### Brain Teasers

# Small Cubes

A solid wooden cuboid is made which measures an exact, and even, number of centimetres along each of its edges. It is then painted black all over its exposed faces and, when the paint is dry, the cuboid is cut up into 1 centimetre cubes. These small cubes are counted and it is then found that the number having NO paint on them at all, is exactly the same as the number which do have some paint on them. What is the LEAST number of small cubes there could be in total?

### Answer

Least is 960 (8 by 10 by 12)Hide Answer Show Answer

## What Next?

View a Similar Brain Teaser...

If you become a registered user you can vote on this brain teaser, keep track of which ones you have seen, and even make your own.

### Solve a Puzzle

## Comments

Hey Mad... doesn't cuboid mean 'like a cube' therefore congruent. ie... equal sides. That's what threw me off as I worked out the formula for a cube, but it is impossible to resolve if it is a cube.

The question was least no. of SMALL cubes. So the answer is 480.

(6x8x10)

(6x8x10)

Sorry. My mistake.

Cuboid = A rectangular parallelepiped.

Fun puzzle, thank you.

Fun puzzle, thank you.

Cool teaser!

The solution should have included the method of determining the answer.

The basic equation that needs to be satisfied for sides of length x, y, z is:

x*y*z/2 = x*y*2 + (x-2)*z*2 + (y-2)*(z-2)*2

Solving the equation is a bit of a mess since it's non-determinant, so apply some analysis to constrain the problem.

The minimum dimension is 8. This is so because a slice through the middle of the cuboid will produce a 1-cube thick rectangular slice of small cubes. The perimeter cubes must be less than the interior cubes since the slices on the ends of the cuboid will contain only painted cubes.

A 6x6 slice has (5+5)*2 = 20 perimeter cubes and 6*6 - 20 = 16 interior cubes.

A 6x8 slice has (5+7)*2 = 24 perimeter cubes and 6*8 - 24 = 24 interior cubes.

An 8x8 slice has (7+7)*2 = 28 perimeter cubes and 8*8 - 28 = 36 interior cubes.

(Although a 6x10 slice also has 28 perimeter cubes, it only has 32 interior cubes, requiring twice as many interior slices, so this configuration can be discarded. Given two rectangles of equal perimeter, the rectangle with the smallest difference between the sides will produce more extra interior cubes and no extra perimeter cubes, so I didn't even bother considering 6x10.)

Next, given two sets of dimensions that produce an answer, the set producing the least number of small cubes is the set with the lowest product of the dimensions.

Using an 8x8x? cuboid, the 8 extra interior cubes from each slice need to total the 8*8*2 exterior cubes on the two ends. This means that there must be 128/8 = 16 interior slices plus the two ends, giving an 8x8x18 cuboid.

An 8x10x? cuboid has interior slices that are 8x10 and have (7+9)*2 = 32 perimeter cubes and 8*10 - 32 = 48 interior cubes. There are 16 extra interior cubes per slice, so 8*10*2/16 = 10 interior slices plus the two ends are needed giving an 8x10x12 cuboid.

8*8*18 = 1152

8*10*12 = 960

At this point there is no possible cuboid with even dimensions that could have fewer small cubes.

It took about a minute or so to work this out in my head. There was some novelty to the question, so about half that time was spent working out the method of solution and half spent working the calculations above.

Again, nice teaser!

The solution should have included the method of determining the answer.

The basic equation that needs to be satisfied for sides of length x, y, z is:

x*y*z/2 = x*y*2 + (x-2)*z*2 + (y-2)*(z-2)*2

Solving the equation is a bit of a mess since it's non-determinant, so apply some analysis to constrain the problem.

The minimum dimension is 8. This is so because a slice through the middle of the cuboid will produce a 1-cube thick rectangular slice of small cubes. The perimeter cubes must be less than the interior cubes since the slices on the ends of the cuboid will contain only painted cubes.

A 6x6 slice has (5+5)*2 = 20 perimeter cubes and 6*6 - 20 = 16 interior cubes.

A 6x8 slice has (5+7)*2 = 24 perimeter cubes and 6*8 - 24 = 24 interior cubes.

An 8x8 slice has (7+7)*2 = 28 perimeter cubes and 8*8 - 28 = 36 interior cubes.

(Although a 6x10 slice also has 28 perimeter cubes, it only has 32 interior cubes, requiring twice as many interior slices, so this configuration can be discarded. Given two rectangles of equal perimeter, the rectangle with the smallest difference between the sides will produce more extra interior cubes and no extra perimeter cubes, so I didn't even bother considering 6x10.)

Next, given two sets of dimensions that produce an answer, the set producing the least number of small cubes is the set with the lowest product of the dimensions.

Using an 8x8x? cuboid, the 8 extra interior cubes from each slice need to total the 8*8*2 exterior cubes on the two ends. This means that there must be 128/8 = 16 interior slices plus the two ends, giving an 8x8x18 cuboid.

An 8x10x? cuboid has interior slices that are 8x10 and have (7+9)*2 = 32 perimeter cubes and 8*10 - 32 = 48 interior cubes. There are 16 extra interior cubes per slice, so 8*10*2/16 = 10 interior slices plus the two ends are needed giving an 8x10x12 cuboid.

8*8*18 = 1152

8*10*12 = 960

At this point there is no possible cuboid with even dimensions that could have fewer small cubes.

It took about a minute or so to work this out in my head. There was some novelty to the question, so about half that time was spent working out the method of solution and half spent working the calculations above.

Again, nice teaser!

Fun teaser! I realised that dimensions would shrink by 2 for unpainted so minimum dimension was 3. Then I noticed that if any dimension was odd, the total was odd so could not be halved. I tried 4x4x4 but the painted cubes were 8 times the unpainted,

By comparing with a 2 dimensional model of a rectangle, I realised that having rectangular cross sections rather than square would reduce the ratio of painted to unpainted.

So then I tried 4x6x8. The painted cubes are 4 times the unpainted so I next went to 6x8x10 which was closer and my third trial was 8x10x12 which worked.

By comparing with a 2 dimensional model of a rectangle, I realised that having rectangular cross sections rather than square would reduce the ratio of painted to unpainted.

So then I tried 4x6x8. The painted cubes are 4 times the unpainted so I next went to 6x8x10 which was closer and my third trial was 8x10x12 which worked.

To post a comment, please create an account and sign in.

## Follow Braingle!