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 = 67Output:35Explanation:

At one’s place, we have digits 5 and 7, their product is 35. At ten’s place we have 6 inN2. AsN1has 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 = 1548Output:48Explanation:

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 <bits/stdc++.h>` `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

`<script>` `// Javascript program to calculate the` `// sum of same placed digits` `// of two numbers` `function` `sumOfProductOfDigits(n1, n2)` `{` ` ` `let sum = 0;` ` ` ` ` `// Loop until one of the numbers` ` ` `// have no digits remaining` ` ` `while` `(n1 > 0 && n2 > 0)` ` ` `{` ` ` `sum += ((n1 % 10) * (n2 % 10));` ` ` `n1 = Math.floor(n1/10);` ` ` `n2 = Math.floor(n2/10);` ` ` `}` ` ` `return` `sum;` `}` `// Driver Code` ` ` `let n1 = 25;` ` ` `let n2 = 1548;` ` ` `document.write(sumOfProductOfDigits(n1, n2));` `// This code is contributed by Mayank Tyagi` `</script>` |

**Output:**

48

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