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Find all numbers having digit product equal to k in 1 to N

Here given code implementation process.

/*
    Java program for
    Find all numbers having digit product equal to k in 1 to N
*/
public class Product
{
	public int findProduct(int num)
	{
		int n = num;
		int product = 1;
		// Calculate digit product
		while (n != 0 && product > 0)
		{
			product *= (n % 10);
			n = n / 10;
		}
		return product;
	}
	public void digitProduct(int n, int k)
	{
		if (n < 0)
		{
			return;
		}
		boolean result = false;
		System.out.print("\n Given k " + k);
		System.out.print("\n Product of number digit from (1 to " + 
                         n + ") is \n");
		for (int i = 0; i <= n; ++i)
		{
			if (findProduct(i) == k)
			{
				System.out.print("  " + i);
				result = true;
			}
		}
		if (result == false)
		{
			// When no result
			System.out.print(" None ");
		}
	}
	public static void main(String[] args)
	{
		Product task = new Product();
		// Test A
		int k = 40;
		int n = 100;
		/*
		    k = 40
		    n = 100
		    ----------
		    5 ✕ 8 = 40
		    8 ✕ 5 = 40
		    -----------
		    [58 84]
		*/
		task.digitProduct(n, k);
		// Test B
		k = 0;
		n = 80;
		/*
		    k = 0
		    n = 80
		    ------------
		    1 ✕ 0 = 0
		    2 ✕ 0 = 0
		    3 ✕ 0 = 0
		    4 ✕ 0 = 0
		    5 ✕ 0 = 0
		    6 ✕ 0 = 0 
		    7 ✕ 0 = 0
		    8 ✕ 0 = 0
		    ------------
		    [10  20  30  40  50  60  70  80]
		*/
		task.digitProduct(n, k);
	}
}

Output

 Given k 40
 Product of number digit from (1 to 100) is
  58  85
 Given k 0
 Product of number digit from (1 to 80) is
  10  20  30  40  50  60  70  80
// Include header file
#include <iostream>

using namespace std;
/*
    C++ program for
    Find all numbers having digit product equal to k in 1 to N
*/
class Product
{
	public: int findProduct(int num)
	{
		int n = num;
		int product = 1;
		// Calculate digit product
		while (n != 0 && product > 0)
		{
			product *= (n % 10);
			n = n / 10;
		}
		return product;
	}
	void digitProduct(int n, int k)
	{
		if (n < 0)
		{
			return;
		}
		bool result = false;
		cout << "\n Given k " << k;
		cout << "\n Product of number digit from (1 to " 
             << n << ") is \n";
		for (int i = 0; i <= n; ++i)
		{
			if (this->findProduct(i) == k)
			{
				cout << "  " << i;
				result = true;
			}
		}
		if (result == false)
		{
			// When no result
			cout << " None ";
		}
	}
};
int main()
{
	Product *task = new Product();
	// Test A
	int k = 40;
	int n = 100;
	/*
	    k = 40
	    n = 100
	    ----------
	    5 ✕ 8 = 40
	    8 ✕ 5 = 40
	    -----------
	    [58 84]
	*/
	task->digitProduct(n, k);
	// Test B
	k = 0;
	n = 80;
	/*
	    k = 0
	    n = 80
	    ------------
	    1 ✕ 0 = 0
	    2 ✕ 0 = 0
	    3 ✕ 0 = 0
	    4 ✕ 0 = 0
	    5 ✕ 0 = 0
	    6 ✕ 0 = 0 
	    7 ✕ 0 = 0
	    8 ✕ 0 = 0
	    ------------
	    [10  20  30  40  50  60  70  80]
	*/
	task->digitProduct(n, k);
	return 0;
}

