Stacking cannonballs – solving maths with code

Stacking cannonballs – solving maths with code

Numberphile have recently done a video looking at the maths behind stacking cannonballs – so in this post I’ll look at the code needed to solve this problem.

Triangular based pyramid.

A triangular based pyramid would have:

1 ball on the top layer

1 + 3 balls on the second layer

1 + 3 + 6 balls on the third layer

1 + 3 + 6 + 10 balls on the fourth layer.

Therefore a triangular based pyramid is based on the sum of the first n triangular numbers.

The formula for the triangular numbers is:

and the formula for the sum of the first n triangular numbers is:

We can simplify this by using the identity for the sum of the first n square numbers and also the identity for the sum of the first n natural numbers:

Therefore:

and the question we want to find out is whether there is triangular based pyramid with a certain number of cannonballs which can be rearranged into a triangular number i.e.:

here n and m can be any natural number. For example if we choose n = 3 and m = 4 we see that we have the following:

Therefore we can have a triangular pyramid of height 3, which has 10 cannonballs. There 10 cannonballs can then be rearranged into a triangular number.

Square based pyramids and above.

For a square based pyramid we would have:

1 ball on the top layer

1 + 4 balls on the second layer

1 + 4 + 9 balls on the third layer

1 + 4 + 9 + 16 balls on the fourth layer.

This is the sum of the first n square numbers.  So the formula for the square numbers is:

and the sum of the first n square numbers is:

For a pentagonal based pyramid we have:

1 ball on the top layer

1 + 5 balls on the second layer

1 + 5 + 12 balls on the third layer

1 + 5 + 12 + 22 balls on the fourth layer.

This is the sum of the first n pentagonal numbers. So the formula for the pentagonal numbers is:

and the formula for the first n pentagonal numbers is:

For a hexagonal based pyramid we have:

The formula for the first n hexagonal numbers:

and the formula for the sum of the first n hexagonal numbers:

For a k-agon based pyramid we have

and the formula for the sum of the first n k-agon numbers:

Therefore the general case is to ask if a k-agonal pyramid can be rearranged into a k-agon number i.e:

Computers to the rescue

We can then use some coding to brute force some solutions by running through large numbers of integers and seeing if any values give a solution.  Here is the Python code.  Type it (taking care with the spacing) into a Python editor and you can run it yourself.

You can then change the k range to check larger k-agons and also change the range for a and b.  Running this we can find the following.  (The first number is the value of k, the second the height of a k-agonal pyramid, the third number the k-agon number and the last number the number of cannonballs used).

Solutions:

3 , 3 , 4 , 10
3 , 8 , 15 , 120
3 , 20 , 55 , 1540
3 , 34 , 119 , 7140
4 , 24 , 70 , 4900
6 , 11 , 22 , 946
8 , 10 , 19 , 1045
8 , 18 , 45 , 5985
10 , 5 , 7 , 175
11 , 25 , 73 , 23725
14 , 6 , 9 , 441
14 , 46 , 181 , 195661
17 , 73 , 361 , 975061
20 , 106 , 631 , 3578401
23 , 145 , 1009 , 10680265
26 , 190 , 1513 , 27453385
29 , 241 , 2161 , 63016921
30 , 17 , 41 , 23001
32 , 298 , 2971 , 132361021
35 , 361 , 3961 , 258815701
38 , 430 , 5149 , 477132085
41 , 204 , 1683 , 55202400
41 , 505 , 6553 , 837244045
43 , 33 , 110 , 245905
44 , 586 , 8191 , 1408778281
50 , 34 , 115 , 314755
88 , 15 , 34 , 48280
145, 162, 1191, 101337426
276,  26,  77, 801801)
322, 28, 86, 1169686
823, 113, 694, 197427385
2378, 103, 604, 432684460
31265, 259, 2407,  90525801730

For example we can see a graphical representation of this.  When k is 6, we have a hexagonal pyramid with height 11 or the 22nd hexagonal number – both of which give a solution of 946.  These are all the solutions I can find – can you find any others?  Leave a comment below if you do find any others and I’ll add them to the list!