2023 AMC 8 详解与视频讲解 | LIVE by Po-Shen Loh

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2023 AMC 8 详解

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所有题目均经美国数学协会（MAA）官方合法授权使用。

1.

求下列表达式的值：

$(8 \times 4 + 2) - (8 + 4 \times 2)$

What is the value of the following expression?

$(8 \times 4 + 2) - (8 + 4 \times 2)$

$0$

$6$

$10$

$18$

$24$

知识点：运算顺序

难度评级：370

小提示：

先按运算顺序计算，再做减法

Follow order of operations before subtracting

大提示：

分别计算两个括号中的表达式

Compute the two parenthesized expressions separately

视频讲解：

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文字解答：

可以按运算顺序化简如下。

$\begin{align*} (32 + 2) - (8 + 8) &= 34 - 16 \\ &= 18 \end{align*}$

所以正确答案是 D。

We can simplify this as follows.

$\begin{align*} (32 + 2) - (8 + 8) &= 34 - 16 \\ &= 18 \end{align*}$

Thus, D is the correct answer.

2.

一张正方形纸如下图所示折叠两次，形成四个相等的部分，然后沿虚线剪开。展开后，纸张会匹配下列哪一个图形？

A square piece of paper is folded twice into four equal quarters, as shown below, then cut along the dashed line. When unfolded, the paper will match which of the following figures?

知识点：折纸对称性

难度评级：660

小提示：

展开时，把剪痕分别关于每条折线反射

Unfold the cut by reflecting it across each fold line

大提示：

两次折叠都展开后，单条剪痕会出现四次

The single cut appears four times after both folds are undone

视频讲解：

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文字解答：

把剪开后的纸展开，就得到下面的图形。

所以正确答案是 E。

We can unfold the cut up paper to achieve the following figure.

Thus, E is the correct answer.

3.

风寒温度衡量人在室外有风时感觉到的寒冷程度。风寒温度可用下面的计算式近似：

$W=T-0.7\cdot S$

其中 $W$ 表示风寒温度， $T$ 表示以华氏度 $(^{\circ}F)$ 计的气温， $S$ 表示以英里每小时计的风速。

若气温是 $36^{\circ}F$ ，风速是每小时 $18$ 英里，下列哪一项最接近估算出的风寒温度？

Wind chill is a measure of how cold people feel when exposed to wind outside. A good estimate for wind chill can be found using this calculation:

$W=T-0.7\cdot S$

Here $W$ represents the wind chill, $T$ represents air temperature measured in degrees Fahrenheit $(^{\circ}F),$ and $S$ represents wind speed measured in miles per hour (mph).

Suppose the air temperature is $36^{\circ}F$ and the wind speed is $18$ mph. Which of the following is closest to the approximate wind chill?

$18$

$23$

$28$

$32$

$35$

知识点：换元法小数估算

难度评级：450

小提示：

代入 $T=36$ 和 $S=18$

Substitute $T=36$ and $S=18$

大提示：

$0.7\cdot18$ 略大于 $12$

$0.7\cdot18$ is a little more than $12$

视频讲解：

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文字解答：

代入公式，风寒温度为

$\begin{align*} 36 - 0.7 \cdot 18 &= 36 - 12.6 \\ &=23.4\text{。} \end{align*}$

最接近的选项是 $23$ 。

所以正确答案是 B。

Using the formula, the wind chill is

$\begin{align*} 36 - 0.7 \cdot 18 &= 36 - 12.6 \\ &=23.4. \end{align*}$

The closest choice is $23.$

Thus, B is the correct answer.

4.

数字 $1$ 到 $49$ 从中心开始按螺旋形排列在一个正方形网格上。下方网格中已经填入了前几个数字。考虑与数字 $7$ 在同一条对角线上的四个阴影方格中将出现的数字。这四个数字中有多少个是质数？

The numbers from $1$ to $49$ are arranged in a spiral pattern on a square grid, beginning at the center. The first few numbers have been entered into the grid below. Consider the four numbers that will appear in the shaded squares, on the same diagonal as the number $7.$ How many of these four numbers are prime?

$0$

$1$

$2$

$3$

$4$

知识点：质数找规律

难度评级：960

小提示：

继续填写螺旋，直到阴影对角线上的数都确定

Continue the spiral until the shaded diagonal is filled

大提示：

分别检查每个阴影数字是否能被小质数整除

Check each shaded number for divisibility by small primes

视频讲解：

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文字解答：

把其余的数填进去，就得到完整的网格。

由此可以看出，阴影方格中只有 $19$ 、 $23$ 和 $47$ 是质数。

所以正确答案是 D。

We can fill in the other numbers to get the complete grid.

From this, we can see that the only prime numbers in the shaded boxes are $19,$ $23,$ and $47.$

Thus, D is the correct answer.

5.

一个湖中有 $250$ 条鳟鱼，还有其他多种鱼。一位海洋生物学家从湖中捕捉并放回 $180$ 条鱼作为样本，其中 $30$ 条被识别为鳟鱼。假设样本中鳟鱼占总鱼数的比例与湖中相同，湖中共有多少条鱼？

A lake contains $250$ trout, along with a variety of other fish. When a marine biologist catches and releases a sample of $180$ fish from the lake, $30$ are identified as trout. Assume that the ratio of trout to the total number of fish is the same in both the sample and the lake. How many fish are there in the lake?

$1250$

$1500$

$1750$

$1800$

$2000$

知识点：比与比例

难度评级：720

小提示：

样本的 $180$ 条鱼中有 $30$ 条鳟鱼

The sample has $30$ trout out of $180$ fish

大提示：

鳟鱼占湖中所有鱼的 $\frac16$

Trout are $\frac16$ of all fish in the lake

视频讲解：

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文字解答：

样本中鳟鱼比例为 $\dfrac{30}{180} = \dfrac{1}{6}$ 。因此湖中所有鱼数是鳟鱼数的 $6$ 倍。

湖中鱼的总数为 $6 \cdot 250 = 1500$ 。

所以正确答案是 B。

Note that $\dfrac{30}{180} = \dfrac{1}{6}.$ This means that a sixth of the fish in the lake are trout, or in other words, the total number of fish is $6$ times the number of trout.

Therefore, there are $6 \cdot 250 = 1500$ fish in the lake.

Thus, B is the correct answer.

6.

