Cube Digit Pairs

There's only ${10 \choose 6} = 210$ distinct cubes, making this problem easy to brute force. We first make a generator for those cubes.

from itertools import combinations

def cubes():
    yield from combinations(range(0, 10), 6)

We also write a simple function for handling the fact that we can flip 6s and 9s.

def extended_set(s):
    s = set(s)
    if 6 in s:
        s.add(9)
    elif 9 in s:
        s.add(6)
        
    return s

Another simple function tests if we can display all the square numbers with a given pair of cubes.

def all_squares_displayable(s, t):
    es = extended_set(s)
    et = extended_set(t)
    
    square_digits = ((0, 1), (0, 4), (0, 9), (1, 6), (2, 5), (3, 6), (4, 9), (6, 4), (8, 1))
    for (m, n) in square_digits:
        if not ((m in es and n in et) or (n in es and m in et)):
            return False
    
    return True

Then we just check all the cube pairs.

arrangements = set()
for s, t in combinations(cubes(), 2):
    if all_squares_displayable(s, t):
        arrangements.add((s, t))

len(arrangements)

1217

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