一档名为“伊奥纳倒置塔（Inverted Tower of Ionah）”的广播游戏节目每天都会进行直播。每天都会邀请一位参与者来玩这个游戏。参与者需要将厚度相同但横截面积逐渐增大的圆盘堆叠起来，以建造一座塔。塔的每一层都由一个圆盘组成。塔的底部是一个底面积为 6 cm$^2$ 的圆盘。往上每一层放置的圆盘底面积都比前一层大 8 cm$^2$。如果塔倒塌，游戏将停止，参与者输掉游戏。 艾伦（Alan）是参加第一天游戏的参与者。

1. Formulate the general expression for the base area of the $n$th disc

The base areas form an arithmetic progression with first term $a = 6$ and common difference $d = 8$.
$A_n = a + (n - 1)d$
$A_n = 6 + (n - 1)8 = 8n - 2$

2. Equate the expression to 140 to check if $n$ is an integer

Assume $8n - 2 = 140$.
$8n = 142$
$n = 17.75$
Since $n$ must be a positive integer, it is not possible for a disc to have a base area of $140 \text{ cm}^2$.

1. Calculate the sum of the base areas of the 22 discs using the formula for the sum of an arithmetic progression.

Let the base area of the $n$-th disc be $A_n$. The sequence of base areas forms an arithmetic progression with first term $a = 6$ and common difference $d = 8$.
The sum of the base areas of the first 22 discs, $S_{22}$, is:
$S_{22} = \frac{22}{2} [2(6) + (22 - 1)8]$
$S_{22} = 11 [12 + 168]$
$S_{22} = 11 [180] = 1980 \text{ cm}^2$

2. Set up an equation using the total volume and the total base area to solve for $h$.

The volume of each disc is given by its base area multiplied by its uniform thickness $h$.
$\text{Total volume} = S_{22} \times h$
$1980h = 4950$
$h = \frac{4950}{1980} = 2.5$

1. Formulate an inequality for the time taken on the $n$-th attempt.

Let $n$ be the number of attempts.
The time taken for each attempt decreases by 5% each time, forming a geometric sequence with first term $a = 3$ and common ratio $r = 1 - 0.05 = 0.95$.
The time taken for the $n$-th attempt is given by $T_n = 3(0.95)^{n-1}$.
We require the time taken to be 2 minutes or less:
$3(0.95)^{n-1} \le 2$

2. Solve the inequality to find the minimum integer value of $n$.

$(0.95)^{n-1} \le \frac{2}{3}$
Taking logarithms on both sides (or using trial and improvement):
$(n-1) \lg 0.95 \le \lg \left(\frac{2}{3}\right)$
Since $\lg 0.95$ is negative, the inequality sign flips when dividing:
$n-1 \ge \frac{\lg(2/3)}{\lg 0.95}$
$n-1 \ge 7.9048\dots$
$n \ge 8.9048\dots$
Since the number of attempts $n$ must be an integer, the minimum value is $9$.

1. Formulate the recurrence relation and expand the first few terms to observe the pattern.

Let $u_n$ be the prize money on Day $n$.
$u_1 = 200$
$u_2 = 0.8u_1 + 150 = 200(0.8) + 150$
$u_3 = 0.8u_2 + 150 = 0.8[200(0.8) + 150] + 150$
$u_3 = 200(0.8)^2 + 150(0.8) + 150$

2. Generalise the pattern to find an expression for $u_n$.

By following the pattern up to the $n$-th day, we have $n-1$ multiplications by $0.8$ for the initial $200$, and a descending sum of powers for the $150$ terms:
$u_n = 200(0.8)^{n-1} + 150(0.8)^{n-2} + 150(0.8)^{n-3} + \dots + 150(0.8) + 150$

3. Identify and evaluate the geometric series component.

The series component $150 + 150(0.8) + \dots + 150(0.8)^{n-2}$ is a geometric progression with first term $a = 150$, common ratio $r = 0.8$, and $n-1$ terms.
Using the sum formula $S_{n-1} = \frac{a(1-r^{n-1})}{1-r}$:
$S_{n-1} = \frac{150[1 - (0.8)^{n-1}]}{1 - 0.8}$
$S_{n-1} = \frac{150}{0.2}[1 - (0.8)^{n-1}] = 750[1 - (0.8)^{n-1}]$

4. Substitute the evaluated sum back and simplify to reach the required expression.

$u_n = 200(0.8)^{n-1} + 750[1 - (0.8)^{n-1}]$
$u_n = 200(0.8)^{n-1} + 750 - 750(0.8)^{n-1}$
$u_n = 750 - 550(0.8)^{n-1}$
(Shown)

1. Express the prize money won by the first and second winners in terms of $k$.

The first winner won on Day $k$, so the prize money won is:
$P_k = 750 - 550(0.8)^{k-1}$

After the first win on Day $k$, the prize resets to $\char36{}200$ on Day $k+1$. The second winner wins on Day $24$. This means the second prize money accumulated for $24 - k$ days.
The second winner's prize money is:
$P_{24-k} = 750 - 550(0.8)^{(24-k)-1} = 750 - 550(0.8)^{23-k}$

2. Formulate the inequality based on the given condition.

We are given that the first winner won at least $\char36{}100$ more than the second winner:
$P_k - P_{24-k} \ge 100$
$\left[ 750 - 550(0.8)^{k-1} \right] - \left[ 750 - 550(0.8)^{23-k} \right] \ge 100$
$550(0.8)^{23-k} - 550(0.8)^{k-1} \ge 100$

3. Solve the inequality to find the valid set of integer values for $k$.

Using a graphing calculator's table function or by testing integer values of $k$:

• When $k = 16$: $550(0.8)^7 - 550(0.8)^{15} \approx 96.01 < 100$
• When $k = 17$: $550(0.8)^6 - 550(0.8)^{16} \approx 128.71 \ge 100$

So $k \ge 17$.
Since the first winner must win before the second winner (who won on Day 24), we must have $k < 24$, giving $k \le 23$.
Thus, $17 \le k \le 23$.

The set of possible values of $k$ is $\{17, 18, 19, 20, 21, 22, 23\}$.