Sum and Product of digits in a number that divide the number

Given a positive integer N. The task is to find the sum and product of digits of the number which evenly divides the number n.

Examples:

Input: N = 12
Output: Sum = 3, product = 2
1 and 2 divide 12. So, their sum is 3 and product is 2.

Input: N = 1012
Output: Sum = 4, product = 2
1, 1 and 2 divide 1012.

Approach: The idea is to find each digit of the number n by modulus 10 and then check whether it divides n or not. Accordingly, add it to the sum and multiply it with the product. Notice that the digit can be 0, so take care of that case.

Below is the implementation of the above approach:  

C++

// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;

// Print the sum and product of digits
//  that divides the number.
void countDigit(int n)
{
    int temp = n, sum = 0, product = 1;
    while (temp != 0) {

        // Fetching each digit of the number
        int d = temp % 10;
        temp /= 10;

        // Checking if digit is greater than 0
        // and can divides n.
        if (d > 0 && n % d == 0) {
            sum += d;
            product *= d;
        }
    }

    cout << "Sum = " << sum;
    cout << "\nProduct = " << product;
}

// Driver code
int main()
{
    int n = 1012;

    countDigit(n);
    return 0;
}

Java

// Java implementation of the
// above approach
import java.lang.*;
import java.util.*;

class GFG
{
// Print the sum and product of
// digits that divides the number.
static void countDigit(int n)
{
    int temp = n, sum = 0, product = 1;
    while (temp != 0)
    {

        // Fetching each digit of
        // the number
        int d = temp % 10;
        temp /= 10;

        // Checking if digit is greater
        // than 0 and can divides n.
        if (d > 0 && n % d == 0)
        {
            sum += d;
            product *= d;
        }
    }

    System.out.print("Sum = " + sum);
    System.out.print("\nProduct = " + product);
}

// Driver code
public static void main(String args[])
{
    int n = 1012;

    countDigit(n);
}
}

// This code is contributed
// by Akanksha Rai(Abby_akku)

Python3

# Python3 implementation of the above approach

# Print the sum and product of digits
# that divides the number.
def countDigit(n):
    temp = n
    sum = 0
    product = 1
    while(temp != 0):

        # Fetching each digit of the number
        d = temp % 10
        temp //= 10

        # Checking if digit is greater
        # than 0 and can divides n.
        if(d > 0 and n % d == 0):
            sum += d
            product *= d

    print("Sum =", sum)
    print("Product =", product)

# Driver code
if __name__=='__main__':

    n = 1012
    countDigit(n)

# This code is contributed
# by Kirti_Mangal

C#

// C# implementation of the
// above approach
using System;

class GFG
{
// Print the sum and product of
// digits that divides the number.
static void countDigit(int n)
{
    int temp = n, sum = 0, product = 1;
    while (temp != 0)
    {

        // Fetching each digit of
        // the number
        int d = temp % 10;
        temp /= 10;

        // Checking if digit is greater
        // than 0 and can divides n.
        if (d > 0 && n % d == 0)
        {
            sum += d;
            product *= d;
        }
    }

    Console.Write("Sum = " + sum);
    Console.Write("\nProduct = " + product);
}

// Driver code
public static void Main()
{
    int n = 1012;

    countDigit(n);
}
}

// This code is contributed
// by Akanksha Rai(Abby_akku)

PHP

<?php
// PHP implementation of the above approach

// Print the sum and product of digits
// that divides the number.
function countDigit($n)
{
    $temp = $n;
    $sum = 0;
    $product = 1;
    while ($temp != 0) {

        // Fetching each digit of the number
        $d = $temp % 10;
        $temp =(int)($temp/10);

        // Checking if digit is greater than 0
        // and can divides n.
        if ($d > 0 && $n % $d == 0) {
            $sum += $d;
            $product *= $d;
        }
    }

    echo "Sum = ".$sum;
    echo "\nProduct = ".$product;
}

// Driver code

    $n = 1012;

    countDigit($n);

// This code is contributed by mits
?>

Javascript

<script>
// java script  implementation of the above approach

// Print the sum and product of digits
// that divides the number.
function countDigit(n)
{
    let temp = n;
    let sum = 0;
    let product = 1;
    while (temp != 0) {

        // Fetching each digit of the number
        let d = temp % 10;
        temp =parseInt(temp/10);

        // Checking if digit is greater than 0
        // and can divides n.
        if (d > 0 && n % d == 0) {
            sum += d;
            product *= d;
        }
    }

    document.write( "Sum = "+sum);
    document.write( "<br>Product = "+product);
}

// Driver code

    let n = 1012;

    countDigit(n);

// This code is contributed by Gottumukkala Bobby
</script>

Output:

Sum = 4
Product = 2

Time Complexity: O(log₁₀n), Auxiliary Space: O(1)