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 10, the output is always n+10. For example for n = 7, the 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 the result is 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++

#include<bits/stdc++.h> 
using namespace std; 
  
// 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; 
      
    // To store digits of result 
    // in reverse order 
    int res[MAX];  
  
    // Case 1: If number is smaller than 10 
    if (n < 10) 
    { 
        cout << 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) 
    { 
        cout << "Not possible"; 
        return; 
    } 
  
    // Print the result array in reverse order 
    for (i = j - 1; i >= 0; i--) 
        cout << res[i]; 
} 
  
// Driver Code 
int main() 
{ 
    findSmallest(7); 
    cout << "\n"; 
  
    findSmallest(36); 
    cout << "\n"; 
  
    findSmallest(13); 
    cout << "\n"; 
  
    findSmallest(100); 
    return 0; 
} 
  
// This code is contributed by Code_Mech 

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 
        # occurrences of current digit in res 
        while n % i == 0: 
            n = n / i 
            res.append(i) 
      
    # If n could not be broken in the 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 
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

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My Personal Notes

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

C++

C

Java

Python

C#

PHP

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