Output

 Given k 40
 Product of number digit from (1 to 100) is
  58  85
 Given k 0
 Product of number digit from (1 to 80) is
  10  20  30  40  50  60  70  80
// Include namespace system
using System;
/*
    Csharp program for
    Find all numbers having digit product equal to k in 1 to N
*/
public class Product
{
	public int findProduct(int num)
	{
		int n = num;
		int product = 1;
		// Calculate digit product
		while (n != 0 && product > 0)
		{
			product *= (n % 10);
			n = n / 10;
		}
		return product;
	}
	public void digitProduct(int n, int k)
	{
		if (n < 0)
		{
			return;
		}
		Boolean result = false;
		Console.Write("\n Given k " + k);
		Console.Write("\n Product of number digit from (1 to " + 
                      n + ") is \n");
		for (int i = 0; i <= n; ++i)
		{
			if (this.findProduct(i) == k)
			{
				Console.Write("  " + i);
				result = true;
			}
		}
		if (result == false)
		{
			// When no result
			Console.Write(" None ");
		}
	}
	public static void Main(String[] args)
	{
		Product task = new Product();
		// Test A
		int k = 40;
		int n = 100;
		/*
		    k = 40
		    n = 100
		    ----------
		    5 ✕ 8 = 40
		    8 ✕ 5 = 40
		    -----------
		    [58 84]
		*/
		task.digitProduct(n, k);
		// Test B
		k = 0;
		n = 80;
		/*
		    k = 0
		    n = 80
		    ------------
		    1 ✕ 0 = 0
		    2 ✕ 0 = 0
		    3 ✕ 0 = 0
		    4 ✕ 0 = 0
		    5 ✕ 0 = 0
		    6 ✕ 0 = 0 
		    7 ✕ 0 = 0
		    8 ✕ 0 = 0
		    ------------
		    [10  20  30  40  50  60  70  80]
		*/
		task.digitProduct(n, k);
	}
}

Output

 Given k 40
 Product of number digit from (1 to 100) is
  58  85
 Given k 0
 Product of number digit from (1 to 80) is
  10  20  30  40  50  60  70  80
package main
import "fmt"
/*
    Go program for
    Find all numbers having digit product equal to k in 1 to N
*/

func findProduct(num int) int {
	var n int = num
	var product int = 1
	// Calculate digit product
	for (n != 0 && product > 0) {
		product *= (n % 10)
		n = n / 10
	}
	return product
}
func digitProduct(n, k int) {
	if n < 0 {
		return
	}
	var result bool = false
	fmt.Print("\n Given k ", k)
	fmt.Print("\n Product of number digit from (1 to ", 
		n, ") is \n")
	for i := 0 ; i <= n ; i++ {
		if findProduct(i) == k {
			fmt.Print("  ", i)
			result = true
		}
	}
	if result == false {
		// When no result
		fmt.Print(" None ")
	}
}
func main() {
	
	// Test A
	var k int = 40
	var n int = 100
	/*
	    k = 40
	    n = 100
	    ----------
	    5 ✕ 8 = 40
	    8 ✕ 5 = 40
	    -----------
	    [58 84]
	*/
	digitProduct(n, k)
	// Test B
	k = 0
	n = 80
	/*
	    k = 0
	    n = 80
	    ------------
	    1 ✕ 0 = 0
	    2 ✕ 0 = 0
	    3 ✕ 0 = 0
	    4 ✕ 0 = 0
	    5 ✕ 0 = 0
	    6 ✕ 0 = 0 
	    7 ✕ 0 = 0
	    8 ✕ 0 = 0
	    ------------
	    [10  20  30  40  50  60  70  80]
	*/
	digitProduct(n, k)
}

Output

 Given k 40
 Product of number digit from (1 to 100) is
  58  85
 Given k 0
 Product of number digit from (1 to 80) is
  10  20  30  40  50  60  70  80
<?php
/*
    Php program for
    Find all numbers having digit product equal to k in 1 to N
*/
class Product
{
	public	function findProduct($num)
	{
		$n = $num;
		$product = 1;
		// Calculate digit product
		while ($n != 0 && $product > 0)
		{
			$product *= ($n % 10);
			$n = (int)($n / 10);
		}
		return $product;
	}
	public	function digitProduct($n, $k)
	{
		if ($n < 0)
		{
			return;
		}
		$result = false;
		echo("\n Given k ".$k);
		echo("\n Product of number digit from (1 to ".$n.
			") is \n");
		for ($i = 0; $i <= $n; ++$i)
		{
			if ($this->findProduct($i) == $k)
			{
				echo("  ".$i);
				$result = true;
			}
		}
		if ($result == false)
		{
			// When no result
			echo(" None ");
		}
	}
}

