Find the smallest number whose digits multiply to a given number n

Given a number ‘n’, find the smallest number ‘p’ such that if we multiply all digits of ‘p’, we get ‘n’. The result ‘p’ should have minimum two digits.

Examples:

Input:  n = 36
Output: p = 49 
// Note that 4*9 = 36 and 49 is the smallest such number

Input:  n = 100
Output: p = 455
// Note that 4*5*5 = 100 and 455 is the smallest such number

Input: n = 1
Output:p = 11
// Note that 1*1 = 1

Input: n = 13
Output: Not Possible

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

For a given n, following are the two cases to be considered.
Case 1: n < 10 When n is smaller than n, the output is always n+10. For example for n = 7, output is 17. For n = 9, output is 19.

Case 2: n >= 10 Find all factors of n which are between 2 and 9 (both inclusive). The idea is to start searching from 9 so that the number of digits in result are minimized. For example 9 is preferred over 33 and 8 is preferred over 24.
Store all found factors in an array. The array would contain digits in non-increasing order, so finally print the array in reverse order.

Following is the implementation of above concept.

C/C++

#include<stdio.h>

// Maximum number of digits in output
#define MAX 50

// prints the smallest number whose digits multiply to n
void findSmallest(int n)
{
    int i, j=0;
    int res[MAX]; // To sore digits of result in reverse order

    // Case 1: If number is smaller than 10
    if (n < 10)
    {
        printf("%d", n+10);
        return;
    }

    // Case 2: Start with 9 and try every possible digit
    for (i=9; i>1; i--)
    {
        // If current digit divides n, then store all
        // occurrences of current digit in res
        while (n%i == 0)
        {
            n = n/i;
            res[j] = i;
            j++;
        }
    }

    // If n could not be broken in form of digits (prime factors of n
    // are greater than 9)
    if (n > 10)
    {
        printf("Not possible");
        return;
    }

    // Print the result array in reverse order
    for (i=j-1; i>=0; i--)
        printf("%d", res[i]);
}

// Driver program to test above function
int main()
{
    findSmallest(7);
    printf("\n");

    findSmallest(36);
    printf("\n");

    findSmallest(13);
    printf("\n");

    findSmallest(100);
    return 0;
}

Java

// Java program to find the smallest number whose 
// digits multiply to a given number n

import java.io.*;

class Smallest
{
    // Function to prints the smallest number whose 
    // digits multiply to n
    static void findSmallest(int n)
    {
        int i, j=0;
        int MAX = 50;
        // To sore digits of result in reverse order
        int[] res = new int[MAX]; 
 
        // Case 1: If number is smaller than 10
        if (n < 10)
        {
            System.out.println(n+10);
            return;
        }
 
        // Case 2: Start with 9 and try every possible digit
        for (i=9; i>1; i--)
        {
            // If current digit divides n, then store all
            // occurrences of current digit in res
            while (n%i == 0)
            {
                n = n/i;
                res[j] = i;
                j++;
            }
        }
 
        // If n could not be broken in form of digits (prime factors of n
        // are greater than 9)
        if (n > 10)
        {
            System.out.println("Not possible");
            return;
        }
 
        // Print the result array in reverse order
        for (i=j-1; i>=0; i--)
            System.out.print(res[i]);
        System.out.println();
    }
    
    // Driver program
    public static void main (String[] args) 
    {
        findSmallest(7);
        findSmallest(36);
        findSmallest(13);
        findSmallest(100);
    }
}

// Contributed by Pramod Kumar

Python

# Python code to find the smallest number 
# whose digits multiply to give n

# function to print the smallest number whose 
# digits multiply to n
def findSmallest(n):
    # Case 1 - If the number is smaller than 10
    if n < 10:
        print n+10
        return
    
    # Case 2 - Start with 9 and try every possible digit
    res = [] # to sort digits
    for i in range(9,1,-1):
        # If current digit divides n, then store all
        # occurences of current digit in res
        while n % i == 0:
            n = n / i
            res.append(i)
    
    # If n could not be broken in form of digits
    # prime factors of  n are greater than 9
    
    if n > 10:
        print "Not Possible"
        return
        
    # Print the number from result array in reverse order
    n = res[len(res)-1]
    for i in range(len(res)-2,-1,-1):
        n = 10 * n + res[i]
    print n
    
# Driver Code to test above function

findSmallest(7)

findSmallest(36)

findSmallest(13)

findSmallest(100)

# This code is contributed by Harshit Agrawal

C#

// C# program to find the smallest number whose 
// digits multiply to a given number n
using System;

class GFG {
    
    // Function to prints the smallest number
    // whose digits multiply to n
    static void findSmallest(int n)
    {
        
        int i, j=0;
        int MAX = 50;
        
        // To sore digits of result in
        // reverse order
        int []res = new int[MAX]; 

        // Case 1: If number is smaller than 10
        if (n < 10)
        {
            Console.WriteLine(n + 10);
            return;
        }

        // Case 2: Start with 9 and try every
        // possible digit
        for (i = 9; i > 1; i--)
        {
            
            // If current digit divides n, then
            // store all occurrences of current
            // digit in res
            while (n % i == 0)
            {
                n = n / i;
                res[j] = i;
                j++;
            }
        }

        // If n could not be broken in form of
        // digits (prime factors of n
        // are greater than 9)
        if (n > 10)
        {
            Console.WriteLine("Not possible");
            return;
        }

        // Print the result array in reverse order
        for (i = j-1; i >= 0; i--)
            Console.Write(res[i]);
            
        Console.WriteLine();
    }
    
    // Driver program
    public static void Main () 
    {
        findSmallest(7);
        findSmallest(36);
        findSmallest(13);
        findSmallest(100);
    }
}

// This code is contributed by nitin mittal.

PHP


<?php
// PHP program to find the 
// smallest number whose 
// digits multiply to a 
// given number n prints the
// smallest number whose digits
// multiply to n

function findSmallest($n)
{
    
    // To sore digits of 
    // result in reverse order
    $i;
    $j = 0;
    $res; 

    // Case 1: If number is 
    // smaller than 10
    if ($n < 10)
    {
        echo $n + 10;
        return;
    }

    // Case 2: Start with 9 and 
    // try every possible digit
    for ($i = 9; $i > 1; $i--)
    {
        
        // If current digit divides 
        // n, then store all
        // occurrences of current 
        // digit in res
        while ($n % $i == 0)
        {
            $n = $n / $i;
            $res[$j] = $i;
            $j++;
        }
    }

    // If n could not be broken 
    // in form of digits 
    // (prime factors of n
    // are greater than 9)
    if ($n > 10)
    {
        echo "Not possible";
        return;
    }

    // Print the result 
    // array in reverse order
    for ($i = $j - 1; $i >= 0; $i--)
        echo $res[$i];
}

    // Driver Code
    findSmallest(7);
    echo "\n";

    findSmallest(36);
    
    echo "\n";

    findSmallest(13);
    echo "\n";

    findSmallest(100);

// This code is contributed by ajit
?>

Output:

17
49
Not possible
455

This article is contributed by Ashish Bansal. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

My Personal Notes

Improved By : nitin mittal, jit_t

Practice Tags :

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