Mixture

I would imagine the proportion equals the ratio of size between the two cases. The chests are 1/22 of each case.

16:6, although the problem was admittedly waring.

http://en.wikipedia.org/wiki/Cube_(algebra)

ekename,

Please show your solution. Thanks.

Randy Dandy

I guessed, japing the name of an eminent mathematician and occasional cubist then posted a spurious attribution to lend credence to my otherwise obfuscatory obtuseness.

If you don't have the common courtesy to respond to answers to your first question, don't expect an answer to this one.

Thank you.

I apologize for my error. Thanks for replying to my question.

The solution isn't so much a formula as the realisation that the equation x^3 + y^3 = 22.z^3
is constrained by the condition that the answer is the sum of variants of 2 sets of cubes.

Main article: Waring's problem

Every positive integer can be written as the sum of nine (or fewer) positive cubes. This upper limit of nine cubes cannot be reduced because, for example, 23 cannot be written as the sum of fewer than nine positive cubes:
23 = 2^3 + 2^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3.


Main article: Fermat's last theorem

The equation x^3 + y^3 = z^3 has no non-trivial (i.e. xyz ≠ 0) solutions in integers. In fact, it has none in Eisenstein integers.[1]

Both of these statements are also true for the equation[2] x^3 + y^3 = 3z^3.


http://en.wikipedia.org/wiki/Cube_(algebra)

Thanks ekename