## Cutting coloured cubes

A carpenter C had a solid wooden cube with whole number dimensions.He painted the entire surface of the cube with red paint.Then he cut the cube into smaller cubes of size 1*1*1 working parallel to the faces of the bigger cube.A certain number of smaller cubes were completely free of paint (x) A certain number of cubes had paint on only one side (y) A certain number number of cubes had paint on 2 sides (z).If (y+z) is 33% of (x) what was the original cube size?

(Proposed by Hrajan Nil)

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Let the original cube have edge length n (integer). Then
x = (n – 2)³, y = 6·(n – 2)², z = 12·(n – 2)
Then there are also 8 small cubes with paint on three sides (those at the original corners).

[ I guess that by 33% you mean 1/3. Then y + z = x/3, which gives
3·[6(n – 2)² + 12(n – 2)] = (n – 2)³. If the binomial powers are expanded we get
3·[6n² – 12n] = n³ – 6n² + 12n – 8, and then n³ – 24n² + 48n – 8 = 0.

This equation can be written (n – 2)(n² – 22·n + 4) = 0.
This equation has just on integer solution, n = 2. – There is one solution very near 22 but not exactly equal.

Aha, there is more to this problem than you may think at first. ]

Suppose 33% mean exactly 33%, i e 33/100. Then we get instead the equation

100·[6(n – 2)² + 12(n – 2)] = 33(n – 2)³. If the binomial powers are expanded we get
100(6n² – 12n) = 33(n³ – 6n² + 12n – 8), and then 33n³ – 798n² + 1596n – 264 = 0.
This equation has the integer solutions n = 2 and n = 22 exactly (and the third solution is n = 2/11).

So the original cube had edge length 2 or 22. However n = 2 gives x = y = z = 0 so I think 22 is meant.

My first attempt with 1/3 instead of 33% is really no part of the solution but I let it be there within [ ] to illustrate how you may or may not think.
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