input())
n = []
for _ in range(a):
    m = int(input())
    n.append(m)
even = 0
odd = 0
for x in range(0, len(n)):
    if n[x] % 2 == 0:
        even = even + n[x]
    else:
        odd = odd + n[x]
print(even)
print(odd)

Make it easy like this

Accepted Answer

Namepace SumOfEvenAndOddDigits

{

class Program

{

static void Main(string[] args)

{

Console.Write("Enter an integer: ");

int number = int.Parse(Console.ReadLine());

int sumEven = 0;

int sumOdd = 0;

// Convert the number to a string to access individual digits

string numString = number.ToString();

// Loop through each character in the string (representing each digit)

foreach (char digitChar in numString)

{

int digit = int.Parse(digitChar.ToString());

if (digit % 2 == 0)

{

// Even digit

sumEven += digit;

}

else

{

// Odd digit

sumOdd += digit;

}

Console.WriteLine("Sum _of_ even digits: " + sumEven);

Console.WriteLine("Sum of _odd digits: " + sumOdd);

}

Accepted Answer

def sum_of_even_and_odd_digits(N):

even_sum = 0

odd_sum = 0

while N > 0:

digit = N % 10 # Extract the last digit

if digit % 2 == 0: # Check if the digit is even

even_sum += digit

else:

odd_sum += digit

N //= 10 # Remove the last digit from the number

return even_sum, odd_sum

# Main program

N = int(input("Enter an integer: "))

even_sum, odd_sum = sum_of_even_and_odd_digits(N)

print("Sum of even digits:", even_sum)

print("Sum of odd digits:", odd_sum)