把数字 $2$ 、 $0$ 、 $2$ 和 $3$ 填入下列表达式中，每个方框放一个数字。这个表达式的最大可能值是多少？

The digits $2,$ $0,$ $2,$ and $3$ are placed in the expression below, one digit per box. What is the maximum possible value of the expression?

$0$

$8$

$9$

$16$

$18$

知识点：指数最优化

难度评级：900

小提示：

避免把 $0$ 放在底数位置

Avoid putting $0$ in a base

大提示：

把 $0$ 用作指数，使其中一个因子变成 $1$

Use $0$ as an exponent so one factor becomes $1$

视频讲解：

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文字解答：

不要把 $0$ 放在底数位置，因为那样会使整个表达式等于 $0$ 。

所以 $0$ 必须放在指数位置。指数为 $0$ 的那一项会自动等于 $1$ 。

把两个 $2$ 中的一个用于 $2^0$ ，就能把另一个 $2$ 和 $3$ 留给另一项。

另一项可以是 $2^3$ 或 $3^2$ 。因为 $3^2$ 更大，所以取 $3^2$ 。

于是最大值为 $2^0 \times 3^2 = 1 \times 9 = 9\text{。}$

所以正确答案是 C。

Note that we do not want $0$ as a base, since that would make the expression equal to $0.$

This means that $0$ must be an exponent. The number whose exponent is $0$ will automatically evaluate to $1.$

Using one of the two copies of $2$ in the factor $2^0$ leaves $2$ and $3$ for the other factor.

The other factor can then be $2^3$ or $3^2.$ Since $3^2$ is larger, we use $3^2.$

This gives us a final value of $2^0 \times 3^2 = 1 \times 9 = 9.$

Thus, C is the correct answer.

7.

一个边平行于 $x$ 轴和 $y$ 轴的矩形，其一对相对顶点为 $(15, 3)$ 和 $(16, 5)$ 。一条直线经过点 $A(0, 0)$ 和 $B(3, 1)$ 。另一条直线经过点 $C(0, 10)$ 和 $D(2, 9)$ 。矩形上有多少个点至少在这两条直线之一上？

A rectangle, with sides parallel to the $x$ -axis and $y$ -axis, has opposite vertices located at $(15, 3)$ and $(16, 5).$ A line is drawn through points $A(0, 0)$ and $B(3, 1).$ Another line is drawn through points $C(0, 10)$ and $D(2, 9).$ How many points on the rectangle lie on at least one of the two lines?

$0$

$1$

$2$

$3$

$4$

知识点：坐标几何一次方程

难度评级：1070

小提示：

写出两条直线的方程

Find the equations of the two lines

大提示：

矩形满足 $15\le x\le16$ 且 $3\le y\le5$

The rectangle has $15\le x\le16$ and $3\le y\le5$

视频讲解：

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文字解答：

画出这两条直线。

从图中可见，只有矩形左上角落在其中一条直线上。

所以正确答案是 B。

We can graph the two lines.

From this, we see that only the top left corner of the rectangle intersects either line.

Thus, B is the correct answer.

8.

洛拉、洛洛、蒂娅和蒂约参加乒乓球比赛。每位选手都与另外三位选手各比赛两次。下方显示了选手们的胜负记录。数字 $1$ 和 $0$ 分别表示胜和负。例如，洛拉赢了五场，输了第四场。蒂约的胜负记录是什么？

Lola, Lolo, Tiya, and Tiyo participated in a ping pong tournament. Each player competed against each of the other three players exactly twice. Shown below are the win-loss records for the players. The numbers $1$ and $0$ represent a win or loss, respectively. For example, Lola won five matches and lost the fourth match. What was Tiyo’s win-loss record?

$000101$

$001001$

$010000$

$010101$

$011000$

知识点：逻辑推理奇偶性

难度评级：1020

小提示：

每场比赛贡献一个胜和一个负

Each match contributes one win and one loss

大提示：

每列记录中必须有两个数等于 $1$ ，另两个数等于 $0$

In each column of the records, two entries must equal $1$ and two must equal $0$

视频讲解：

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文字解答：

每一列代表同一轮的两场比赛，因此恰好包含两个胜场记录 $1$ 和两个负场记录 $0$ 。

逐列填入使该列达到两个 $1$ 和两个 $0$ 所需的数，得到蒂约的胜负记录为 $000101$ 。

所以正确答案是 A。

Each column represents one round of two matches, so every column must contain exactly two $1$ ’s and two $0$ ’s.

Completing each column to contain two $1$ entries and two $0$ entries gives Tiyo’s record $000101.$

Thus, A is the correct answer.

9.

玛莱卡在山上滑雪。下图显示了她沿雪道滑行时，相对于山脚的海拔高度（米）。她总共有多少秒处在 $4$ 米到 $7$ 米之间的海拔高度？

Malaika is skiing on a mountain. The graph below shows her elevation, in meters, above the base of the mountain as she skis along a trail. In total, how many seconds does she spend at an elevation between $4$ and $7$ meters?

$6$

$8$

$10$

$12$

$14$

知识点：数据与图表解读

难度评级：1020

小提示：

数出图像位于两个高度之间的时间区间

Count the time intervals where the graph is between the two heights

大提示：

图中可见的三个分段分别计算

Split the count into the three separate intervals visible on the graph

视频讲解：

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文字解答：

她第一次达到 $7$ 米是在 $2$ 秒时。

随后她在 $4$ 秒后低于 $4$ 米，这一段贡献 $4 - 2 = 2$ 秒。

接着她在 $6$ 秒时重新高于 $4$ 米，并在 $10$ 秒时再次达到 $7$ 米。

这一段贡献 $10 - 6 = 4$ 秒。最后她在 $12$ 秒时低于 $7$ 米。

再到 $14$ 秒时她低于 $4$ 米，最后一段贡献 $14 - 12 = 2$ 秒。

三个时间段的总长度为 $2 + 4 + 2 = 8$ 所以正确答案是 B。

The first time that she hits an elevation of $7$ meters is at $2$ seconds.

She then dips below $4$ meters after $4$ seconds. This adds $4 - 2 = 2$ seconds to the total answer.

Malaika then goes above $4$ meters at $6$ seconds. She hits $7$ meters again at $10$ seconds.

This adds $10 - 6 = 4$ more seconds to the total. She finally dips below $7$ meters for the last time at $12$ seconds.

She then falls below $4$ meters at $14$ seconds, finally adding $14 - 12 = 2$ seconds to the total time.