function main()
{
	$task = new Product();
	// Test A
	$k = 40;
	$n = 100;
	/*
	    k = 40
	    n = 100
	    ----------
	    5 ✕ 8 = 40
	    8 ✕ 5 = 40
	    -----------
	    [58 84]
	*/
	$task->digitProduct($n, $k);
	// Test B
	$k = 0;
	$n = 80;
	/*
	    k = 0
	    n = 80
	    ------------
	    1 ✕ 0 = 0
	    2 ✕ 0 = 0
	    3 ✕ 0 = 0
	    4 ✕ 0 = 0
	    5 ✕ 0 = 0
	    6 ✕ 0 = 0 
	    7 ✕ 0 = 0
	    8 ✕ 0 = 0
	    ------------
	    [10  20  30  40  50  60  70  80]
	*/
	$task->digitProduct($n, $k);
}
main();

Output

 Given k 40
 Product of number digit from (1 to 100) is
  58  85
 Given k 0
 Product of number digit from (1 to 80) is
  10  20  30  40  50  60  70  80
/*
    Node JS program for
    Find all numbers having digit product equal to k in 1 to N
*/
class Product
{
	findProduct(num)
	{
		var n = num;
		var product = 1;
		// Calculate digit product
		while (n != 0 && product > 0)
		{
			product *= (n % 10);
			n = parseInt(n / 10);
		}
		return product;
	}
	digitProduct(n, k)
	{
		if (n < 0)
		{
			return;
		}
		var result = false;
		process.stdout.write("\n Given k " + k);
		process.stdout.write("\n Product of number digit from (1 to " + 
                             n + ") is \n");
		for (var i = 0; i <= n; ++i)
		{
			if (this.findProduct(i) == k)
			{
				process.stdout.write("  " + i);
				result = true;
			}
		}
		if (result == false)
		{
			// When no result
			process.stdout.write(" None ");
		}
	}
}

function main()
{
	var task = new Product();
	// Test A
	var k = 40;
	var n = 100;
	/*
	    k = 40
	    n = 100
	    ----------
	    5 ✕ 8 = 40
	    8 ✕ 5 = 40
	    -----------
	    [58 84]
	*/
	task.digitProduct(n, k);
	// Test B
	k = 0;
	n = 80;
	/*
	    k = 0
	    n = 80
	    ------------
	    1 ✕ 0 = 0
	    2 ✕ 0 = 0
	    3 ✕ 0 = 0
	    4 ✕ 0 = 0
	    5 ✕ 0 = 0
	    6 ✕ 0 = 0 
	    7 ✕ 0 = 0
	    8 ✕ 0 = 0
	    ------------
	    [10  20  30  40  50  60  70  80]
	*/
	task.digitProduct(n, k);
}
main();

Output

 Given k 40
 Product of number digit from (1 to 100) is
  58  85
 Given k 0
 Product of number digit from (1 to 80) is
  10  20  30  40  50  60  70  80
#    Python 3 program for
#    Find all numbers having digit product equal to k in 1 to N
class Product :
	def findProduct(self, num) :
		n = num
		product = 1
		#  Calculate digit product
		while (n != 0 and product > 0) :
			product *= (n % 10)
			n = int(n / 10)
		
		return product
	
	def digitProduct(self, n, k) :
		if (n < 0) :
			return
		
		result = False
		print("\n Given k ", k, end = "")
		print("\n Product of number digit from (1 to ",
              n ,") is ")
		i = 0
		while (i <= n) :
			if (self.findProduct(i) == k) :
				print("  ", i, end = "")
				result = True
			
			i += 1
		
		if (result == False) :
			#  When no result
			print(" None ", end = "")
		
