一家大米零售商承诺，服务学习项目的参与者每跑一公里，零售商就捐赠一碗大米。捐赠将以整碗为单位，基于每个人在 28 天结束时的累计跑步距离。多个人的跑步距离不会合并计算。例如，如果 A 跑了 18.8 km，B 跑了 11.2 km，零售商将总共捐赠 29 碗。两位这样的参与者，运动员 A 和 B，将根据各自制定的计划，在总共 28 天内累计他们的跑步距离。

• 运动员 A 计划在第一天跑 5 km，之后每天跑的距离比前一天增加固定的 0.65 km。
• 运动员 B 计划在第一天跑 7 km，之后每天跑的距离比前一天增加 4%。

1. Express the cumulative distance of Athlete A after $n$ days.

Let $n$ be the number of days.
Athlete A's daily distance forms an arithmetic progression with $a = 5$ and $d = 0.65$.
$S_A = \frac{n}{2} [2(5) + (n-1)(0.65)] = \frac{n}{2} (10 + 0.65n - 0.65) = \frac{n}{2} (9.35 + 0.65n)$

2. Express the cumulative distance of Athlete B after $n$ days.

Athlete B's daily distance forms a geometric progression with $a = 7$ and $r = 1.04$.
$S_B = \frac{7(1.04^n - 1)}{1.04 - 1} = \frac{7(1.04^n - 1)}{0.04} = 175(1.04^n - 1)$

3. Find the least number of days for $S_A$ to exceed $S_B$.

We require $S_A > S_B$:
$\frac{n}{2} (9.35 + 0.65n) > 175(1.04^n - 1)$
Using the table function on a graphing calculator:
When $n = 13$, $S_A = 115.7$ km, $S_B \approx 116.39$ km ($S_A < S_B$).
When $n = 14$, $S_A = 129.15$ km, $S_B \approx 128.04$ km ($S_A > S_B$).

The least number of days required is 14.

1. Calculate the cumulative distance run by Athlete A over 28 days

Athlete A's distance forms an arithmetic progression with $a = 5$ and $d = 0.65$.
$S_A = \frac{28}{2}[2(5) + (28-1)(0.65)]$
$= 14[10 + 17.55] = 385.7 \text{ km}$

2. Calculate the cumulative distance run by Athlete B over 28 days

Athlete B's distance forms a geometric progression with $a = 7$ and $r = 1.04$.
$S_B = \frac{7(1.04^{28} - 1)}{1.04 - 1}$
$= \frac{7(1.99870...)}{0.04} \approx 349.77 \text{ km}$

3. Determine the total number of bowls donated by summing the complete bowls from each athlete

Based on the example given in the problem, fractions of a kilometre do not count towards the bowls (truncate/floor the distance).
$\text{Bowls from Athlete A} = 385$
$\text{Bowls from Athlete B} = 349$
$\text{Total bowls} = 385 + 349 = 734$

1. Express the total distance run over 28 days in terms of the first day's distance, $a$.

Let the distance run on day 1 be $a$ km.
The distances form an arithmetic progression with $n = 28$ days and common difference $d = 0.65$ km.
$S_{28} = \frac{28}{2}[2a + (28 - 1)(0.65)]$
$S_{28} = 14(2a + 17.55)$

2. Set up an inequality for the total distance to be at least 400 km and solve for $a$.

$14(2a + 17.55) \ge 400$
$2a + 17.55 \ge \frac{400}{14}$
$2a \ge \frac{200}{7} - 17.55$
$2a \ge 11.0214...$
$a \ge 5.5107...$

3. Convert the minimum distance to metres and round to the nearest metre.

Minimum distance in km = $5.5107...$ km
Minimum distance in metres = $5.5107... \times 1000 = 5510.7...$ m
Rounding to the nearest metre, the minimum distance required is $5511$ m.

1. Find the first day where Athlete B's planned distance exceeds 10 km.

Athlete B's planned distance on day $n$ forms a geometric progression where $a = 7$ and $r = 1.04$.
We want to find $n$ such that:
$7(1.04)^{n-1} > 10$
$1.04^{n-1} > \frac{10}{7}$
$(n-1) \ln 1.04 > \ln \left( \frac{10}{7} \right)$
$n - 1 > 9.094 \implies n > 10.094$
The planned distance strictly exceeds $10 \text{ km}$ starting from day 11. Therefore, Athlete B runs his original planned distances for the first $10$ days.

2. Calculate the total cumulative distance run by Athlete B over the 28 days.

Distance for the first 10 days:
$S_{10} = \frac{7(1.04^{10} - 1)}{1.04 - 1} \approx 84.0427 \text{ km}$
Distance for the remaining 18 days ($28 - 10$):
$18 \times 10 = 180 \text{ km}$
Total distance run:
$84.0427 + 180 = 264.0427 \text{ km}$

3. Determine the number of complete bowls contributed.

Donations are based on the complete cumulative distance of the individual:
$\lfloor 264.0427 \rfloor = 264$