Equalising numbers

I would like to equalise a set of 3 (or maybe more) numbers.

For example I would like
6,4,2
to become
4,4,4

or

9,0,5
to become
5,4,5

or

0,3,0
to become
1,1,1

If I had 2 numbers the the sum is simple. I'd subtract the lower number from the higher. Then half that number and add it to the lower and subtract from the higher.
8,6 would become 7,7

But how can I equalise 3 or more numbers?

How can I do this?

Thanks!

It looks to me like you are basically trying to take a set of numbers and find the average.

Average(6,4,2)=4
average(8,6)=7
average(0,3,0)=1
average(9,0,5)=4.67, which you have rounded some up/some down to get 5,4,5, so that the three numbers add up to the same quantity.

Sound like the right idea?

Yeah it's sort of an average/median I suppose!
But
average(9,0,5)=4.67
What formula would I use to change 9,0,5 to 5,4,5 ???

That's the hard part, isn't it.

roundup and rounddown can be used to find the two numbers that will be present: roundup(4.67,0)=5 rounddown(4.67,0)=4.
I haven't tested this rigorously, but I think you can use the fractional result (the remainder if you remember from elementary math) to tell you how many of the roundup values you need. the remainder can be returned using the mod function: mod(14,3)=2, so you need two 5's and one 4.

In another case, use the five numbers: 1,2,3,4,6
sum(1,2,3,4,6)=16
count(1,2,3,4,6)=5
roundup(average(1,2,3,4,6),0)=4
rounddown(average(1,2,3,4,6),0)=3
mod(sum(1,2,3,4,6),count(1,2,3,4,6))=1
so you need one 4 and (5-1)=four 3's.

The algorithm should work in the other cases where average is an integer, because roundup and rounddown will return the same value, and mod will return 0.

At this point, do you need a routine to populate 5 cells with 3's and 4's, or will it be enough to say that the answer is one 4 and four 3's?