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


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


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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