**The Problem**

The volume of a modified 4D Menger Sponge, which is created by successively subtracting volumes from the interior of a 4D tesseract of side length of 2, can be expressed as

\[\dfrac{a}{b} {\pi}^{c},\]

where \(a,b\) and \(c\) are positive integers with \(a,b\) coprime.

Find the sum \(a+b+c\).

The method of successively subtracting volumes from the interior shall now be explained

**Preliminaries**

Given an arbitrary odd number \(2n+1\), we have the set of digits \((0,1,2,3,\ldots , 2n)\) in base \(2n+1\). Then we construct all possible distinct sets of the form

\((n,n,a,b)\)

where \(a,b\) are digits from that first set, and need not be different, i.e.,\(a\) may be the same as \(b\). Then for each such distinct set, we list all possible distinct permutations of it. **For example**, let the odd number be \(5\), where \(n=2\), and we have the set of digits \((0, 1, 2, 3, 4)\). We then will have the following distinct sets

\((2,2,0,0)\)

\((2,2,0,1)\)

\((2,2,0,2)\)

\((2,2,0,3)\)

\((2,2,0,4)\)

\((2,2,1,1)\)

\((2,2,1,2)\)

\((2,2,1,3)\)

\((2,2,1,4)\)

\((2,2,2,2)\)

\((2,2,2,3)\)

\((2,2,2,4)\)

\((2,2,3,3)\)

\((2,2,3,4)\)

\((2,2,4,4)\)

and then for example, for the set \((2,2,0,0)\), we have the following permutations

\((2,2,0,0)\)

\((2,0,2,0)\)

\((2,0,0,2)\)

\((0,2,0,2)\)

\((0,0,2,2)\)

\((0,2,2,0)\)

Thus, we will have a total of \(113\) such sets with this example

**Subtraction of Volumes**

Given a tesseract, **for example**, we can subdivide it into \({(2n+1)}^{4}={5}^{4}=625\) smaller tesseracts, of which the 4D coordinates of each can be described as \((a,b,c,d)\). We subtract the 113 smaller tesseracts as per the list of 4D coordinates. If we sort the list by the first tuple, that is

\((0,a, b, c)\)

\((1, a, b, c)\)

\((2, a, b, c)\)

\((3, a, b, c)\)

\((4, a, b, c)\)

where \(a, b, c\) are arbitrary digits, then the "layers" \(0, 1, 3, 4\) will look like this, the green blocks being subtracted

while "layer" \(2\) will look like this, the green blocks being subtracted

Because there's already so many occurrences of the digit \(2\), it shouldn't be hard to see that the "middle layer" of \(2\) is going to have a lot of green blocks being subtracted from it.

Even though these graphics look like 3D graphics, they are still a valid projection of 4D tesseracts, we can imagine each green block as a tesseract each with its own coordinates.

**Method of Successive Subtractions**

The classic 3D Menger Sponge is created from a 3D cube by letting \((2n+1)=3\) for all recursive steps, and the sets are of the form

\((1, 1, a)\)

where \(a=(0, 1, 2)\), always. After the first round, \( \dfrac {7}{27} \) of the volume of a 3D cube is subtracted. Then for the 2nd round, EACH of the remaining cubes has \( \dfrac {7}{27} \) of their volumes removed. Repeat for the 3rd round and thereafter. We can see that starting with an intial volume of \({2}^{3}=8\) we are reducing it by a factor of \( \dfrac {20}{27} \) successively. Since the infinite product of this fixed fraction \(<1\) has the limit of \(0\), the classic 3D Menger Sponge has a volume of \(0\) as the limit. We are going to do things slightly differently here.

Here, instead of letting \( (2n+1)=3\) for all recursive steps, the value is going to be \((2n+1)\) where \(n\) stands for the \(n\)th recursive step. That is, in the case of the modified 4D Menger Sponge, which has the volume \({2}^{4}=16\), the first round will reduce it by a factor of

\( \dfrac {81-33}{81}=\dfrac{48}{81} \).

The 2nd round will reduce it by a factor of

\( \dfrac {625-113}{625}=\dfrac{512}{625} \)

And so on. The infinite product has a limit value which is not zero. What is the final volume?

**Note**

Here's how a 3D Menger Sponge looks like, after a few recursions, to help with visualizing this problem. Try to generalize from 3D to 4D. This one, as already noted, will have zero volume.

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