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# Add N digits to A such that it is divisible by B after each addition?

Given a, b and n. And we have to consider the following conditions and find the optimal solution to add n digits to a such that it is divisible by b after every iteration.

Add a digit to

**a**in such**a**way that after adding it,**a**is divisible by**b**.Print the smallest value of

**a**possible after n iterations of step1.Print

**fail**if the operation fails.

check divisibility after every digit addition.

**Input**

a=5 b=4 n=4

**Output**

52000

### Explanation

The first digit to be added from **0 **to **9**, if none of the digits make **a **divisible by **b **then the answer is **-1** which means the if **n **digits are added in **a**. a never be divided by **b **. else add the first digit that satisfies the condition and then add **0 **after that **(n-1)** times because if **a **is divisible by **b **then **a*10,** **a*100,** … will also be divisible by **b**.

## Example

#include <iostream> using namespace std; int main() { int a = 5, b = 4, n = 4; int num = a; for (int i = 0; i <= 9; i++) { int temp = a * 10 + i; if (temp % b == 0) { a = temp; break; } } if (num == a) { a = -1; } for (int j = 0; j < n - 1; j++) { a *= 10; } if(a>-1) { cout <<a; } else { cout <<”fail”; } return 0; }

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