	

def main() :
	task = Product()
	#  Test A
	k = 40
	n = 100
	#    k = 40
	#    n = 100
	#    ----------
	#    5 ✕ 8 = 40
	#    8 ✕ 5 = 40
	#    -----------
	#    [58 84]
	task.digitProduct(n, k)
	#  Test B
	k = 0
	n = 80
	#    k = 0
	#    n = 80
	#    ------------
	#    1 ✕ 0 = 0
	#    2 ✕ 0 = 0
	#    3 ✕ 0 = 0
	#    4 ✕ 0 = 0
	#    5 ✕ 0 = 0
	#    6 ✕ 0 = 0 
	#    7 ✕ 0 = 0
	#    8 ✕ 0 = 0
	#    ------------
	#    [10  20  30  40  50  60  70  80]
	task.digitProduct(n, k)

if __name__ == "__main__": main()

Output

 Given k  40
 Product of number digit from (1 to  100 ) is
   58   85
 Given k  0
 Product of number digit from (1 to  80 ) is
   10   20   30   40   50   60   70   80
#    Ruby program for
#    Find all numbers having digit product equal to k in 1 to N
class Product 
	def findProduct(num) 
		n = num
		product = 1
		#  Calculate digit product
		while (n != 0 && product > 0) 
			product *= (n % 10)
			n = n / 10
		end

		return product
	end

	def digitProduct(n, k) 
		if (n < 0) 
			return
		end

		result = false
		print("\n Given k ", k)
		print("\n Product of number digit from (1 to ", n ,") is \n")
		i = 0
		while (i <= n) 
			if (self.findProduct(i) == k) 
				print("  ", i)
				result = true
			end

			i += 1
		end

		if (result == false) 
			#  When no result
			print(" None ")
		end

	end

end

def main() 
	task = Product.new()
	#  Test A
	k = 40
	n = 100
	#    k = 40
	#    n = 100
	#    ----------
	#    5 ✕ 8 = 40
	#    8 ✕ 5 = 40
	#    -----------
	#    [58 84]
	task.digitProduct(n, k)
	#  Test B
	k = 0
	n = 80
	#    k = 0
	#    n = 80
	#    ------------
	#    1 ✕ 0 = 0
	#    2 ✕ 0 = 0
	#    3 ✕ 0 = 0
	#    4 ✕ 0 = 0
	#    5 ✕ 0 = 0
	#    6 ✕ 0 = 0 
	#    7 ✕ 0 = 0
	#    8 ✕ 0 = 0
	#    ------------
	#    [10  20  30  40  50  60  70  80]
	task.digitProduct(n, k)
end

main()

Output

 Given k 40
 Product of number digit from (1 to 100) is 
  58  85
 Given k 0
 Product of number digit from (1 to 80) is 
  10  20  30  40  50  60  70  80
/*
    Scala program for
    Find all numbers having digit product equal to k in 1 to N
*/
class Product()
{
	def findProduct(num: Int): Int = {
		var n: Int = num;
		var product: Int = 1;
		// Calculate digit product
		while (n != 0 && product > 0)
		{
			product *= (n % 10);
			n = n / 10;
		}
		return product;
	}
	def digitProduct(n: Int, k: Int): Unit = {
		if (n < 0)
		{
			return;
		}
		var result: Boolean = false;
		print("\n Given k " + k);
		print("\n Product of number digit from (1 to " + n + ") is \n");
		var i: Int = 0;
		while (i <= n)
		{
			if (findProduct(i) == k)
			{
				print("  " + i);
				result = true;
			}
			i += 1;
		}
		if (result == false)
		{
			// When no result
			print(" None ");
		}
	}
}
object Main
{
	def main(args: Array[String]): Unit = {
		var task: Product = new Product();
		// Test A
		var k: Int = 40;
		var n: Int = 100;
		/*
		    k = 40
		    n = 100
		    ----------
		    5 ✕ 8 = 40
		    8 ✕ 5 = 40
		    -----------
		    [58 84]
		*/
		task.digitProduct(n, k);
		// Test B
		k = 0;
		n = 80;
		/*
		    k = 0
		    n = 80
		    ------------
		    1 ✕ 0 = 0
		    2 ✕ 0 = 0
		    3 ✕ 0 = 0
		    4 ✕ 0 = 0
		    5 ✕ 0 = 0
		    6 ✕ 0 = 0 
		    7 ✕ 0 = 0
		    8 ✕ 0 = 0
		    ------------
		    [10  20  30  40  50  60  70  80]
		*/
		task.digitProduct(n, k);
	}
}

