ENIGMA 1570 — Set of Cubes

by Richard England

(1) Find the set of perfect cubes that between them use each of the digits 0 to 9 at least once whose sum is as small as possible. What is the sum of your set of cubes?

(2) Find the set of perfect cubes that between them use each of the digits 0 to 9 exactly once whose sum is as small as possible. What is the sum of your set of cubes?

Answer to part 2 from Chris

Take your table of cubes and for each digit list the first 5 cubes that contain that digit and have no repeated digits. (see below)

Note that 0, 1 and 8 can be ignored at this stage as they are1-digit cubes.

Note that each cube containing 7 also contains 2.

Note also that 3, 5 and 7 have sparse lists (they get big quickly).
In the 3 list only 4913 and 19683 do not contain 2.
In the 5 list only 6859 does not contain 2.
But 9 is common to all three cubes.

Therefore extend the 7 list (42875, 54872, 205379).
Extend the 3 list (205379) and the 5 list (54872, 205379)
The first two cubes (42875, 54872) cannot be used as they contain 4 and 8, which are also present in the 3 and 5 lists above.

Consider 205379 (593). Digits needed are 1, 4, 6, 8.
64 is a cube, as are 1 and 8 - therefore this is a valid solution. As it is the smallest cube containing a 7 which does not conflict the problem, it is also the part of the minimum sum.
The minimum sum is 205,452.

Table of valid cubes (first five and extensions)
2125, 216, 512, 729, 1728
34913, 5832, 13824, 19683, 24389205379
464, 4096, 4913, 10648, 13824
5125, 512, 5832, 6859, 4287554872, 205379
664, 216, 4096, 6859, 9261
727, 729, 1728, 2197, 3276842875, 54872, 205379
9729, 2197, 4096, 4913, 6859