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++
// 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
// 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)> |
Python3
# 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#
// 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 // 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 ?> |
Javascript
<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)