贝蒂（Betty）需要为派对装饰一面长度为 5 米的墙。她从墙的最左端开始安装挂钩，并将每个挂钩依次编号为 “1”、“2”、“3” 等。第 $1^{st}$ 和第 $2^{nd}$ 个挂钩之间的间距为 50 厘米，此后每个相邻挂钩之间的间距比前一个间距短 2 厘米。这将持续进行，直到她因空间不足而无法放置下一个挂钩为止。 贝蒂还将一卷 6 米长的丝带剪成不同长度的片段。剪下的第一段长度为 80 厘米，之后剪下的每一段都比前一段短 10 %。

1. Identify the spacings as an arithmetic progression and write the formula for the sum of the first $n-1$ spacings.

The spacings form an arithmetic progression with first term $a = 50$ and common difference $d = -2$.
Let $n$ be the total number of hooks attached. The distance of the $n^{\text{th}}$ hook from the extreme left is the sum of the first $n-1$ spacings:
$S_{n-1} = \frac{n-1}{2}[2(50) + (n-1-1)(-2)]$
$S_{n-1} = \frac{n-1}{2}[100 - 2n + 4] = (n-1)(52 - n)$

2. Set up an inequality restricting the total distance to the length of the wall.

The total length of the wall is 5 metres, which is $500\text{ cm}$.
Therefore, we require:
$S_{n-1} \le 500$
$(n-1)(52 - n) \le 500$
$52n - n^2 - 52 + n \le 500$
$n^2 - 53n + 552 \ge 0$

3. Solve the quadratic inequality to find the bounds for $n$.

Solving $n^2 - 53n + 552 = 0$:
$n = \frac{53 \pm \sqrt{(-53)^2 - 4(1)(552)}}{2}$
$n = \frac{53 \pm \sqrt{2809 - 2208}}{2}$
$n = \frac{53 \pm \sqrt{601}}{2}$
$n \approx 14.24 \quad \text{or} \quad n \approx 38.76$
For the inequality $n^2 - 53n + 552 \ge 0$, we have $n \le 14.24$ or $n \ge 38.76$.
Since the spacings must be positive ($k^{\text{th}}$ spacing $= 52 - 2k > 0 \implies k < 26$), we consider the valid range $n \le 14.24$.

4. Determine the final integer answer and verify the remaining space.

The maximum integer $n$ satisfying the condition is $n = 14$.
Checking the distances:
$S_{13} = 13 \times (52 - 14) = 494 \text{ cm} \le 500 \text{ cm}$
$S_{14} = 14 \times (52 - 15) = 518 \text{ cm} > 500 \text{ cm}$
After attaching the $14^{\text{th}}$ hook, the remaining space on the wall is $500 - 494 = 6 \text{ cm}$.
The required spacing for the next hook would be $50 - 13(2) = 24 \text{ cm}$. Since $6 \text{ cm} < 24 \text{ cm}$, Betty cannot place the $15^{\text{th}}$ hook.
Thus, she can attach exactly 14 hooks.

1. Identify the parameters of the geometric progression for the lengths of the cut ribbon pieces.

Let the length of the $k^{\text{th}}$ piece cut off be $L_k$ cm.
$L_k$ forms a geometric progression with first term $a = 80$ and common ratio $r = 1 - 0.10 = 0.9$.

2. Set up the inequality for the remaining length of the ribbon.

Initial length of the ribbon is $6 \text{ m} = 600 \text{ cm}$.
The remaining length after $n$ cuts is $(600 - S_n)$ cm.
We require the remaining length to be less than $1 \text{ m}$ ($100 \text{ cm}$):
$600 - S_n < 100$
$S_n > 500$

3. Substitute the formula for the sum of a finite geometric series and simplify.

$\frac{80(1 - 0.9^n)}{1 - 0.9} > 500$
$\frac{80(1 - 0.9^n)}{0.1} > 500$
$800(1 - 0.9^n) > 500$
$1 - 0.9^n > \frac{500}{800}$
$0.9^n < 0.375$

4. Solve for $n$ using logarithms and find the smallest integer value.

Taking the natural logarithm on both sides:
$n \ln 0.9 < \ln 0.375$
Since $\ln 0.9$ is a negative value, the inequality sign flips when dividing:
$n > \frac{\ln 0.375}{\ln 0.9}$
$n > 9.309$
The smallest integer value of $n$ is $10$.

1. Express the length of the $n^{\text{th}}$ piece of ribbon and the distance between the $n^{\text{th}}$ and $(n+1)^{\text{th}}$ hooks.

The length of the $n^{\text{th}}$ piece of ribbon follows a geometric progression with first term $a = 80$ and common ratio $r = 0.9$:
$L_n = 80(0.9)^{n-1}$

The distance between the $n^{\text{th}}$ and $(n+1)^{\text{th}}$ hook follows an arithmetic progression with first term $a = 50$ and common difference $d = -2$:
$D_n = 50 + (n-1)(-2) = 52 - 2n$

2. Formulate the inequality required for the ribbon to be hung.

For the ribbon to be able to hang between the two hooks, its length must be greater than or equal to the distance between the hooks (assuming negligible length is needed for tying):
$80(0.9)^{n-1} \geq 52 - 2n$

1. Formulate the inequality comparing the length of the $n$-th ribbon to the spacing between the $n$-th and $(n+1)$-th hooks.

From part (iii), the length of the $n$-th piece of ribbon must be greater than or equal to the spacing between the $n$-th and $(n+1)$-th hooks.
$80(0.9)^{n-1} \ge 50 + (n-1)(-2)$
$80(0.9)^{n-1} \ge 52 - 2n$

2. Find the maximum integer value of $n$ that satisfies the inequality.

Using systematic substitution (or the table function on a graphing calculator):
When $n = 9$,
$\text{Ribbon length} = 80(0.9)^8 \approx 34.43 \text{ cm}$
$\text{Spacing} = 52 - 2(9) = 34 \text{ cm}$
Since $34.43 \ge 34$, the piece can be hung.

When $n = 10$,
$\text{Ribbon length} = 80(0.9)^9 \approx 30.99 \text{ cm}$
$\text{Spacing} = 52 - 2(10) = 32 \text{ cm}$
Since $30.99 < 32$, this piece is too short to be hung.

Thus, exactly 9 pieces of ribbon can be successfully hung.

3. Determine the corresponding number of hooks with ribbons on them.

Since 9 pieces of ribbon are hung continuously spanning adjacent hooks (from the 1st to the 2nd, the 2nd to the 3rd, and so on up to the 9th to the 10th), the number of hooks involved is $n + 1$.
$\text{Number of hooks} = 9 + 1 = 10$