Rounding a number to the nearest 3?

Is it possible to round a number to the nearest 3? like

8.999 -> 9

2.5 -> 3

11.9 -> 12

I cant find this kinda thing online

Lets say inside a script I have an int string the int strings value is 58 The script would round this to the nearest 3 (57)

of if the int strings value was 58.5 it would round to nearest 3 [This is in the middle of two 3s so it would be rounded up (60)]

Thanks to everybody who answered ( I shouldve explained better) but I got an answer now.

You can easily do this using string.format. In your case, you can use zero as the argument for the n parameter Here's an example implementation:

local function roundToNPlaces(number, n)
	local roundedNumber = tonumber(string.format("%."..n.."f", number))
	if roundedNumber then
		return roundedNumber
	else
		return 0
	end
end

After your edits, I'm able to see what you mean. In future, try to better phrase your questions so that people understand (the examples especially made it confusing).

While it's true that you might not be able to find this stuff for Lua, if you look it up, there are actually many websites (this website is an example) with the simple solution:

-- if you want to find the closest above
print(3*math.ceil(13/3))
-- if you want to find the closest below
print(3*math.floor(13/3))
-- if you want to find the closest in general
print(3*math.floor(13/3 + 0.5))

####I hope this helps. Have a great day scripting!

Modulus

Well, the easiest way of doing this would be using the modulus operator, which essentially gets the remainder of quotient of the number before it and the number after it.

For example:

9%4

would return 1 as 4 goes into 9 twice, 4*2 = 8, and 9-8 is 1

An example of this would be:

function RoundToNumber (num,roundNum)
	if num%roundNum  >= roundNum/2 then
		return num - num%roundNum + roundNum
	else
		return num - num%roundNum
	end
end

this would round to the nearest provided roundNum.

It essentially rounds up if the remainder of quotient of the number and the number you want to round to is greater than half of the number you want to round to.

Visualizing with numbers

By adapting a more general function like that, we can get this result:

function RoundTo3 (num)
	if num%3  >= 3/2 then
		return num - num%3 + 3--round up
	else
		return num - num%3--round down
	end
end

Hopefully now you can visualize it more clearly.

The reason that we use modulo here is that it lets us get to a number between 0 and 3, which then can help us deduce whether it is closer to the upper multiple of 3, (result >=1.5), or the lower multiple of 3 (result < 1.5)

An example of this function would be:

RoundTo3(67)

as 67%3 is 1 (as the multiple of 3 below it is 66), it rounds down to 66:

so: the function returns 67 - 67%3, or 67 -1, which is 66

Another example would be:

RoundTo3(23)

as 23%3 is 2 (as the multiple of 3 below it is 21), it rounds up to 24:

so: the function returns 23 - 23%3 + 3 = 23 - 2 + 3 = 21 + 3 which is 24

Hopefully this helped, be sure to accept if it did!