How to solve Centrifuge problem

Question

I have a balancing problem which I am not sure if I understand completely correct.

Centrifuge has 36 holes. If in a hole only one test tube can be placed and all the test tubes are with equal weight.

What are the number of test tubes which can be used to balance the Centrifuge?

The answer that I have is:

The Centrifuge can work with even number of test tubes. (2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34 and 36)

And cannot work with odd number of test tubes (1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35)

The reasoning behind that is the following: If a test tube has equal weight as all the others in order to balance it each and every test tube should be placed in front of another.

Is this a correct approach?

Answer 1

My answer is

Every set except 1 and 35

Even numbers
It is obviously possible to place one pair of tubes on opposite sides.
Then you can add another pair of tubes at any free opposite places.
The centre of gravity remains in the centre of the wheel.
And so on, this covers $2$ to $36$ even numbers.

Odd numbers
Three tubes can be placed symmetrically like this where marked $3$:



As before, you can add another pair at any free opposite places, eg marked $5$.
Again, the centre of gravity remains in the centre of the wheel.
And so on, place them in 15 pairs of places – all except the original set of $3$.
Which leaves 3 holes while cannot be filled.
This gives $3$, $5$ ... $33$ odd numbers.

So all combinations are possible except $1$ and $35$.
You can't balance a single tube, or a single hole.

Answer 2

Thank you all. Weather Vane, your drawing explains it really good and I can visualize it in my head.

I tried to do the math behind the conjunction: You can balance $k$ identical test tubes, $1 ≤ k ≤ n$, in an $n$-hole centrifuge if and only if both $k$ and $n-k$ can be expressed as a sum of prime divisors of n.

So the first thing that we need to to is to find the divisors of $n$. If >!$n$ is $36$ its divisors are: $$\text{1 2 3 4 6 9 12 18 and 36}$$ The prime divisors are: $$\text{2 and 3}$$ So now for all numbers both $n$ and $n - k$ should be made by summing only $2$ and $3$. It is obvious that for 1 and 35 we cannot do it because for: $$n=1 \text{ cannot be made of prime divisors of 36}$$ $$=> n - k = 35 \text{ sum of prime 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 2}$$ $$\dots$$ $$n=35 \text{ sum of prime 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 2}$$ $$n=1 \text{ cannot be made of prime divisors of 36}$$ for all the others we can make the numbers by summing 2 and 3. $$n = 2 \text{ sum of prime 2}$$ $$ => n - k = 34 \text{ sum of prime 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 2 + 2}$$ $$\dots$$ $$n = 3 \text{ sum of prime 3}$$ $$ => n - k = 33 \text{ sum of prime 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3}$$ $$\dots$$ $$n = 4 \text{ sum of prime 2 + 2}$$ $$ => n - k = 32 \text{ sum of prime 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 2}$$ etc for all the other numbers.