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

Here we will see how to generate a number A by adding N digits with it, and while adding new digits in each stage it will be divisible by another number B. Let us consider we are going to make a 5-digit number by adding 4 extra digits with it. We will check the divisibility by 7. The number will start from 8. So at first it will append 4 with it, so the number will be 84, that is divisible by 7. Then add 0 with the number so it will remain divisible by 7. If the number cannot be generated, it will return -1.

Algorithm

addNDigits(a, b, n)

begin
   num := a
   for all number x from 0 to 9, do
      temp := a * 10 + x
      if temp mod b is 0, then
         a := temp
         break
      end if
   done
   if num = a, then
      return -1
   end if
   add remaining 0’s with a
   return a.
end

Example

 Live Demo

#include<iostream>
using namespace std;
int add_n_digits(int a, int b, int n) {
   int num = a;
   for (int i = 0; i <= 9; i++) { //test by adding all digits (0-9)
      int tmp = a * 10 + i;
      if (tmp % b == 0) {
         a = tmp; //update a after adding
         break;
      }
   }
   if (num == a) //if no digit is added, return -1
      return -1;
   for (int j = 0; j < n - 1; j++) //after getting divisible number, add 0s
      a *= 10;
   return a;
}
main() {
   int a, b, n;
   cout << "Enter A, B and N: ";
   cin >> a >> b >> n;
   int res = add_n_digits(a, b, n);
   if(res == -1) {
      cout << "Unable to get this type of number";
   } else {
      cout << "Result is " << res;
   }
}

Output

Enter A, B and N: 8 7 4
Result is 84000

Output

Enter A, B and N: 10 11 5
Unable to get this type of number

Arnab Chakraborty

Published on 24-Jul-2019 12:54:28

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