The desired total is $2 + 4 + 2 = 8$ Thus, B is the correct answer.

10.

哈罗德做了一个李子派去野餐。他只吃了这个派的 $\frac{1}{4}$ ，把剩下的留给朋友。一只驼鹿经过，吃掉了哈罗德留下部分的 $\frac{1}{3}$ 。之后，一只豪猪又吃掉了驼鹿留下部分的 $\frac{1}{3}$ 。豪猪离开后，原来的派还剩多少？

Harold made a plum pie to take on a picnic. He was able to eat only $\frac{1}{4}$ of the pie, and he left the rest for his friends. A moose came by and ate $\frac{1}{3}$ of what Harold left behind. After that, a porcupine ate $\frac{1}{3}$ of what the moose left behind. How much of the original pie still remained after the porcupine left?

$\dfrac{1}{12}$

$\dfrac{1}{6}$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{5}{12}$

知识点：分数

难度评级：900

小提示：

始终用原来整个派的分数来表示剩余量

Work with the fraction of the original pie remaining

大提示：

哈罗德吃完后剩 $\frac34$ ；驼鹿吃完后再乘 $\frac23$

After Harold, $\frac34$ remains; after the moose, multiply by $\frac23$

视频讲解：

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文字解答：

哈罗德把整个派的 $1 - \frac{1}{4} = \frac{3}{4}$ 留给了朋友。

驼鹿吃掉其中 $\frac{1}{3} \cdot \frac{3}{4} = \frac{1}{4}$ ，所以剩下 $\frac{3}{4} - \frac{1}{4} = \frac{1}{2}$ 。

豪猪再吃掉剩余的 $\frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6}$ ，所以最后剩下 $\frac{1}{2} - \frac{1}{6} = \frac{1}{3}$ 。

所以正确答案是 D。

Harold left $1 - \frac{1}{4} = \frac{3}{4}$ of the pie for his friends.

The moose ate $\frac{1}{3} \cdot \frac{3}{4} = \frac{1}{4}$ of the pie, leaving $\frac{3}{4} - \frac{1}{4} = \frac{1}{2}$ of the pie.

Finally, the porcupine ate $\frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6}$ of the pie. This leaves $\frac{1}{2} - \frac{1}{6} = \frac{1}{3}$ of the pie.

Thus, D is the correct answer.

11.

NASA 的“毅力号”探测车于 $2020$ 年 7 月 $30$ 日发射。它行进 $292{,}526{,}838$ 英里后，约 $6.5$ 个月后降落在火星的杰泽罗陨石坑。下列哪一项最接近探测车的平均星际速度，单位为英里每小时？

NASA’s Perseverance Rover was launched on July $30,$ $2020.$ After traveling $292{,}526{,}838$ miles, it landed on Mars in Jezero Crater about $6.5$ months later. Which of the following is closest to the Rover’s average interplanetary speed in miles per hour?

$6{,}000$

$12{,}000$

$60{,}000$

$120{,}000$

$600{,}000$

知识点：路程、速度与时间估算单位换算

难度评级：1100

小提示：

把 $6.5$ 个月近似为约 $200$ 天

Approximate $6.5$ months as about $200$ days

大提示：

把每天英里数除以 $24$ ，得到每小时英里数

Convert miles per day to miles per hour by dividing by $24$

视频讲解：

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文字解答：

把距离近似为 $300{,}000{,}000$ 英里，并把 $6.5$ 个月近似为 $30 \times 6.5 = 195 \approx 200$ 天。

于是每天约行进 $300{,}000{,}000 \div 200 = 1{,}500{,}000$ 英里。

再除以 $24$ ，得到约 $62{,}500$ 英里每小时，最接近 $60{,}000$ 。

所以正确答案是 C。

We can round the distance to $300{,}000{,}000$ miles and approximate $6.5$ months as $30 \times 6.5 = 195 \approx 200$ days.

This gives $300{,}000{,}000 \div 200 = 1{,}500{,}000$ miles per day.

Dividing by $24$ gives about $62{,}500$ miles per hour, which is closest to $60{,}000.$

Thus, C is the correct answer.

12.

下图显示一个大的未涂色圆，内部有若干较小的未涂色圆和阴影圆。大未涂色圆内部的面积中，阴影部分占几分之几？

The figure below shows a large unshaded circle with a number of smaller unshaded and shaded circles in its interior. What fraction of the interior of the large unshaded circle is shaded?

$\dfrac{1}{4}$

$\dfrac{11}{36}$

$\dfrac{1}{3}$

$\dfrac{19}{36}$

$\dfrac{5}{9}$

知识点：圆面积面积比

难度评级：1370

小提示：

圆面积与半径的平方成正比

Circle areas are proportional to the square of the radius

大提示：

用最小圆的面积作为单位来数阴影面积

Count shaded area in units of the smallest circle’s area

视频讲解：

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文字解答：

不妨设每个最小圆的半径为 $1$ 。

这样有 $3$ 个阴影单位圆，总面积为 $3 \cdot 1^2 \pi = 3 \pi$ 。

另外还有一个半径为 $4$ 的阴影圆，其中挖去了两个半径为 $2$ 的未涂色圆。

这部分额外阴影面积为 $\begin{align*} 4^2 \pi - 2 \cdot 2^2 \pi &= 16 \pi - 8 \pi \\ &= 8 \pi\text{。} \end{align*}$

阴影总面积为 $8\pi + 3\pi = 11\pi\text{。}$ 大未涂色圆的面积为 $6^2\pi = 36\pi\text{。}$ 所以所求分数为 $\dfrac{11\pi}{36\pi} = \dfrac{11}{36}\text{。}$

所以正确答案是 B。

Let each smallest circle have radius $1.$

This means that there are $3$ shaded unit circles, which total to $3 \cdot 1^2 \pi = 3 \pi$ area.

There is also a shaded circle with radius $4$ with two unshaded circles of radius $2$ inside.

This gives us an extra shaded area of $\begin{align*} 4^2 \pi - 2 \cdot 2^2 \pi &= 16 \pi - 8 \pi \\ &= 8 \pi. \end{align*}$

The total shaded area is $8\pi + 3\pi = 11\pi.$ The area of the large unshaded circle is $6^2\pi = 36\pi.$ Therefore, the desired fraction is $\dfrac{11\pi}{36\pi} = \dfrac{11}{36}.$

Thus, B is the correct answer.

