Problem 1407

### Part (a): Prove that $t > 300$

1. **Construct a set $T$ of rectangles:**
Consider rectangles with dimensions $1 \times 6$, $2 \times 5$, $3 \times 4$, $4 \times 3$, $5 \times 2$, and $6 \times 1$. These rectangles are chosen such that no rectangle is a subrectangle of another.

2. Count the number of such rectangles:
- For $1 \times 6$ rectangles, there are $10 \times 5 = 50$ possible positions.
- For $2 \times 5$ rectangles, there are $9 \times 6 = 54$ possible positions.
- For $3 \times 4$ rectangles, there are $8 \times 7 = 56$ possible positions.
- For $4 \times 3$ rectangles, there are $7 \times 8 = 56$ possible positions.
- For $5 \times 2$ rectangles, there are $6 \times 9 = 54$ possible positions.
- For $6 \times 1$ rectangles, there are $5 \times 10 = 50$ possible positions.

3. Sum the number of rectangles:
$$50 + 54 + 56 + 56 + 54 + 50 = 320$$

4. Conclusion for part (a):
Since $320 > 300$, we have $t \geq 320$, proving that $t > 300$.

### Part (b): Prove that $t < 600$

1. Consider the edges of the board:
Let $E$ be an edge of the $10 \times 10$ table. For each rectangle in $T$, define its "base" as the side parallel to $E$ that is closest to $E$.

2. Count the possible bases:
- For each row, there are $10$ possible positions for the base.
- The number of rows is $10$, and the number of possible bases in each row is the sum of the first $10$ natural numbers:
$$10 + 9 + 8 + \cdots + 1 = \frac{10 \times 11}{2} = 55$$

3. Total number of possible bases:
$$55 \times 10 = 550$$

4. Conclusion for part (b):
Since it is impossible for two rectangles to share the same base without one being a subrectangle of the other, we have $t \leq 550$. Therefore, $t < 600$.

$\blacksquare$