# Numbered Billiard Balls

Ten billiard balls are stacked in the form of a regular tetrahedron, six in the bottom layer, three in the middle layer, and one in the top layer or cuspid. All the balls are numbered (with positive integers) differently, and the number on each ball in the top and middle layers is the sum of the three numbers on the balls on which each of these balls rests.

What is the least possible number the ball on the apex can have?

What is the least possible number it can have if there are four layers and 20 balls altogether?

• What are the numbers on the balls? It's obviously not 0-16, like regular billiards. – Hugh Nov 20 '18 at 4:14
• I'm guessing you want all the numbers to be non-negative, otherwise the answer is negative infinity. And I'm guessing you want all the numbers to be integers, otherwise we could pick a collection of numbers infinitesimally greater than 0 and get the total to be something that's also indistinguisable from zero. I can't guess if you want to allow one ball to be numbered 0. – MattClarke Nov 20 '18 at 6:02
• @MattClarke: Yes, positive integers. I have edited question accordingly. – Bernardo Recamán Santos Nov 20 '18 at 11:24
• Oh, positive integers only... darn it! You wrote "billiard balls", and in billiards there is sometimes a ball number 0 so I assumed I could use 0. Ah well – Hugh Nov 20 '18 at 14:46

I will assume the balls are meant to be numbered using positive integers (i.e. 1 is the smallest number used).

For the size 3 tetrahedron:

If the bottom layer is:

  a b c
d e
f
Then the apex ball is $$(a+b+d)+(b+c+e)+(d+e+f) = (a+c+f)+2(b+d+e)$$.
The smallest this could possibly be using distinct positive integers is if $$\{b,d,e\}=\{1,2,3\}$$ and $$\{a,c,f\}=\{4,5,6\}$$.
This would makes the apex ball equal to $$15+2*6=27$$. The only thing to check is that they can actually be arranged so that the middle layer numbers don't duplicate any other number. It can be done like this:
  4 1 5     7 9     27
2 3       11
6

The size 4 tetrahedron can be analysed in a similar way.

If the bottom layer is:

  a b c d
e f g
h i
j
The apex becomes $$(a+d+j)+3(b+c+e+g+h+i)+6f$$. So the minimum could be when $$f=1$$, $$\{b,c,e,g,h,i\}=\{2,3,4,5,6,7\}$$ and $$\{a,d,j\}=\{8,9,10\}$$, giving an apex number of $$114$$. Unfortunately this is impossible. Wherever you place the $$2$$, there will be a number in the middle layer that is $$1+2+x$$ where $$x\le7$$, which is a duplicate. The best I've been able to do is to increase the corners to $$\{a,d,j\}=\{9,11,13\}$$. This gives an apex of $$120$$.
  9 2 5 11   18 8 19    38 37     120
7 1 3      12 10      45
4 6        23
13

• Ah, darn it. I assumed 0 could be used since the question says "billiard balls" and there's a billiard numbered zero. Oh well! – Hugh Nov 20 '18 at 14:00

Partial Answer — What is the least possible number the ball on the cuspid can have?

The question has since been edited to clarify which numbered balls can be used. I assumed you could use zero, but I guess you can't. Oh well!

Assuming that there's a ball numbered 0, the minimum value for the cuspid is 20, constructed like so:

  6
2 1      9
5 0 3    7 4    20

For this proof, I'll use $$b_n$$ to represent the ball numbered $$n$$.
First, notice that the balls in the middle layer cannot be $$b_0$$, $$b_1$$, or $$b_2$$; as $$0$$, $$1$$, and $$2$$ cannot be represented as the sum of three unique positive integers. The smallest number with this property is $$3$$, represented as $$0 + 1 + 2$$.

To try and minimize the value of the cuspid, let's pick the next two smallest numbered for our middle layer. This makes our middle layer $$b_3$$, $$b_4$$, and $$b_5$$, with a cuspid of $$b_{12}$$.

But, notice that $$b_3$$, $$b_4$$, and $$b_5$$ cannot make up the middle layer. To see this, let's figure out what numbered balls balls $$3$$, $$4$$, and $$5$$ themselves can sit on. Namely, let's list the partitions of 3, 4, and 5, which are:
— size three
— composed of no duplicates

The only partition of three which satisfies these conditions is $$0 + 1 + 2$$. That means, $$b_3$$ can only be placed on $$b_0$$, $$b_1$$, and $$b_2$$.
The only partition of four which satisfies these conditions is $$0 + 1 + 3$$. That means, $$b_4$$ can only be placed on $$b_0$$, $$b_1$$, and $$b_3$$.
The only partitions of five which satisfy these conditions are $$0 + 1 + 4$$ and $$0 + 2 + 3$$. That means, $$b_5$$ can only be placed on [$$b_0$$, $$b_1$$, and $$b_4$$] or on [$$b_0$$, $$b_2$$, and $$b_3$$].

