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Smallest number divisible by n and has at-least k trailing zeros
  • Difficulty Level : Hard
  • Last Updated : 25 Jun, 2018

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

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

Method 2 : The efficient approach is to calculate the LCM of 10k 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 10k) 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 :

30000

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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