Maths Olympiad Prep

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Combinatorics Difficulty 6.2 National olympiad Find the answer

Vova has a square grid 72×7272\times 72. Unfortunately, nn cells are stained with coffee. Determine if Vova always can cut out a clean square 3×33\times 3 without its central cell, if
a) n=699n=699;
b) n=750n=750.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Consider a square grid of size 72×7272 \times 72. We need to determine if Vova can always cut out a clean 3×33 \times 3 square without its central cell, given that some cells are stained.

### Part a) When n=699n = 699

1. Calculate Total Cells:
72×72=5184 72 \times 72 = 5184
There are 5184 cells in total.

2. Calculate Clean Cells:
Clean cells=5184699=4485 \text{Clean cells} = 5184 - 699 = 4485

3. **Determine Clean 3×33 \times 3 Squares:**
- Each 3×33 \times 3 square without the central cell contains 8 cells.
- For Vova to be unable to cut out such a square, all possible arrangements of 3×33 \times 3 squares must contain at least 1 stained cell.

4. **Calculate Total 3×33 \times 3 Configurations:**
(722)×(722)=70×70=4900 (72 - 2) \times (72 - 2) = 70 \times 70 = 4900
There are 4900 possible 3×33 \times 3 square arrangements.

5. Compare Clean Cells and Required Conditions:
- The number of clean cells (4485) is more than half of the total possible 3×33 \times 3 square configurations (2450 if each were to perfectly avoid repeats).
- Since 4485 clean cells is more than half, Vova can always find a 3×33 \times 3 configuration with only clean cells (fewer than 49004900 is required due to overlapping).

Thus, for n=699 n = 699 , Vova can always cut out a clean 3×33 \times 3 square without its central cell.

### Part b) When n=750n = 750

1. Calculate Clean Cells:
Clean cells=5184750=4434 \text{Clean cells} = 5184 - 750 = 4434

2. **Compare with the Requirement for Clean 3×33 \times 3 Squares:**
- The reasoning is similar to part (a).
- However, with 4434 clean cells, ensuring a clean 3×33 \times 3 square for all 4900 possible positions without overlapping via excess becomes untenable.

From combinatorial reasoning and applications of the pigeonhole principle, it's clear that with 4434 clean cells, intersections among the 3×33\times3 blocks will eventually prohibit finding one clean block for every arrangement.

Thus, for n=750 n = 750 , Vova cannot always cut out a clean 3×33 \times 3 square without its central cell.

Therefore, the final answers are:
a) Yes, b) No \boxed{\text{a) Yes, b) No}}

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.