1. **Problem statement:** We have a $32 \times 32 \times 32$ cube made of unit cubes, all six faces painted. We consider all subcubes inside it (of any side length from 1 to 32). We select one subcube at random from the collection $C$ of all subcubes. We want the expected number of its sides that are completely painted. 2. **Total number of subcubes:** For side length $k$ ($1 \le k \le 32$), the number of subcubes of side $k$ is $ (32 - k + 1)^3 = (33 - k)^3$. Total subcubes: $$\sum_{k=1}^{32} (33 - k)^3 = \sum_{m=1}^{32} m^3 = \left(\frac{32 \cdot 33}{2}\right)^2 = 16^2 \cdot 33^2 = 256 \cdot 1089 = 278784.$$ 3. **Painting of subcube sides:** Each subcube has 6 faces. A face of the subcube is painted if it lies on the outer surface of the big cube (since only the outer faces are painted). 4. **Counting subcubes by number of painted faces:** The number of painted faces of a subcube depends on its position inside the big cube. - 0 painted faces: subcube fully inside, not touching any outer face. - 1 painted face: subcube touches exactly one outer face. - 2 painted faces: subcube touches exactly two outer faces (an edge). - 3 painted faces: subcube touches exactly three outer faces (a corner). No subcube can have more than 3 painted faces because the cube has 3 dimensions. 5. **For a subcube of side length $k$, define:** - Number of positions with 3 painted faces (corners): 8 (corners of the big cube) - Number of positions with 2 painted faces (edges but not corners): $$12 \times ((33 - k) - 2) = 12 (31 - k)$$ - Number of positions with 1 painted face (faces but not edges): $$6 \times ((33 - k) - 2)^2 = 6 (31 - k)^2$$ - Number of positions with 0 painted faces (interior): $$((33 - k) - 2)^3 = (31 - k)^3$$ 6. **Check sum for each $k$: ** $$8 + 12(31-k) + 6(31-k)^2 + (31-k)^3 = (33-k)^3$$ 7. **Expected number of painted faces for subcubes of side $k$: ** $$E_k = \frac{8 \times 3 + 12(31-k) \times 2 + 6(31-k)^2 \times 1 + (31-k)^3 \times 0}{(33-k)^3} = \frac{24 + 24(31-k) + 6(31-k)^2}{(33-k)^3}$$ Simplify numerator: $$24 + 24(31-k) + 6(31-k)^2 = 6(31-k)^2 + 24(31-k) + 24$$ 8. **Overall expected value:** $$E = \frac{\sum_{k=1}^{32} E_k \times (33-k)^3}{\sum_{k=1}^{32} (33-k)^3} = \frac{\sum_{k=1}^{32} \left(6(31-k)^2 + 24(31-k) + 24\right)}{\sum_{k=1}^{32} (33-k)^3}$$ Since $E_k$ was weighted by $(33-k)^3$ in denominator, multiplying back cancels denominator, so numerator is sum of numerators. Rewrite sums with $m = 31 - k$: When $k=1$, $m=30$; when $k=32$, $m=-1$ (ignore negative, sum from $m=0$ to $30$): $$\sum_{m=0}^{30} (6m^2 + 24m + 24)$$ Calculate each sum: $$\sum_{m=0}^{30} m^2 = \frac{30 \cdot 31 \cdot 61}{6} = 9455$$ $$\sum_{m=0}^{30} m = \frac{30 \cdot 31}{2} = 465$$ $$\sum_{m=0}^{30} 1 = 31$$ So numerator: $$6 \times 9455 + 24 \times 465 + 24 \times 31 = 56730 + 11160 + 744 = 68634$$ Denominator: $$\sum_{k=1}^{32} (33-k)^3 = \sum_{m=1}^{32} m^3 = \left(\frac{32 \cdot 33}{2}\right)^2 = 278784$$ 9. **Final expected value:** $$E = \frac{68634}{278784}$$ Simplify fraction by dividing numerator and denominator by 6: $$\frac{68634 \div 6}{278784 \div 6} = \frac{11439}{46464}$$ Divide numerator and denominator by 3: $$\frac{11439 \div 3}{46464 \div 3} = \frac{3813}{15488}$$ No further common factors. **Answer:** $$\boxed{\frac{3813}{15488}}$$