Smallest K digit number divisible by X

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Integers X and K are given. The task is to find the smallest K-digit number divisible by X. 

Examples : 

Input : X = 83, K = 5
Output : 10043
10040 is the smallest 5 digit
number that is multiple of 83.

Input : X = 5, K = 2
Output : 10

A simple solution is to try all numbers starting from the smallest K digit number 
(which is 100…(K-1)times) and return the first number divisible by X.

An efficient solution would be :  

Compute MIN : smallest K-digit number (1000...(K-1)times)
If, MIN % X is 0, ans = MIN
else, ans = (MIN + X) - ((MIN + X) % X))
This is because there will be a number in 
range [MIN...MIN+X] divisible by X.

// C++ code to find smallest K-digit number
// divisible by X
#include <bits/stdc++.h>
using namespace std;

// Function to compute the result
int answer(int X, int K)
{
    // Computing MIN
    int MIN = pow(10, K - 1);

    // MIN is the result
    if (MIN % X == 0)
        return MIN;

    // returning ans
    return ((MIN + X) - ((MIN + X) % X));
}

// Driver Code
int main()
{
    // Number whose divisible is to be found
    int X = 83;

    // Max K-digit divisible is to be found
    int K = 5;

    cout << answer(X, K);
}

// Java code to find smallest K-digit
// number divisible by X

import java.io.*;
import java.lang.*;

class GFG {
    public static double answer(double X, double K)
    {
        double i = 10;
        // Computing MIN
        double MIN = Math.pow(i, K - 1);

        // returning ans
        if (MIN % X == 0)
            return (MIN);
        else
            return ((MIN + X) - ((MIN + X) % X));
    }

    public static void main(String[] args)
    {

        // Number whose divisible is to be found
        double X = 83;
        double K = 5;

        System.out.println((int)answer(X, K));
    }
}

// Code contributed by Mohit Gupta_OMG <(0_o)>

# Python code to find smallest K-digit  
# number divisible by X

def answer(X, K):
    
    # Computing MAX
    MIN = pow(10, K-1)
    
    if(MIN % X == 0):
        return (MIN)
    
    else:
        return ((MIN + X) - ((MIN + X) % X))
    

X = 83; 
K = 5; 

print(answer(X, K)); 

# Code contributed by Mohit Gupta_OMG <(0_o)>

// C# code to find smallest K-digit
// number divisible by X
using System;

class GFG {

    // Function to compute the result
    public static double answer(double X, double K)
    {

        double i = 10;

        // Computing MIN
        double MIN = Math.Pow(i, K - 1);

        // returning ans
        if (MIN % X == 0)
            return MIN;
        else
            return ((MIN + X) - ((MIN + X) % X));
    }

    // Driver code
    public static void Main()
    {

        // Number whose divisible is to be found
        double X = 83;
        double K = 5;

        Console.WriteLine((int)answer(X, K));
    }
}

// This code is contributed by vt_m.

<?php
// PHP code to find smallest 
// K-digit number divisible by X

// Function to compute 
// the result
function answer($X, $K)
{
    // Computing MIN
    $MIN = pow(10, $K - 1);

    // MIN is the result
    if ($MIN % $X == 0)
        return $MIN;

    // returning ans
    return (($MIN + $X) - 
           (($MIN + $X) % $X));
}

// Driver Code

// Number whose divisible
// is to be found
$X = 83;

// Max K-digit divisible
// is to be found
$K = 5;

echo answer($X, $K);

// This code is contributed by ajit
?>

<script>

// Javascript code to find smallest 
// K-digit number divisible by X

// Function to compute 
// the result
function answer(X, K)
{
    
    // Computing MIN
    let MIN = Math.pow(10, K - 1);

    // MIN is the result
    if (MIN % X == 0)
        return MIN;

    // returning ans
    return ((MIN + X) - 
           ((MIN + X) % X));
}

// Driver Code

// Number whose divisible
// is to be found
let X = 83;

// Max K-digit divisible
// is to be found
let K = 5;

document.write(answer(X, K));
 
// This code is contributed by sravan kumar

</script>

Output : 

10043

Time Complexity: O(logk)

Auxiliary Space: O(1)