13.

在自行车赛路线中，起点和终点之间均匀分布着 $7$ 个补水站，如下图所示。另外还有 $2$ 个维修站也均匀分布在起点和终点之间。第 $3$ 个补水站位于第 $1$ 个维修站之后 $2$ 英里处。这场比赛全长多少英里？

Along the route of a bicycle race, $7$ water stations are evenly spaced between the start and finish lines, as shown in the figure below. There are also $2$ repair stations evenly spaced between the start and finish lines. The $3$ rd water station is located $2$ miles after the $1$ st repair station. How long is the race in miles?

$8$

$16$

$24$

$48$

$96$

知识点：分数比与比例

难度评级：1240

小提示：

七个补水站把赛程分成 $8$ 个相等间隔

Seven water stations divide the race into $8$ equal gaps

大提示：

两个维修站把赛程分成 $3$ 个相等间隔

Two repair stations divide the race into $3$ equal gaps

视频讲解：

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文字解答：

第 $3$ 个补水站在全程的 $\frac{3}{8}$ 处，因为补水站把全程分成 $8$ 个相等间隔。

第一个维修站在全程的 $\frac{1}{3}$ 处。两站之间的距离占全程的 $\dfrac{3}{8} - \dfrac{1}{3} = \dfrac{1}{24}\text{。}$ 这段距离是 $2$ 英里，所以全程为 $24 \cdot 2 = 48$ 英里。

所以正确答案是 D。

The $3$ rd water station is located $\frac{3}{8}$ of the way along the race (the water stations split the race up into $8$ equal spaces).

The first repair station is located $\frac{1}{3}$ of the way along the race. The distance between the stations is $\dfrac{3}{8} - \dfrac{1}{3} = \dfrac{1}{24}$ of the race length. This distance is $2$ miles, so the race is $24 \cdot 2 = 48$ miles long.

Thus, D is the correct answer.

14.

尼古拉计划给朋友安东寄一个包裹，安东是集邮爱好者。为了支付邮资，尼古拉想用很多邮票贴满包裹。假设他有 $5$ 分、 $10$ 分和 $25$ 分邮票，每种正好 $20$ 张。尼古拉最多能用多少张邮票凑成正好 $\$7.10$ 的邮资？

（注意： $\$7.10$ 表示 $7$ 美元 $10$ 美分。一美元等于 $100$ 美分。）

Nicolas is planning to send a package to his friend Anton, who is a stamp collector. To pay for the postage, Nicolas would like to cover the package with a large number of stamps. Suppose he has a collection of $5$ -cent, $10$ -cent, and $25$ -cent stamps, with exactly $20$ of each type. What is the greatest number of stamps Nicolas can use to make exactly $\$7.10$ in postage?

(Note: The amount $\$7.10$ corresponds to $7$ dollars and $10$ cents. One dollar is worth $100$ cents.)

$45$

$46$

$51$

$54$

$55$

知识点：最优化模运算

难度评级：1480

小提示：

尽量使用低面值邮票

Use as many low-value stamps as possible

大提示：

所有 $5$ 分和 $10$ 分邮票合计 $300$ 分，剩余不是 $25$ 的倍数

All $5$ - and $10$ -cent stamps total $300$ cents, leaving a nonmultiple of $25$

视频讲解：

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文字解答：

设 $a$ 、 $b$ 、 $c$ 分别为 $5$ 分、 $10$ 分、 $25$ 分邮票的张数，并设 $n=a+b+c$ 。金额方程除以 $5$ 得 $a+2b+5c=142\text{，}$ 所以 $n+b+4c=142$ 。

假设 $n\ge56$ ，则 $b+4c\le86$ 。又因为 $a=142-2b-5c\le20\text{，}$ 所以 $2b+5c\ge122$ 。合并这两个不等式可得 $c\le16$ ，但这样 $2b+5c\le2(20)+5(16)=120$ ，矛盾。因此最多只能使用 $55$ 张邮票。

取 $a=19$ 、 $b=19$ 、 $c=17$ 时可以达到这个上界： $19(5)+19(10)+17(25)=710\text{。}$ 此时邮票总数为 $19+19+17=55$ 。

所以正确答案是 E。

Let $a,$ $b,$ $c$ be the numbers of $5$ -, $10$ -, and $25$ -cent stamps, and let $n=a+b+c.$ Dividing the value equation by $5$ gives $a+2b+5c=142,$ so $n+b+4c=142.$

Suppose $n\ge56.$ Then $b+4c\le86.$ Also $a=142-2b-5c\le20,$ so $2b+5c\ge122.$ Combining these inequalities gives $c\le16,$ but then $2b+5c\le2(20)+5(16)=120,$ a contradiction. Thus at most $55$ stamps can be used.

The bound is attainable with $a=19,$ $b=19,$ $c=17:$ $19(5)+19(10)+17(25)=710.$ Hence the greatest possible number of stamps is $19+19+17=55.$

Thus, E is the correct answer.

15.

维斯瓦姆每天步行半英里去学校。他的路线由 $10$ 个等长街区组成，每走一个街区需要 $1$ 分钟。今天，在走了 $5$ 个街区后，维斯瓦姆发现他必须绕路，要走 $3$ 个等长街区而不是一个街区才能到达下一个街角。从他开始绕路时起，他必须以多少英里每小时的速度步行，才能按平常时间到达学校？

Viswam walks half a mile to get to school each day. His route consists of $10$ city blocks of equal length and he takes one minute to walk each block. Today, after walking $5$ blocks, Viswam discovers that he has to make a detour, walking $3$ blocks of equal length instead of $1$ block to reach the next corner. From the time he starts his detour, at what speed, in miles per hour, must Viswam walk in order to arrive at school at his usual time?

