Rounding up a number to nearest multiple of 5

Question

Does anyone know how to round up a number to its nearest multiple of 5? I found an algorithm to round it to the nearest multiple of 10 but I can\'t find this one.

T

Answer 1

Round a given number to the nearest multiple of 5.

public static int round(int n)
  while (n % 5 != 0) n++;
  return n; 
}

Answer 2

CODE:

    public class MyMath
    {
        public static void main(String[] args) {
            runTests();
        }
        public static double myFloor(double num, double multipleOf) {
            return ( Math.floor(num / multipleOf) * multipleOf );
        }
        public static double myCeil (double num, double multipleOf) {
            return ( Math.ceil (num / multipleOf) * multipleOf );
        }

        private static void runTests() {
            System.out.println("myFloor (57.3,  0.1) : " + myFloor(57.3, 0.1));
            System.out.println("myCeil  (57.3,  0.1) : " + myCeil (57.3, 0.1));
            System.out.println("");
            System.out.println("myFloor (57.3,  1.0) : " + myFloor(57.3, 1.0));
            System.out.println("myCeil  (57.3,  1.0) : " + myCeil (57.3, 1.0));
            System.out.println("");
            System.out.println("myFloor (57.3,  5.0) : " + myFloor(57.3, 5.0));
            System.out.println("myCeil  (57.3,  5.0) : " + myCeil (57.3, 5.0));
            System.out.println("");
            System.out.println("myFloor (57.3, 10.0) : " + myFloor(57.3,10.0));
            System.out.println("myCeil  (57.3, 10.0) : " + myCeil (57.3,10.0));
        }
    }

OUTPUT:There is a bug in the myCeil for multiples of 0.1 too ... no idea why.

    myFloor (57.3,  0.1) : 57.2
    myCeil  (57.3,  0.1) : 57.300000000000004

    myFloor (57.3,  1.0) : 57.0
    myCeil  (57.3,  1.0) : 58.0

    myFloor (57.3,  5.0) : 55.0
    myCeil  (57.3,  5.0) : 60.0

    myFloor (57.3, 10.0) : 50.0
    myCeil  (57.3, 10.0) : 60.0

Answer 3

if (n % 5 == 1){
   n -= 1;
} else if (n % 5 == 2) {
   n -= 2;
} else if (n % 5 == 3) {
   n += 2;
} else if (n % 5 == 4) {
   n += 1;
}