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Smallest number whose product with N has sum of digits equal to that of N

  • Last Updated : 17 Aug, 2021

Given an integer N, the task is to find the smallest positive integer, which when multiplied by N, has sum of digits equal to the sum of digits of N.

Examples:

Input: N = 4
Output: 28
Explanation: 

  • Sum of digits of N = 4
  • 4 * 28 = 112
  • Sum of digits = 1 + 1 + 2 = 4, which is equal to sum of digits of N.

Input: N = 3029
Output: 37
Explanation: 

  • Sum of digits of N = 3 + 0 + 2 + 9 = 14
  • 3029 * 37 = 112073
  • Sum of digits = 1 + 1 + 2 + 0 + 7 + 3 = 14, which is equal to sum of digits of N.
 

Approach: Follow the steps to solve the problem:



  1. Since N can be large, take the input of N as a string. Calculate the sum of digits of N and store it in a variable, say S.
  2. Since the answer needs to exceed 10, starting from number 11, multiply it with N and store it in a variable, say res.
  3. Calculate the sum of digits of res and check if the sum of digits of res is equal to the S or not. If found to be true, then print the integer.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <iostream>
using namespace std;
 
// Function to find the minimum
// integer having sum of digits
// of a number multiplied by n
// equal to sum of digits of n
void find_num(string n)
{
   
    // Initialize answer
    int ans = 0;
    int sumOfDigitsN = 0;
   
    // Find sum of digits of N
    for(int c = 0; c < n.length(); c++)
    {
        sumOfDigitsN += n - '0';
         
    }
    int x=11;
       
    while(true)
    {
        // Multiply N with x
        int newNum = x * stoi(n);
        int tempSumDigits = 0;
        string temp = to_string(newNum);
       
        // Sum of digits of the new number
        for(int c = 0; c < temp.length(); c++)
        {
            tempSumDigits += temp - '0';
        }
       
        // If condition satisfies
        if (tempSumDigits == sumOfDigitsN)
        {
            ans = x;
            break;
        }
        //increase x
          x++;
    }
   
    // Print answer
    cout << ans << endl;
}
 
// Driver Code
int main()
{
    string N = "3029";
   
    // Function call
    find_num(N);
    return 0;
}
 
// This code is contributed by Sunil Sakariya

Java




// Java program for the above approach
 
class GFG {
 
    // Function to find the minimum
    // integer having sum of digits
    // of a number multiplied by n
    // equal to sum of digits of n
    static void find_num(String n)
    {
        // Initialize answer
        int ans = 0;
 
        // Convert string to
        // character array
        char[] digitsOfN
            = n.toCharArray();
 
        int sumOfDigitsN = 0;
 
        // Find sum of digits of N
        for (char c : digitsOfN) {
 
            sumOfDigitsN
                += Integer.parseInt(
                    Character.toString(c));
        }
 
        for (int x = 11; x > 0; x++) {
 
            // Multiply N with x
            int newNum
                = x * Integer.parseInt(n);
 
            int tempSumDigits = 0;
 
            char[] temp
                = Integer.toString(
                             newNum)
                      .toCharArray();
 
            // Sum of digits of the new number
            for (char c : temp) {
                tempSumDigits
                    += Integer.parseInt(
                        Character.toString(c));
            }
 
            // If condition satisfies
            if (tempSumDigits == sumOfDigitsN) {
                ans = x;
                break;
            }
        }
 
        // Print answer
        System.out.println(ans);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        String N = "3029";
 
        // Function call
        find_num(N);
    }
}

Python3




# Python3 program for the above approach
 
# Function to find the minimum
# integer having sum of digits
# of a number multiplied by n
# equal to sum of digits of n
def find_num(n):
     
    # Initialize answer
    ans = 0
 
    # Convert string to
    # character array
    digitsOfN = str(n)
 
    sumOfDigitsN = 0
 
    # Find sum of digits of N
    for c in digitsOfN:
        sumOfDigitsN += int(c)
 
    for x in range(11, 50):
         
        # Multiply N with x
        newNum = x * int(n)
 
        tempSumDigits = 0
 
        temp = str(newNum)
 
        # Sum of digits of the new number
        for c in temp:
            tempSumDigits += int(c)
        #print(tempSumDigits,newNum)
 
        # If condition satisfies
        if (tempSumDigits == sumOfDigitsN):
            ans = x
            break
 
    # Print answer
    print(ans)
 
# Driver Code
if __name__ == '__main__':
     
    N = "3029"
 
    # Function call
    find_num(N)
 
# This code is contributed by mohit kumar 29

C#




// C# program for the above approach 
using System;
using System.Globalization;
 
class GFG{
  
// Function to find the minimum
// integer having sum of digits
// of a number multiplied by n
// equal to sum of digits of n
static void find_num(string n)
{
     
    // Initialize answer
    int ans = 0;
 
    // Convert string to
    // character array
    char[] digitsOfN = n.ToCharArray();
 
    int sumOfDigitsN = 0;
 
    // Find sum of digits of N
    foreach(char c in digitsOfN)
    {
        sumOfDigitsN += Int32.Parse(
                Char.ToString(c));
    }
 
    for(int x = 11; x > 0; x++)
    {
         
        // Multiply N with x
        int newNum = x * Int32.Parse(n);
 
        int tempSumDigits = 0;
         
        string str = newNum.ToString();
        char[] temp = str.ToCharArray();
 
        // Sum of digits of the new number
        foreach(char c in temp)
        {
            tempSumDigits += Int32.Parse(
                    Char.ToString(c));
        }
 
        // If condition satisfies
        if (tempSumDigits == sumOfDigitsN)
        {
            ans = x;
            break;
        }
    }
 
    // Print answer
    Console.WriteLine(ans);
}
 
// Driver Code
public static void Main()
{
    string N = "3029";
     
    // Function call
    find_num(N);
}
}
 
// This code is contributed by susmitakundugoaldanga

Javascript




<script>
 
// Javascript program for the above approach
 
// Function to find the minimum
// integer having sum of digits
// of a number multiplied by n
// equal to sum of digits of n
function find_num(n)
{
   
    // Initialize answer
    var ans = 0;
    var sumOfDigitsN = 0;
   
    // Find sum of digits of N
    for(var c = 0; c < n.length; c++)
    {
        sumOfDigitsN += n - '0';
         
    }
    var x=11;
       
    while(true)
    {
        // Multiply N with x
        var newNum = x * parseInt(n);
        var tempSumDigits = 0;
        var temp = (newNum.toString());
       
        // Sum of digits of the new number
        for(var c = 0; c < temp.length; c++)
        {
            tempSumDigits += temp - '0';
        }
       
        // If condition satisfies
        if (tempSumDigits == sumOfDigitsN)
        {
            ans = x;
            break;
        }
        //increase x
          x++;
    }
   
    // Print answer
    document.write( ans );
}
 
// Driver Code
var N = "3029";
 
// Function call
find_num(N);
 
</script>
Output: 
37

 

Time Complexity: O(N)
Auxiliary Space: O(log N)

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