$4$

$4.2$

$4.5$

$4.8$

$5$

知识点：路程、速度与时间单位换算

难度评级：1410

小提示：

一个街区是 $\frac1{20}$ 英里

One block is $\frac1{20}$ mile

大提示：

从绕路点开始，他要在原来的 $5$ 分钟内走 $7$ 个街区

From the detour point, he must walk $7$ blocks in the usual $5$ minutes

视频讲解：

Click to load, then click again to play

文字解答：

半英里对应 $10$ 个街区，所以一个街区长 $\dfrac{1}{2} \div 10 = \dfrac{1}{20}$ 英里。

从绕路开始，维斯瓦姆需要走 $7 \cdot \dfrac{1}{20} = \dfrac{7}{20}$ 英里。

正常情况下，从这个位置到学校需要 $5$ 分钟。现在他需要在这 $5$ 分钟内走完 $\dfrac{7}{20}$ 英里。

注意 $5$ 分钟等于 $\dfrac{5}{60} = \dfrac{1}{12}$ 小时，所以所需速度为 $\begin{align*} \dfrac{\frac{7}{20}}{\frac{1}{12}} &= 12 \cdot \dfrac{7}{20} \\&= \dfrac{21}{5} \\ &= 4.2 \end{align*}$ 英里每小时。

所以正确答案是 B。

If half a mile is the same as $10$ blocks, then one block is $\dfrac{1}{2} \div 10 = \dfrac{1}{20}$ miles.

Starting from the detour, Viswam has to walk $7 \cdot \dfrac{1}{20} = \dfrac{7}{20}$ miles.

Normally, from this spot Viswam would take $5$ minutes to walk to school. Now he has to travel $\dfrac{7}{20}$ miles in $5$ minutes.

Note that $5$ minutes is $\dfrac{5}{60} = \dfrac{1}{12}$ hours. This means his speed must be $\begin{align*} \dfrac{\frac{7}{20}}{\frac{1}{12}} &= 12 \cdot \dfrac{7}{20} \\&= \dfrac{21}{5} \\ &= 4.2 \end{align*}$ miles per hour.

Thus, B is the correct answer.

16.

字母 $P$ 、 $Q$ 、 $R$ 按下图所示的规律填入一个 $20 \times 20$ 表格。完成后的表格中会出现多少个 $P$ 、 $Q$ 和 $R$ ？

The letters $P,$ $Q,$ and $R$ are entered into a $20 \times 20$ table according to the pattern shown below. How many $P$ s, $Q$ s, and $R$ s will appear in the completed table?

$132$ 个 $P$ ， $134$ 个 $Q$ ， $134$ 个 $R$

$132$ $P$ s, $134$ $Q$ s, $134$ $R$ s

$133$ 个 $P$ ， $133$ 个 $Q$ ， $134$ 个 $R$

$133$ $P$ s, $133$ $Q$ s, $134$ $R$ s

$133$ 个 $P$ ， $134$ 个 $Q$ ， $133$ 个 $R$

$133$ $P$ s, $134$ $Q$ s, $133$ $R$ s

$134$ 个 $P$ ， $132$ 个 $Q$ ， $134$ 个 $R$

$134$ $P$ s, $132$ $Q$ s, $134$ $R$ s

$134$ 个 $P$ ， $133$ 个 $Q$ ， $133$ 个 $R$

$134$ $P$ s, $133$ $Q$ s, $133$ $R$ s

知识点：找规律模运算

难度评级：1540

小提示：

每行每列的规律每 $3$ 个格子重复一次

The pattern repeats every $3$ entries along each row and column

大提示：

因为 $20=3\cdot6+2$ ，每行中有两个字母会多出现一次

Since $20=3\cdot6+2$ , two letters get one extra appearance in each line

视频讲解：

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文字解答：

因为 $20 = 3 \cdot 6 + 2$ ，每列底部的 $2$ 个字母会比第三个字母多出现一次。

所以每列中第三个字母出现 $6$ 次，另外 $2$ 个字母各出现 $7$ 次。

在 $20$ 列中，有 $7$ 列的 $P$ 只出现 $6$ 次，其余 $13$ 列中则出现 $7$ 次。 $R$ 的情况相同。

因此 $P$ 和 $R$ 都各出现 $7\cdot6+13\cdot7=42+91=133$ 次。

因此 $Q$ 出现了 $\begin{align*} 20 \cdot 20 - 2 \cdot 133 &=400 - 266 \\ &= 134 \end{align*}$ 次。

所以正确答案是 C。

Since $20 = 3 \cdot 6 + 2,$ the bottom $2$ letters in each column will occur one more time than the third letter.

This means that the third letter in each column will occur $6$ times, whereas the other $2$ will appear $7$ times.

Across the $20$ columns, $P$ is the letter that appears $6$ times in $7$ columns and appears $7$ times in the other $13$ columns. The same is true of $R.$

Thus $P$ and $R$ each appear $7\cdot6+13\cdot7=42+91=133$ times.

$Q$ will therefore appear $\begin{align*} 20 \cdot 20 - 2 \cdot 133 &=400 - 266 \\ &= 134 \end{align*}$ times.

Thus, C is the correct answer.

17.

一个正八面体有八个等边三角形面，每个顶点处有四个面相交。俊将通过折叠下面这张纸做出图中所示的正八面体。哪个编号的面最终会在阴影区域 $Q$ 的右侧？

A regular octahedron has eight equilateral triangle faces with four faces meeting at each vertex. Jun will make the regular octahedron shown in the figure by folding the piece of paper below. Which numbered face will end up to the right of the shaded region $Q?$

$1$

$2$

$3$

$4$

$5$

知识点：展开图（立体几何）多面体

难度评级：1580

小提示：

跟踪哪些面会折到阴影面 $Q$ 周围

Track which faces fold around the shaded face $Q$

大提示：

面 $2$ 、 $3$ 、 $4$ 、 $5$ 形成与 $Q$ 相对的那一半

Faces $2,$ $3,$ $4,$ $5$ form the opposite half from $Q$

视频讲解：

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文字解答：

折叠后，编号 $2$ 、 $3$ 、 $4$ 和 $5$ 的面形成八面体的下半部分，因此剩余四个面构成上半部分。

候选面是 $1$ 、 $6$ 和 $7$ 。从展开图可以看出， $6$ 会在 $Q$ 的左侧，而 $7$ 又在 $6$ 的左侧。

因此，剩下的面 $1$ 必须在 $Q$ 的右侧。

所以正确答案是 A。

Begin by observing that when folded, the faces labelled $2,$ $3,$ $4,$ and $5$ form the bottom half of the octahedron. As such, the remaining four faces must make up the top half of the octahedron.

From here, we have narrowed down our possibilities to $1,$ $6,$ and $7.$ We can see that $6$ will be the face to the left of the shaded region $Q.$ This also gives us that $7$ is to the left of $6.$

Therefore, we know that the only remaining face, $1,$ must be to the right of the shaded region $Q.$

Thus, A is the correct answer.

