Sunday Teaser 2648 – Painted Cubes

by Graham Smithers

I found some painted cubes whose sides were varying whole numbers of centimetres.

After choosing some of these cubes whose edges were consecutive integers, I cut them all into smaller cubes with edge lengths of one centimetre.

I found that half of the resulting cubes had no painted faces.

How many of the small cubes had no painted faces?

One Comment Leave one →

brian gladman permalink

A very easy one this week.


# the total volume of cubes with sides m .. m + n - 1
def sum_cubes(m, n):
  return (((m + n - 1) * (m + n)) ** 2 - ((m - 1) * m) ** 2) // 4

# m is the edge length of the smallest cube
for m in range(3, 20):
  # n is the number of cubes
  for n in range(1, 20):
    # the number of inner unit cubes
    inner = sum_cubes(m - 2, n)
    # is half the total number of unit cubes
    if 2 * inner == sum_cubes(m, n):
      fs = 'There are {:d} small cubes with no painted faces.'
      print(fs.format(inner))

# the total volume of cubes with sides m .. m + n - 1

def sum_cubes(m, n):

return (((m + n - 1) * (m + n)) ** 2 - ((m - 1) * m) ** 2) // 4

# m is the edge length of the smallest cube

for m in range(3, 20):

# n is the number of cubes

for n in range(1, 20):

# the number of inner unit cubes

inner = sum_cubes(m - 2, n)

# is half the total number of unit cubes

if 2 * inner == sum_cubes(m, n):

fs = 'There are {:d} small cubes with no painted faces.'

print(fs.format(inner))

An alternative solution requires a bit of analysis. If we let \(m\) be the edge size of the smallest cube in the sequence, and \(n\) be the number of cubes, then the number of unit cubes is given by \[C(p,n)=(np/4)(4p^2+n^2-1)\] where \(p=m+(n-1)/2\), the average side length of the cubes.

For the inner cubes the side lengths are 2 smaller and we are told that the number of cubes with no painted faces is one half of the total number – that is \(C(p,n)=2C(p-2, n)\). This simplifies to give:\[n^2= -\frac{(4p^3-48p^2+95p-60)}{(p-4)}\] Here we can solve for \(n\) and \(p\) by stepping \(p\) in half integers greater than 4. So if we let \(q=2p\) we obtain \[n^2= -\frac{(q^3-24q^2+95q-120)}{(q-8)}\] which is the form used in the program below.


from itertools import count

fs = 'There are {:d} small cubes with no painted faces.'

for q in count(9):
  t = -(q ** 3 - 24 * q ** 2 + 95 * q - 120)
  if t < 0:
    break
  n = round((t / (q - 8)) ** 0.5)
  if (q - 8) * n ** 2 == t:
    no_cubes = n * q * (q ** 2 + n ** 2 - 1) // 16
    print(fs.format(no_cubes))

from itertools import count

fs = 'There are {:d} small cubes with no painted faces.'

for q in count(9):

t = -(q ** 3 - 24 * q ** 2 + 95 * q - 120)

if t < 0:

break

n = round((t / (q - 8)) ** 0.5)

if (q - 8) * n ** 2 == t:

no_cubes = n * q * (q ** 2 + n ** 2 - 1) // 16

print(fs.format(no_cubes))

Sunday Teaser 2648 – Painted Cubes

by Graham Smithers

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