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
An efficient solution would be :
Compute MIN : smallest K-digit number (1000...K-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.
// 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)> |
Output:
10043
To understand Math.pow() function, please refer point 18 of the article :
https://www.geeksforgeeks.org/java-lang-math-class-java-set-2/amp/
Please refer complete article on Smallest K digit number divisible by X for more details!
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