18.

蚱蜢格蕾塔坐在池塘中一长排睡莲叶上。从任意一片睡莲叶出发，格蕾塔可以向右跳 $5$ 片，或向左跳 $3$ 片。格蕾塔至少要跳多少次，才能到达从起点向右 $2023$ 片的睡莲叶？

Greta Grasshopper sits on a long line of lily pads in a pond. From any lily pad, Greta can jump $5$ pads to the right or $3$ pads to the left. What is the fewest number of jumps Greta must make to reach the lily pad located $2023$ pads to the right of her starting position?

$405$

$407$

$409$

$411$

$413$

知识点：丢番图方程模运算最优化

难度评级：1740

小提示：

先用 $404$ 次向右跳到 $2020$

Start with $404$ jumps to the right, which reaches $2020$

大提示：

额外跳跃的净位移必须为 $3$

The extra jumps must have net displacement $3$

视频讲解：

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文字解答：

设向右跳 $r$ 次，向左跳 $l$ 次，则需要 $5r-3l=2023\text{。}$ 模 $5$ 化简得 $-3l\equiv3\pmod5$ ，所以 $l\equiv4\pmod5$ 。

为了最少跳跃，取满足条件的最小 $l$ ，也就是 $l=4$ 。此时 $5r-12=2023$ ，所以 $r=407$ 。

总跳数为 $407+4=411$ 。

所以正确答案是 D。

Let $r$ be the number of right jumps and $l$ be the number of left jumps. We need $5r-3l=2023.$ Reducing modulo $5$ gives $-3l\equiv3\pmod5$ , so $l\equiv4\pmod5$ .

To minimize the total number of jumps, use the smallest possible $l$ , namely $l=4$ . Then $5r-12=2023$ , so $r=407$ .

The fewest number of jumps is $407+4=411$ .

Thus, D is the correct answer.

19.

一个等边三角形放在一个更大的等边三角形内部，使得它们之间的区域可分成三个全等的梯形，如下图所示。内三角形边长是大三角形边长的 $\frac{2}{3}$ 。一个梯形的面积与内三角形的面积之比是多少？

An equilateral triangle is placed inside a larger equilateral triangle so that the region between them can be divided into three congruent trapezoids, as shown below. The side length of the inner triangle is $\frac{2}{3}$ the side length of the larger triangle. What is the ratio of the area of one trapezoid to the area of the inner triangle?

$1 : 3$

$3 : 8$

$5 : 12$

$7 : 16$

$4 : 9$

知识点：面积比相似

难度评级：1410

小提示：

相似三角形面积按边长比例的平方缩放

Areas of similar triangles scale as the square of side lengths

大提示：

内三角形面积是大三角形面积的 $\left(\frac23\right)^2$

The inner triangle has $\left(\frac23\right)^2$ of the large triangle’s area

视频讲解：

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文字解答：

内三角形边长是大三角形的 $\frac{2}{3}$ ，所以面积是大三角形的 $\left(\frac{2}{3}\right)^2 = \frac{4}{9}$ 。

三个梯形总面积是大三角形的 $1 - \frac{4}{9} = \frac{5}{9}$ 。

一个梯形面积是大三角形的 $\frac{5}{9} \div 3 = \frac{5}{27}$ 。

它与内三角形的面积比为 $\dfrac{\frac{5}{27}}{\frac{4}{9}} = \dfrac{5}{27} \cdot \dfrac{9}{4} = \dfrac{5}{12}\text{。}$

所以正确答案是 C。

Since the inner triangle’s side length is $\frac{2}{3}$ the side length of the outer triangle, its area is $\left(\frac{2}{3}\right)^2 = \frac{4}{9}$ the area of the outer triangle.

This means that the three trapezoids are $1 - \frac{4}{9} = \frac{5}{9}$ the area of the outer triangle.

Therefore, one trapezoid is $\frac{5}{9} \div 3 = \frac{5}{27}$ the area of the outer triangle.

This makes the ratio of the areas of one trapezoid and the inner triangle $\dfrac{\frac{5}{27}}{\frac{4}{9}} = \dfrac{5}{27} \cdot \dfrac{9}{4} = \dfrac{5}{12}.$

Thus, C is the correct answer.

20.

在列表 $3$ 、 $3$ 、 $8$ 、 $11$ 、 $28$ 中插入两个整数，使其极差变为原来的两倍。众数和中位数保持不变。这两个新增数字的和最大可能是多少？

Two integers are inserted into the list $3,$ $3,$ $8,$ $11,$ $28$ to double its range. The mode and median remain unchanged. What is the maximum possible sum of the two additional numbers?

$56$

$57$

$58$

$60$

$61$

知识点：极差中位数（数据）众数最优化

难度评级：1600

小提示：

原极差是 $25$ ，所以新极差必须是 $50$

The original range is $25$ , so the new range must be $50$

大提示：

为了保持中位数为 $8$ ，需在 $8$ 下方加一个数，在 $8$ 上方加一个数

To keep the median $8$ , add one number below $8$ and one above $8$

视频讲解：

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文字解答：

原来的极差是 $28-3=25$ ，所以新极差必须是 $50$ 。要保持中位数为 $8$ ，新增的一个数 $x$ 必须小于 $8$ ，另一个数 $y$ 必须大于 $8$ 。

若 $x\ge3$ ，最小值仍为 $3$ ，所以新的最大值必须是 $53$ 。取 $x=3$ 会改变众数，因此 $x\le7$ 。所以 $x+y\le7+53=60$ 。

若 $x<3$ 且 $y>28$ ，则 $y-x=50$ ，所以 $x+y=2x+50\le54$ 。若 $y\le28$ ，最大值仍为 $28$ ，这会迫使 $x=-22$ ，得到的和更小。因此没有任何情况能超过 $60$ 。

取 $x=7$ 和 $y=53$ 时，众数和中位数都不变，极差为 $50$ ，所以最大和是 $60$ 。

所以正确答案是 D。

The original range is $28-3=25,$ so the new range must be $50.$ To keep the median $8,$ one added number $x$ must be less than $8$ and the other, $y,$ must be greater than $8.$

If $x\ge3,$ the minimum remains $3,$ so the new maximum must be $53.$ The value $x=3$ would change the mode, so $x\le7.$ Hence $x+y\le7+53=60.$

If $x<3$ and $y>28,$ then $y-x=50,$ so $x+y=2x+50\le54.$ If $y\le28,$ the maximum remains $28,$ forcing $x=-22$ and giving an even smaller sum. Therefore no case exceeds $60.$

The values $x=7$ and $y=53$ preserve the mode and median and give range $50,$ so the maximum sum is $60.$

Thus, D is the correct answer.

