Cuboid Layers

The minimum number of cubes to cover every visible face on a cuboid measuring $3 \times 2 \times 1$ is twenty-two.

If we then add a second layer to this solid it would require forty-six cubes to cover every visible face, the third layer would require seventy-eight cubes, and the fourth layer would require one-hundred and eighteen cubes to cover every visible face.
However, the first layer on a cuboid measuring $5 \times 1 \times 1$ also requires twenty-two cubes; similarly the first layer on cuboids measuring $5 \times 3 \times 1$, $7 \times 2 \times 1$, and $11 \times 1 \times 1$ all contain forty-six cubes.
We shall define $C(n)$ to represent the number of cuboids that contain $n$ cubes in one of its layers. So $C(22) = 2$, $C(46) = 4$, $C(78) = 5$, and $C(118) = 8$.
It turns out that $154$ is the least value of $n$ for which $C(n) = 10$.
Find the least value of $n$ for which $C(n) = 1000$.

This problem essentially boils down to understanding interior volume and surface area. When adding an additional layer to a rectangular prism (a cuboid), you’re increasing the surface area that needs to be covered. The surface area of a rectangular prism can be calculated with the formula 2lw + 2lh + 2wh, where l, w, and h are the length, width, and height of the prism, respectively.

If we increase each dimension by one (which is equivalent to adding an additional layer), the surface area becomes 2(l+1)(w+1) + 2(l+1)(h+1) + 2(w+1)(h+1).

Now, observe that the difference between the new surface area and the initial one is equal to the number of additional cubes required to cover the new layer. This difference comes out to be 2(l+w+h) + 4.

So the problem can be rephrased as: for a given n, find the number of integer solutions (l,w,h) such that 2(l+w+h) + 4 = n, and l≥w≥h≥1.

The cuboids with equal dimensions are counted more than once. For this reason, you have to track the number of repeated counts for each surface area and subtract these from the total count.

It would be easier to use a program to solve this problem. A simple Python solution follows:

“`python
count = [0]*20000
for h in range(1, 200):
for w in range(h, (20000-4)//(2*h)+1):
p = 2*(h + w) + 4
for l in range(w, (20000-p)//(2*w+2*h)+1):
count[p + 2*l] += 1
idx = 0
while count[idx] < 1000: idx += 2 print(idx) ``` You're iterating over possible values of h and w (taking into account that n must be even), and then calculate the corresponding l. You increment the count for each sum of areas. To get the final answer, you continue until you find the first n with count[n] greater than or equal to 1000. This program should give you the least value of n for which C(n) = 1000. It may take a while for it to run, as the problem is computationally heavy.

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