## Task A-2.5.

Ivica made a large cube with a side length of $n$ from $n^3$ unit cubes and then painted some of the six faces of the large cube, while leaving others unpainted. When he disassembled the large cube, he found that exactly 1000 unit cubes had no painted faces. Show that this is indeed possible and determine the number of faces of the large cube that Ivica painted.
## Solution.

After painting some sides of a cube composed of $n^{3}$ smaller cubes, the number of unpainted smaller cubes is certainly less than $n^{3}$ and greater than $(n-2)^{3}$.

Thus, $(n-2)^{3}<1000<n^{3}$, so it must be that $(n-1)^{3}=10^{3}$, i.e., $n=11$.

Now let's count the unpainted smaller cubes in all cases, depending on the number of painted sides of the larger cube.

If 1 side of the larger cube is painted, the number of unpainted smaller cubes is $11 \cdot 10 \cdot 10=1210$.

If 2 sides of the larger cube are painted, the number of unpainted smaller cubes is $11 \cdot 10 \cdot 10=1100$ (if the painted sides are adjacent) or $11 \cdot 11 \cdot 9=1089$ (if the painted sides are not adjacent).

If 3 sides of the larger cube are painted, the number of unpainted smaller cubes can be $10 \cdot 10 \cdot 10=1000$ (if all three painted sides share a common vertex) or $11 \cdot 10 \cdot 9=990$ (if the three painted sides do not share a common vertex).

If 4 sides of the larger cube are painted, the number of unpainted smaller cubes is $10 \cdot 10 \cdot 9=900$ (if the two unpainted sides are adjacent) or $11 \cdot 9 \cdot 9=891$ (if the two unpainted sides are not adjacent).

If 5 sides of the larger cube are painted, the number of unpainted smaller cubes is $10 \cdot 9 \cdot 9=810$.

We see that only in the case where three sides of the larger cube are painted can the number of unpainted unit cubes be 1000, so we conclude that Ivica painted three sides of the larger cube.

Note: It is not necessary to calculate the exact number of painted cubes in each case, but to eliminate certain cases.

## COUNTY COMPETITION IN MATHEMATICS

## 3rd grade - high school - A variant

February 15, 2013.

## IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE PROBLEM, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH APPROPRIATELY.