21.

阿丽娜把数字 $1$ 、 $2$ 、 $\cdots$ 、 $9$ 分别写在卡片上，每张卡片写一个数。她想把这些卡片分成 $3$ 组，每组 $3$ 张，使每组数字之和相同。有多少种分法？

Alina writes the numbers $1,$ $2,$ $\cdots,$ $9$ on separate cards, one number per card. She wishes to divide the cards into $3$ groups of $3$ cards so that the sum of the numbers in each group will be the same. In how many ways can this be done?

$0$

$1$

$2$

$3$

$4$

知识点：分拆与有序分拆分类讨论

难度评级：1690

小提示：

每组和必须为 $15$

Each group must have sum $15$

大提示：

先看包含 $9$ 的那一组

Look at the group containing $9$

视频讲解：

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文字解答：

所有数字的总和为 $\dfrac{9 \cdot 10}{2} = 45$ ，所以每组和必须为 $45 \div 3 = 15$ 。

含 $9$ 的组中，另外两个数和为 $6$ 。可能是 $1$ 和 $5$ ，或 $2$ 和 $4$ 。

情况 $1$ ：一组是 $1$ 、 $5$ 和 $9$ 。

考虑含 $8$ 的那组，另外两个数必须和为 $7$ 。唯一选择是 $3$ 和 $4$ 。

剩下的一组是 $2$ 、 $6$ 和 $7$ ，其和为 $15$ ，所以这种情况给出一种分法。

情况 $2$ ：一组是 $2$ 、 $4$ 和 $9$ 。

同理，含 $8$ 的那组另外两个数和为 $7$ 。唯一选择是 $1$ 和 $6$ 。

最后一组是 $3$ 、 $5$ 和 $7$ ，其和为 $15$ ，这又给出一种分法。

所有情况都已检查，共有 $2$ 种分法。

所以正确答案是 C。

The sum of all the numbers is $\dfrac{9 \cdot 10}{2} = 45.$ This means that the sum of each group is $45 \div 3 = 15.$

Consider the group with $9$ in it. The other two numbers must add to $6.$ Therefore, the other cards in this group are $1$ and $5$ or $2$ and $4.$

Case $1:$ One group is $1,$ $5,$ and $9.$

Consider the group with $8$ in it. The other numbers must add to $7.$ The only option is $3$ and $4$ with the remaining cards.

The other group is then $2,$ $6,$ and $7.$ This adds to $15,$ so this case contributes one possibility.

Case $2:$ One group is $2,$ $4,$ and $9.$

Consider the group with $8$ in it. As above, the other numbers have to add to $7.$ The only option is $1$ and $6.$

The final group is $3,$ $5,$ and $7,$ which adds to $15.$ This is another configuration.

We have gone through all the cases, which revealed that there are only $2$ possible groupings.

Thus, C is the correct answer.

22.

在一个正整数序列中，从第三项开始，每一项都是前两项的乘积。该序列的第六项是 $4000$ 。第一项是多少？

In a sequence of positive integers, each term after the second is the product of the previous two terms. The sixth term in the sequence is $4000.$ What is the first term?

$1$

$2$

$4$

$5$

$10$

知识点：递推质因数分解

难度评级：1790

小提示：

用第一项 $x$ 和第二项 $y$ 写出前几项

Write the first several terms using first term $x$ and second term $y$

大提示：

第六项是 $x^3y^5$

The sixth term is $x^3y^5$

视频讲解：

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文字解答：

设第一项为 $x$ ，第二项为 $y$ 。

于是，序列的前几项为 $x, y, xy, xy^2, x^2y^3, x^3y^5, \cdots$ 因此 $x^3y^5 = 4000$ 。

将 $4000$ 分解质因数，得到 $4000 = 2^5 \cdot 5^3\text{。}$ 因为 $x$ 和 $y$ 都是正整数，所以能整除 $4000$ 的完全五次方只有 $1$ 和 $32$ 。

若 $y = 1$ ，则 $x^3 = 4000$ ，但 $4000$ 不是完全立方数，不可能成立。因此 $y^5 = 32$ 且 $y = 2\text{。}$

这迫使 $x$ 等于 $5$ 。

所以正确答案是 D。

Let $x$ be the first term and $y$ be the second.

Then we get the following sequence: $x, y, xy, xy^2, x^2y^3, x^3y^5, \cdots$ Thus $x^3y^5 = 4000.$

Factoring $4000,$ gives $4000 = 2^5 \cdot 5^3.$ Because $x$ and $y$ are positive integers, the only fifth powers that divide $4000$ are $1$ and $32.$

If $y = 1,$ then $x^3 = 4000,$ which is impossible because $4000$ is not a perfect cube. Therefore, $y^5 = 32$ and $y = 2.$

This forces $x$ to equal $5.$

Thus, D is the correct answer.

23.

一个 $3 \times 3$ 网格中的每个小方格随机填入右下方所示 $4$ 种阴影和非阴影瓷砖之一。

这个铺法在某个较小的 $2 \times 2$ 网格中包含一个大的阴影菱形的概率是多少？下面是这样一个铺法示例。

Each square in a $3 \times 3$ grid is randomly filled with one of the $4$ shaded-and-unshaded tiles shown below on the right.

What is the probability that the tiling will contain a large shaded diamond in one of the smaller $2 \times 2$ grids? Below is an example of such a tiling.

$\dfrac{1}{1024}$

$\dfrac{1}{256}$

$\dfrac{1}{64}$

$\dfrac{1}{16}$

$\dfrac{1}{4}$

知识点：基本概率乘法原理

难度评级：1840

小提示：

先选择哪个 $2\times2$ 网格包含大菱形

Choose which $2\times2$ grid contains the large diamond

大提示：

一旦选定 $2\times2$ 网格，其中四块瓷砖方向就被确定

Once a $2\times2$ grid is chosen, its four tile orientations are forced

视频讲解：

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文字解答：

总铺法数为 $4^9$ 。可形成大阴影菱形的 $2\times2$ 小网格有 $4$ 个。

选定其中一个 $2\times2$ 小网格后，它内部四块瓷砖的方向就被确定了，其余 $5$ 个方格可以任意填。所以每选定一个这样的 $2\times2$ 小网格，都对应 $4^5$ 种铺法。

两个不同的 $2\times2$ 小网格不能同时形成大阴影菱形，因为重叠方格会要求不同方向。因此有利铺法数为 $4\cdot4^5=4^6$ 。

所求概率为 $\dfrac{4^6}{4^9} = \dfrac{1}{4^3} = \dfrac{1}{64}\text{。}$

所以正确答案是 C。

There are $4^9$ possible tilings. There are $4$ possible $2\times2$ grids where a large shaded diamond could appear.

