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

Another Method or logic to rounding up a number to nearest multiple of 5

double num = 18.0;
    if (num % 5 == 0)
        System.out.println("No need to roundoff");
    else if (num % 5 < 2.5)
        num = num - num % 5;
    else
        num = num + (5 - num % 5);
    System.out.println("Rounding up to nearest 5------" + num);

output :

Rounding up to nearest 5------20.0

Answer 2

Just pass your number to this function as a double, it will return you rounding the decimal value up to the nearest value of 5;

if 4.25, Output 4.25

if 4.20, Output 4.20

if 4.24, Output 4.20

if 4.26, Output 4.30

if you want to round upto 2 decimal places,then use

DecimalFormat df = new DecimalFormat("#.##");
roundToMultipleOfFive(Double.valueOf(df.format(number)));

if up to 3 places, new DecimalFormat("#.###")

if up to n places, new DecimalFormat("#.nTimes #")

 public double roundToMultipleOfFive(double x)
            {

                x=input.nextDouble();
                String str=String.valueOf(x);
                int pos=0;
                for(int i=0;i<str.length();i++)
                {
                    if(str.charAt(i)=='.')
                    {
                        pos=i;
                        break;
                    }
                }

                int after=Integer.parseInt(str.substring(pos+1,str.length()));
                int Q=after/5;
                int R =after%5;

                if((Q%2)==0)
                {
                    after=after-R;
                }
                else
                {
                    after=after+(5-R);
                }

                       return Double.parseDouble(str.substring(0,pos+1).concat(String.valueOf(after))));

            }

Answer 3

int getNextMultiple(int num , int multipleOf) {
    int nextDiff = multipleOf - (num % multipleOf);
    int total = num + nextDiff;
    return total;
}

Answer 4

I think I have it, thanks to Amir

double round( double num, int multipleOf) {
  return Math.floor((num + multipleOf/2) / multipleOf) * multipleOf;
}

Here's the code I ran

class Round {
    public static void main(String[] args){
        System.out.println("3.5 round to 5: " + Round.round(3.5, 5));
        System.out.println("12 round to 6: " + Round.round(12, 6));
        System.out.println("11 round to 7: "+ Round.round(11, 7));
        System.out.println("5 round to 2: " + Round.round(5, 2));
        System.out.println("6.2 round to 2: " + Round.round(6.2, 2));
    }

    public static double round(double num, int multipleOf) {
        return Math.floor((num +  (double)multipleOf / 2) / multipleOf) * multipleOf;
    }
}

And here's the output

3.5 round to 5: 5.0
12 round to 6: 12.0
11 round to 7: 14.0
5 round to 2: 6.0
6.2 round to 2: 6.0

Answer 5

int roundUp(int num) {
    return ((num / 5) + (num % 5 > 0 ? 1 : 0)) * 5;
}

Answer 6

 int roundToNearestMultiple(int num, int multipleOf){
        int floorNearest = ((int) Math.floor(num * 1.0/multipleOf)) * multipleOf;
        int ceilNearest = ((int) Math.ceil(num  * 1.0/multipleOf)) * multipleOf;
        int floorNearestDiff = Math.abs(floorNearest - num);
        int ceilNearestDiff = Math.abs(ceilNearest - num);
        if(floorNearestDiff <= ceilNearestDiff) {
            return floorNearest;
        } else {
            return ceilNearest;
        } 
    }