Smallest number greater than or equal to N which is divisible by its non-zero digits

Given an integer N, the task is to find the smallest number greater than or equal to N such that it is divisible by all of its non-zero digits.

Examples:

Input: N = 31
Output: 33
Explanation: 33 is the smallest number satisfying the given condition. 
At Unit’s place: 33%3 = 0
At One’s place: 33%3 = 0

Input: N = 30
Output: 30
Explanation: 30 is the smallest number satisfying the given condition. 
At One’s place: 30%3 = 0

Approach: Smallest number which is divisible by all digits from 1 to 9 is equal to the LCM of (1, 2, 3, 4, 5, 6, 7, 8, 9) = 2520. Therefore, the multiples of 2520 are also divisible by all digits from 1 to 9 implying that (N + 2520) will always satisfy the condition. Therefore, iterate in the range [N, 2520 + N] and check for the smallest number satisfying the given condition. Follow the steps below to solve the problem:

• Initialize ans as 0 to store the smallest number greater than or equal to N such that it is divisible by all its non-zero digits.
• Iterate over the range [N, N + 2520] using the variable i.
  • Initialize a variable possible as 1 to check if the current number i satisfies the given condition or not.
  • Get all non-zero digits of i and check if i is divisible by each of them. If found to be true, then update possible to 1, and update ans as i, and break out of the loop.
• After the above steps, print the value of ans as the result.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the smallest number
// greater than or equal to N such that
// it is divisible by its non-zero digits
void findSmallestNumber(int n)
{

    // Iterate in range[N, N + 2520]
    for (int i = n; i <= (n + 2520); ++i) {

        // To check if the current number
        // satisfies the given condition
        bool possible = 1;

        // Store the number in a temporary
        // variable
        int temp = i;

        // Loop until temp > 0
        while (temp) {

            // Check only for non zero digits
            if (temp % 10 != 0) {

                // Extract the current digit
                int digit = temp % 10;

                // If number is divisible
                // by current digit or not
                if (i % digit != 0) {

                    // Otherwise, set
                    // possible to 0
                    possible = 0;

                    // Break out of the loop
                    break;
                }
            }

            // Divide by 10
            temp /= 10;
        }

        if (possible == 1) {
            cout << i;
            return;
        }
    }
}

// Driver Code
int main()
{
    int N = 31;

    // Function Call
    findSmallestNumber(N);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG{

// Function to find the smallest number
// greater than or equal to N such that
// it is divisible by its non-zero digits
static void findSmallestNumber(int n)
{

    // Iterate in range[N, N + 2520]
    for(int i = n; i <= (n + 2520); ++i)
    {

        // To check if the current number
        // satisfies the given condition
        int possible = 1;

        // Store the number in a temporary
        // variable
        int temp = i;

        // Loop until temp > 0
        while (temp != 0)
        {

            // Check only for non zero digits
            if (temp % 10 != 0)
            {

                // Extract the current digit
                int digit = temp % 10;

                // If number is divisible
                // by current digit or not
                if (i % digit != 0)
                {

                    // Otherwise, set
                    // possible to 0
                    possible = 0;

                    // Break out of the loop
                    break;
                }
            }

            // Divide by 10
            temp /= 10;
        }

        if (possible == 1)
        {
            System.out.println(i);
            return;
        }
    }
}

// Driver code
public static void main(String[] args)
{
    int N = 31;

    // Function Call
    findSmallestNumber(N);
}
}

// This code is contributed by susmitakundugoaldanga

Python3

# Python3 program for the above approach

# Function to find the smallest number
# greater than or equal to N such that
# it is divisible by its non-zero digits
def findSmallestNumber(n):

    # Iterate in range[N, N + 2520]
    for i in range(n, n + 2521):

        # To check if the current number
        # satisfies the given condition
        possible = 1

        # Store the number in a temporary
        # variable
        temp = i

        # Loop until temp > 0
        while (temp):

            # Check only for non zero digits
            if (temp % 10 != 0):

                # Extract the current digit
                digit = temp % 10

                # If number is divisible
                # by current digit or not
                if (i % digit != 0):

                    # Otherwise, set
                    # possible to 0
                    possible = 0

                    # Break out of the loop
                    break

            # Divide by 10
            temp //= 10

        if (possible == 1):
            print(i, end = "")
            return

# Driver Code
if __name__ == "__main__" :

    N = 31

    # Function Call
    findSmallestNumber(N)

# This code is contributed by AnkThon

C#

// C# program for the above approach
using System;

class GFG{

// Function to find the smallest number
// greater than or equal to N such that
// it is divisible by its non-zero digits
static void findSmallestNumber(int n)
{

    // Iterate in range[N, N + 2520]
    for(int i = n; i <= (n + 2520); ++i)
    {

        // To check if the current number
        // satisfies the given condition
        int possible = 1;

        // Store the number in a temporary
        // variable
        int temp = i;

        // Loop until temp > 0
        while (temp != 0)
        {

            // Check only for non zero digits
            if (temp % 10 != 0)
            {

                // Extract the current digit
                int digit = temp % 10;

                // If number is divisible
                // by current digit or not
                if (i % digit != 0)
                {

                    // Otherwise, set
                    // possible to 0
                    possible = 0;

                    // Break out of the loop
                    break;
                }
            }

            // Divide by 10
            temp /= 10;
        }

        if (possible == 1)
        {
            Console.Write(i);
            return;
        }
    }
}

// Driver code
public static void Main(String[] args)
{
    int N = 31;

    // Function Call
    findSmallestNumber(N);
}
}

// This code is contributed by shivanisinghss2110

Javascript

<script>
// javascript program for the above approach

// Function to find the smallest number
// greater than or equal to N such that
// it is divisible by its non-zero digits
function findSmallestNumber(n)
{

    // Iterate in range[N, N + 2520]
    for(i = n; i <= (n + 2520); ++i)
    {

        // To check if the current number
        // satisfies the given condition
        var possible = 1;

        // Store the number in a temporary
        // variable
        var temp = i;

        // Loop until temp > 0
        while (temp != 0)
        {

            // Check only for non zero digits
            if (temp % 10 != 0)
            {

                // Extract the current digit
                var digit = temp % 10;

                // If number is divisible
                // by current digit or not
                if (i % digit != 0)
                {

                    // Otherwise, set
                    // possible to 0
                    possible = 0;

                    // Break out of the loop
                    break;
                }
            }

            // Divide by 10
            temp  = parseInt(temp / 10);
        }

        if (possible == 1)
        {
            document.write(i);
            return;
        }
    }
}

// Driver code
var N = 31;

// Function Call
findSmallestNumber(N);

// This code is contributed by shikhasingrajput
</script>

Output:

33

Time Complexity: O(2520*log₁₀N)
Auxiliary Space: O(1)