Smallest K digit number divisible by X (original) (raw)
Last Updated : 22 Jun, 2022
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
JavaScript
`
Output :
10043
Time Complexity: O(logk)
Auxiliary Space: O(1)