Smallest Number

I did the solution using JAVA and here's my code:

public class smallestnumber {

public static void main(String[] args) {

    int divisor = 10;       //assigns the initial value of the divisor
    int remainder = 9;    //assigns the  initial value of the remainder
    int value = 19;    //initial value of the number divided with 10 which has a remainder of 9 (this is obvious)

    for(int c = divisor; c >= 2; c--){ // starts the loop from divisor of 10 down to 2

        value = checkValue(divisor, remainder, value); // calls a method which returns the answer for the current divisor
        divisor--; // decreases the divisor by one
        remainder--; // decreases the remainder by one
    } // end of loop
    System.out.println(value); // prints the final answer **just sum up all the digits** --2519 w/c is 17--
}

private static int checkValue(int divisor, int remainder, int value) { // start of the method which checks the answer if divisible by the divisor and has a remainder equal to the specified remainder

    int add = checkAdd(divisor); // checks the number to be added to the previous answer
    int val = value+add; // sums up the previous answer with the new added value;
    return(((value % divisor) != remainder)? checkValue(divisor, remainder, val): value); // if every condition is met, it returns the answer, else it continues to increment the value;
} // end of method

private static int checkAdd(int divisor) {// start of method for check how much to add

    int val = 1; // initial value
    for(int c = divisor; c <= 10; c++){// loops from the divisor up to 10

        if(divisor == 10) // if the divisor is equal to 10 then assign the value to the variable val
            val = 10;
        else            //else multiply the current divisor with the current value of the variable val
            val = val * c;
    }

    return (((divisor == 10)? val : (val/divisor)));// returns the final value of the variable val
}

}

The number must be in the form: N=RK-1,where R is the least common multiple of 2,3,4,5,6,7,8 and 9. K=1,2,3,4.... R=2520 N=2520-1: N=2519 Hence, the sum of digits is 2+5+1+9=17