Smallest number divisible by n and has at-least k trailing zeros

Two integers n and k are given. Our task is to print K-rounding of n. K-rounding is the minimum positive integer X, such that x ends with k or more zeros and is divisible by n.

Examples :

Input :  n = 30, k = 3.
Output : 3000
3000 is the smallest number that
has at-least k 0s and is divisible
by n.

Input : n = 375, k = 4.
Output : 30000

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Method 1 :
The brute force approach is to start with result = 10^k. Check if result is divided by n. If yes, it’s the answer, else increase it by 10^k

Method 2 : The efficient approach is to calculate the LCM of 10^k and n.
Suppose, n = 375, k = 4.
result = 10000.
Now, LCM of 375 and 10000 is the lowest number divided by both of them.
It will contain k or more zeros (because it is multiple of 10^k) and will be a multiple of n as well.

Below is the implementation :

C++

// CPP code to print K-rounded value of n 
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to compute the rounded value 
long long getRounding(long long n, long long k) 
{ 
    long long rounding = pow(10, k); 
  
    // Computing GCD 
    long long result = __gcd(rounding, n); 
  
    // Returning LCM (GCD * LCM = n * k) 
    return ((rounding * n) / result); 
} 
  
// Driver Code 
int main() 
{ 
  
    long long n = 375, k = 4; 
  
    // Function call 
    cout << getRounding(n, k); 
  
    return 0; 
} 

Java

// JAVA Code For Smallest number divisible by 
// n and has at-least k trailing zeros 
import java.util.*; 
  
class GFG { 
      
     // Function to find gcd 
     static long gcd(long a, long b) 
        { 
            // Everything divides 0  
            if (a == 0 || b == 0) 
               return 0; 
            
            // base case 
            if (a == b) 
                return a; 
            
            // a is greater 
            if (a > b) 
                return gcd(a-b, b); 
            return gcd(a, b-a); 
        } 
  
    // Function to compute the rounded value 
    public static long getRounding(long n, long k) 
    { 
        long rounding = (long)Math.pow(10, k); 
       
        // Computing GCD 
        long result = gcd(rounding, n); 
       
        // Returning LCM (GCD * LCM = n * k) 
        return ((rounding * n) / result); 
    } 
      
    /* Driver program to test above function */
    public static void main(String[] args)  
    { 
        long n = 375, k = 4; 
           
        // Function call 
        System.out.println( getRounding(n, k)); 
          
    } 
} 
    
// This code is contributed by Arnav Kr. Mandal. 

Python3

# python Code For Smallest number  
# divisible by n and has 
# at-least k trailing zeros 
  
# Function to find gcd 
def gcd(a, b): 
      
    # Everything divides 0  
    if (a == 0 or b == 0): 
        return 0
              
    # base case 
    if (a == b): 
        return a 
              
    # a is greater 
    if (a > b): 
        return gcd(a - b, b) 
          
    return gcd(a, b - a) 
          
# Function to compute the  
# rounded value 
def getRounding(n, k): 
      
    rounding = pow(10, k); 
  
    # Computing GCD 
    result = gcd(rounding, n) 
  
    # Returning LCM (GCD * LCM 
    # = n * k) 
    return ((rounding * n) / result) 
  
# Driver Code 
  
n = 375
k = 4
  
# Function call 
print( int(getRounding(n, k))) 
  
# This code is contributed by Sam007 

C#

// C# Code For Smallest number  
// divisible by n and has 
// at-least k trailing zeros 
using System; 
  
class GFG { 
      
    // Function to find gcd 
    static long gcd(long a, long b) 
        { 
              
            // Everything divides 0  
            if (a == 0 || b == 0) 
            return 0; 
              
            // base case 
            if (a == b) 
                return a; 
              
            // a is greater 
            if (a > b) 
                return gcd(a - b, b); 
            return gcd(a, b - a); 
        } 
  
    // Function to compute the rounded value 
    public static long getRounding(long n, long k) 
    { 
        long rounding = (long)Math.Pow(10, k); 
      
        // Computing GCD 
        long result = gcd(rounding, n); 
      
        // Returning LCM (GCD * LCM = n * k) 
        return ((rounding * n) / result); 
    } 
      
    // Driver Code 
    public static void Main()  
    { 
        long n = 375, k = 4; 
          
        // Function call 
        Console.Write( getRounding(n, k)); 
          
    } 
} 
      
// This code is contributed by Nitin Mittal. 

PHP

<?php 
// PHP Code For Smallest number  
// divisible by n and has 
// at-least k trailing zeros 
function gcd($a, $b) 
{ 
      
    // Everything divides 0  
    if ($a == 0 || $b == 0) 
    return 0; 
      
    // base case 
    if ($a == $b) 
        return $a; 
      
    // a is greater 
    if ($a > $b) 
        return gcd($a - $b, $b); 
    return gcd($a, $b - $a); 
} 
  
// Function to compute  
// the rounded value 
function getRounding($n, $k) 
{ 
    $rounding = intval(pow(10, $k)); 
  
    // Computing GCD 
    $result = gcd($rounding, $n); 
  
    // Returning LCM (GCD * LCM = n * k) 
    return intval(($rounding * $n) /  
                   $result); 
} 
  
// Driver code 
$n = 375; 
$k = 4; 
  
// Function call 
echo getRounding($n, $k); 
  
// This code is contributed by Sam007 
?> 

Output :

This article is contributed by Rohit Thapliyal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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