## Task 2 - 130622

Four opaque cubes with edge lengths $a_{1}=24 \mathrm{~cm}, a_{2}=12 \mathrm{~cm}, a_{3}=6 \mathrm{~cm}$ and $a_{4}=3 \mathrm{~cm}$ are to be stacked on top of each other on an opaque tabletop, with the largest at the bottom, the next largest on top of it, and so on, until the smallest cube is on top, where each cube rests completely on the top face of the one below it (or on the tabletop), without protruding beyond this surface.

Determine the total area of the visible (i.e., not covered) surface parts of these cubes!
Each of the dice has exactly 6 faces. Of these, the face on which each stands is not visible. Moreover, the second-largest die covers a part of the top face of the largest die with its base. The same applies to the third-largest and the smallest of the four dice. No other non-visible partial areas occur.

Therefore, the total area sought is obtained by subtracting the sum of the areas of one face each of the second-largest, third-largest, and smallest dice from the sum of the areas of five faces each of the four dice.

Due to $5 \cdot(242+122+62+32)-(122+62+32)=3636$, the sought total area of the visible surface parts of the four dice is $3636 \mathrm{~cm}^{2}$.