Now, notice that it is impossible to construct this, for a few reasons.
1. $$b_4$$ can only be placed on $$b_0$$, $$b_1$$, and $$b_3$$. This would mean that there are two $$b_3$$'s. One in the middle layer, since we've assumed so; one in the bottom layer supporting $$b_4$$. But, since there is only one of each ball, this is not possible.
2. $$b_3$$ and $$b_4$$ both require to be placed on $$b_0$$ and $$b_1$$. But, because of the topology of our tetrahedron, there is only one ball supporting any two given balls in the middle layer. Therefore, this is not possible.

Therefore, the middle layer cannot be made up of $$b_3$$, $$b_4$$, and $$b_5$$, and the cuspid must be $$b_{n}, n > 12$$.

Above, I've outlined two main candidates that indicate that a given configuration is impossible.
1. $$b_k$$ must be directly supported by $$b_m$$, but we've assumed that $$b_m$$ is in the middle layer. Since we cannot have two $$b_m$$'s, this is impossible.
2. $$b_y$$ and $$b_z$$ both must be supported by $$b_t$$ and $$b_u$$. Since $$b_t$$ and $$b_u$$ cannot both directly support $$b_y$$ and $$b_z$$, this is impossible.

For example, $$b_4$$ and $$b_6$$ cannot simultaneously exist in the middle layer. Of the valid partitions of six, notice that
$$0 + 1 + 5$$ shares a "pair" with $$0 + 1 + 3$$, $$(0, 1)$$
$$0 + 2 + 4$$ requires to be placed on $$b_4$$, but cannot.
$$1 + 2 + 3$$ shares a "pair" with $$0 + 1 + 3$$, $$(1, 3)$$

Now that we have these two rules, we can:
i. assume that the cuspid has some given value ($$13$$, $$14$$, $$15$$, etc...)
ii. work through the possible middle-layer values and prove that is is or it isn't possible.

Assuming a cuspid of $$b_{13}$$; the possible middle-layers are:
$$6 + 4 + 3$$, which is not possible because of the pair $$(3, 4)$$.

Assuming a cuspid of $$b_{14}$$; the possible middle-layers are:
$$7 + 4 + 3$$, which is not possible because of the pair $$(3, 4)$$.
$$6 + 5 + 3$$, which is not possible because of the pair $$(3, 5)$$.

Assuming a cuspid of $$b_{15}$$; the possible middle-layers are:
$$8 + 4 + 3$$, which is not possible because of the pair $$(3, 4)$$.
$$7 + 5 + 3$$, which is not possible because of the pair $$(3, 5)$$.
$$6 + 5 + 4$$, which is not possible because of the pair $$(4, 6)$$.

Assuming a cuspid of $$b_{16}$$; the possible middle-layers are:
$$9 + 4 + 3$$, which is not possible because of the pair $$(3, 4)$$.
$$8 + 5 + 3$$, which is not possible because of the pair $$(3, 5)$$.
$$7 + 6 + 3$$, which is not possible because of the pair $$(3, 6)$$.
$$7 + 5 + 4$$, which is not possible because of the pair $$(4, 5)$$.

Assuming a cuspid of $$b_{17}$$; the possible middle-layers are:
$$10 + 4 + 3$$, which is not possible because of the pair $$(3, 4)$$.
$$9 + 5 + 3$$, which is not possible because of the pair $$(3, 5)$$.
$$8 + 6 + 3$$, which is not possible because of the pair $$(3, 6)$$.
$$8 + 5 + 4$$, which is not possible because of the pair $$(4, 5)$$.
$$7 + 6 + 4$$, which is not possible because of the pair $$(4, 6)$$.

Assuming a cuspid of $$b_{18}$$; the possible middle-layers are:
$$11 + 4 + 3$$, which is not possible because of the pair $$(3, 4)$$.
$$10 + 5 + 3$$, which is not possible because of the pair $$(3, 5)$$.
$$9 + 6 + 3$$, which is not possible because of the pair $$(3, 6)$$.
$$9 + 5 + 4$$, which is not possible because of the pair $$(4, 5)$$.
$$8 + 7 + 3$$, which is not possible because of the pair $$(3, 7)$$.
$$8 + 6 + 4$$, which is not possible because of the pair $$(4, 6)$$.
$$7 + 6 + 5$$, which is not possible because of the combination of pairs $$(5, 6)$$ and $$(6, 7)$$.

It is not hard to continue this logic to find:
$$b_{20}$$ = $$b_{9}$$ + $$b_{7}$$ + $$b_{4}$$
$$b_{9}$$ = $$b_{6}$$ + $$b_{2}$$ + $$b_{1}$$
$$b_{7}$$ = $$b_{0}$$ + $$b_{2}$$ + $$b_{5}$$
$$b_{4}$$ = $$b_{0}$$ + $$b_{1}$$ + $$b_{3}$$

And I'm sure you could extend this logic for the second part.