Output

 Given k 40
 Product of number digit from (1 to 100) is
  58  85
 Given k 0
 Product of number digit from (1 to 80) is
  10  20  30  40  50  60  70  80
/*
    Swift 4 program for
    Find all numbers having digit product equal to k in 1 to N
*/
class Product
{
	func findProduct(_ num: Int) -> Int
	{
		var n: Int = num;
		var product: Int = 1;
		// Calculate digit product
		while (n  != 0 && product > 0)
		{
			product *= (n % 10);
			n = n / 10;
		}
		return product;
	}
	func digitProduct(_ n: Int, _ k: Int)
	{
		if (n < 0)
		{
			return;
		}
		var result: Bool = false;
		print("\n Given k ", k, terminator: "");
		print("\n Product of number digit from (1 to ", n ,") is ");
		var i: Int = 0;
		while (i <= n)
		{
			if (self.findProduct(i) == k)
			{
				print("  ", i, terminator: "");
				result = true;
			}
			i += 1;
		}
		if (result == false)
		{
			// When no result
			print(" None ", terminator: "");
		}
	}
}
func main()
{
	let task: Product = Product();
	// Test A
	var k: Int = 40;
	var n: Int = 100;
	/*
	    k = 40
	    n = 100
	    ----------
	    5 ✕ 8 = 40
	    8 ✕ 5 = 40
	    -----------
	    [58 84]
	*/
	task.digitProduct(n, k);
	// Test B
	k = 0;
	n = 80;
	/*
	    k = 0
	    n = 80
	    ------------
	    1 ✕ 0 = 0
	    2 ✕ 0 = 0
	    3 ✕ 0 = 0
	    4 ✕ 0 = 0
	    5 ✕ 0 = 0
	    6 ✕ 0 = 0 
	    7 ✕ 0 = 0
	    8 ✕ 0 = 0
	    ------------
	    [10  20  30  40  50  60  70  80]
	*/
	task.digitProduct(n, k);
}
main();

Output

 Given k  40
 Product of number digit from (1 to  100 ) is
   58   85
 Given k  0
 Product of number digit from (1 to  80 ) is
   10   20   30   40   50   60   70   80
/*
    Kotlin program for
    Find all numbers having digit product equal to k in 1 to N
*/
class Product
{
	fun findProduct(num: Int): Int
	{
		var n: Int = num;
		var product: Int = 1;
		// Calculate digit product
		while (n != 0 && product > 0)
		{
			product *= (n % 10);
			n = n / 10;
		}
		return product;
	}
	fun digitProduct(n: Int, k: Int): Unit
	{
		if (n < 0)
		{
			return;
		}
		var result: Boolean = false;
		print("\n Given k " + k);
		print("\n Product of number digit from (1 to " + 
              n + ") is \n");
		var i: Int = 0;
		while (i <= n)
		{
			if (this.findProduct(i) == k)
			{
				print("  " + i);
				result = true;
			}
			i += 1;
		}
		if (result == false)
		{
			// When no result
			print(" None ");
		}
	}
}
fun main(args: Array < String > ): Unit
{
	val task: Product = Product();
	// Test A
	var k: Int = 40;
	var n: Int = 100;
	/*
	    k = 40
	    n = 100
	    ----------
	    5 ✕ 8 = 40
	    8 ✕ 5 = 40
	    -----------
	    [58 84]
	*/
	task.digitProduct(n, k);
	// Test B
	k = 0;
	n = 80;
	/*
	    k = 0
	    n = 80
	    ------------
	    1 ✕ 0 = 0
	    2 ✕ 0 = 0
	    3 ✕ 0 = 0
	    4 ✕ 0 = 0
	    5 ✕ 0 = 0
	    6 ✕ 0 = 0 
	    7 ✕ 0 = 0
	    8 ✕ 0 = 0
	    ------------
	    [10  20  30  40  50  60  70  80]
	*/
	task.digitProduct(n, k);
}

Output

 Given k 40
 Product of number digit from (1 to 100) is
  58  85
 Given k 0
 Product of number digit from (1 to 80) is
  10  20  30  40  50  60  70  80
Last updated on by Kalkicode




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