# Sum of the products of same placed digits of two numbers

• Last Updated : 01 Mar, 2021

Given two positive integers N1 and N2, the task is to find the sum of the products of the same placed digits of the two numbers.
Note: For numbers of unequal length, the preceding digits of the smaller number needs to be treated as 0.
Examples:

Input: N1 = 5, N2 = 67
Output: 35
Explanation:
At one’s place, we have digits 5 and 7, their product is 35. At ten’s place we have 6 in N2. As N1 has no digit at ten’s place, 6 will be multiplied with 0, leading to no effect on the sum. Hence, the calculated sum is 35.
Input: N1 = 25, N2 = 1548
Output: 48
Explanation:
Sum = 5 * 8 + 2 * 4 + 0 * 5 + 0 * 1 = 48.

Approach:
To solve the problem mentioned above, we need to follow the steps below:

• Extract the rightmost digits of the two numbers and multiply them and add their product to sum.
• Now remove the digit.
• Keep repeating the above two steps until one of them is reduced to 0. Then, print the final value of sum calculated.

Below is the implementation of the above approach:

## C++

 `// C++ program to calculate the``// sum of same placed digits``// of two numbers``#include ``using` `namespace` `std;` `int` `sumOfProductOfDigits(``int` `n1, ``int` `n2)``{``    ``int` `sum = 0;``    ` `    ``// Loop until one of the numbers``    ``// have no digits remaining``    ``while` `(n1 > 0 && n2 > 0)``    ``{``        ``sum += ((n1 % 10) * (n2 % 10));``        ``n1 /= 10;``        ``n2 /= 10;``    ``}``    ``return` `sum;``}` `// Driver Code``int` `main()``{``    ``int` `n1 = 25;``    ``int` `n2 = 1548;` `    ``cout << sumOfProductOfDigits(n1, n2);``}` `// This code is contributed by grand_master`

## Java

 `// Java program to calculate the``// sum of same placed digits of``// two numbers` `class` `GFG {` `    ``// Function to find the sum of the``    ``// products of their corresponding digits``    ``static` `int` `sumOfProductOfDigits(``int` `n1,``                                    ``int` `n2)``    ``{``        ``int` `sum = ``0``;``        ``// Loop until one of the numbers``        ``// have no digits remaining``        ``while` `(n1 > ``0` `&& n2 > ``0``) {``            ``sum += ((n1 % ``10``) * (n2 % ``10``));``            ``n1 /= ``10``;``            ``n2 /= ``10``;``        ``}` `        ``return` `sum;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String args[])``    ``{` `        ``int` `n1 = ``25``;``        ``int` `n2 = ``1548``;` `        ``System.out.println(``            ``sumOfProductOfDigits(n1, n2));``    ``}``}`

## Python3

 `# Python3 program to calculate the``# sum of same placed digits``# of two numbers` `def` `sumOfProductOfDigits(n1, n2):` `    ``sum1 ``=` `0``;``    ` `    ``# Loop until one of the numbers``    ``# have no digits remaining``    ``while` `(n1 > ``0` `and` `n2 > ``0``):` `        ``sum1 ``+``=` `((n1 ``%` `10``) ``*` `(n2 ``%` `10``));``        ``n1 ``=` `n1 ``/``/` `10``;``        ``n2 ``=` `n2 ``/``/` `10``;``        ` `    ``return` `sum1;` `# Driver Code``n1 ``=` `25``;``n2 ``=` `1548``;` `print``(sumOfProductOfDigits(n1, n2));` `# This code is contributed by Nidhi_biet`

## C#

 `// C# program to calculate the``// sum of same placed digits of``// two numbers``using` `System;``class` `GFG{` `// Function to find the sum of the``// products of their corresponding digits``static` `int` `sumOfProductOfDigits(``int` `n1,``                                ``int` `n2)``{``    ``int` `sum = 0;``    ` `    ``// Loop until one of the numbers``    ``// have no digits remaining``    ``while` `(n1 > 0 && n2 > 0)``    ``{``        ``sum += ((n1 % 10) * (n2 % 10));``        ``n1 /= 10;``        ``n2 /= 10;``    ``}``    ``return` `sum;``}` `// Driver Code``public` `static` `void` `Main()``{``    ``int` `n1 = 25;``    ``int` `n2 = 1548;` `    ``Console.WriteLine(``            ``sumOfProductOfDigits(n1, n2));``}``}` `// This code is contributed by 29AjayKumar`

## Javascript

 ``
Output:
`48`

My Personal Notes arrow_drop_up