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.neveropen.co.uk/java-lang-math-class-java-set-2/
Please refer complete article on Smallest K digit number divisible by X for more details!
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