After one of these $2\times2$ grids is chosen, the four tile orientations inside it are forced, and the other $5$ squares can be filled in any way. This gives $4^5$ tilings for each chosen $2\times2$ grid.

Two different $2\times2$ grids cannot both contain a large shaded diamond, because their overlapping squares would require incompatible tile orientations. Therefore the number of favorable tilings is $4\cdot4^5=4^6.$

The desired probability is then $\dfrac{4^6}{4^9} = \dfrac{1}{4^3} = \dfrac{1}{64}.$

Thus, C is the correct answer.

24.

等腰三角形 $ABC$ 中， $AB$ 和 $BC$ 长度相等。在下面两幅图中，分别画出与 $\overline{AC}$ 平行的线段，使 $\triangle ABC$ 的阴影部分面积相同。两个未阴影部分的高分别为 $11$ 和 $5$ 个单位。 $\triangle ABC$ 的高 $h$ 是多少？

Isosceles triangle $ABC$ has equal side lengths $AB$ and $BC.$ In the figures below, segments are drawn parallel to $\overline{AC}$ so that the shaded portions of $\triangle ABC$ have the same area. The heights of the two unshaded portions are $11$ and $5$ units, respectively. What is the height $h$ of $\triangle ABC?$

$14.6$

$14.8$

$15$

$15.2$

$15.4$

知识点：相似面积比

难度评级：1930

小提示：

阴影面积相等给出两个未阴影三角形面积之间的方程

Equal shaded areas give an equation between the two unshaded triangle areas

大提示：

相似三角形面积按高度比例的平方缩放

Similar triangle areas scale as the square of their heights

视频讲解：

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文字解答：

设 $\triangle ABC$ 的面积为 $a$ 。左图中的未阴影三角形高为 $11$ ，并且与整个三角形相似，所以它的面积是 $a\left(\frac{11}{h}\right)^2$ 。因此，左图的阴影面积是 $a\left(1-\left(\frac{11}{h}\right)^2\right)$ 。

右图中的阴影三角形高为 $h-5$ ，并且与整个三角形相似，所以它的面积是 $a\left(\frac{h-5}{h}\right)^2$ 。令两图的阴影面积相等，再约去 $a$ ，得到 $1-\left(\dfrac{11}{h}\right)^2 =\left(\dfrac{h-5}{h}\right)^2\text{。}$

化简得 $h^2 - 121 = h^2 - 10h + 25\text{，}$ 也就是 $10h = 146\text{。}$

所以 $h=14.6$ 。

所以正确答案是 A。

Let $a$ be the area of $\triangle ABC.$ The unshaded triangle in the left figure has height $11$ and is similar to the full triangle, so its area is $a\left(\frac{11}{h}\right)^2.$ Therefore, the shaded area there is $a\left(1-\left(\frac{11}{h}\right)^2\right).$

In the right figure, the shaded triangle has height $h-5$ and is similar to the full triangle, so its area is $a\left(\frac{h-5}{h}\right)^2.$ Equating the shaded areas and canceling $a$ gives $1-\left(\dfrac{11}{h}\right)^2 =\left(\dfrac{h-5}{h}\right)^2.$

Simplifying yields $h^2 - 121 = h^2 - 10h + 25.$ This simplifies to $10h = 146,$

so $h=14.6.$

Thus, A is the correct answer.

25.

十五个整数 $a_1$ 、 $a_2$ 、 $a_3$ 、 $\cdots$ 、 $a_{15}$ 按顺序排列在数轴上。这些整数等距排列，并满足以下条件：

$1 \leq a_1 \leq 10\text{，}$

$13 \leq a_2 \leq 20\text{，}$

以及

$241 \leq a_{15} \leq 250\text{。}$

$a_{14}$ 的各位数字之和是多少？

Fifteen integers $a_1,$ $a_2,$ $a_3,$ $\cdots,$ $a_{15}$ are arranged in order on a number line. The integers are equally spaced and have the property that

$1 \leq a_1 \leq 10,$

$13 \leq a_2 \leq 20,$

and

$241 \leq a_{15} \leq 250.$

What is the sum of the digits of $a_{14}?$

$8$

$9$

$10$

$11$

$12$

知识点：等差数列极限情形界定数字

难度评级：1950

小提示：

设公差为 $d$

Let $d$ be the common difference

大提示：

用 $a_1$ 和 $a_{15}$ 的最大范围来限制 $d$

Use the widest possible bounds for $a_1$ and $a_{15}$ to force $d$

视频讲解：

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文字解答：

设公差为 $d$ 。将 $a_1$ 取其最大可能值 $10$ ，将 $a_{15}$ 取其最小可能值 $241$ ，可得 $d \geq \dfrac{241 - 10}{14} = 16.5\text{。}$ 因为所有数都是整数，所以 $d$ 至少为 $17$ 。

将 $a_1$ 取其最小可能值 $1$ ，将 $a_{15}$ 取其最大可能值 $250$ ，可得 $d \leq \dfrac{250 - 1}{14} \approx 17.8\text{。}$ 因为所有数都是整数，所以 $d$ 至多为 $17$ 。因此 $d=17$ 。

注意 $17 \cdot 14 = 238$ 。由于 $a_{15}$ 至少为 $241$ ，所以 $a_1$ 至少为 $3$ 。

另一方面，如果 $a_1$ 大于 $3$ ，那么 $a_2=a_1+17$ 就会大于 $20$ ，不符合条件。

所以 $a_1 = 3$ ，且 $d = 17$ 。于是 $a_{14} = 3 + 13 \cdot 17 = 224\text{。}$

数字和为 $2 + 2 + 4 = 8\text{。}$

所以正确答案是 A。

Let $d$ be the common difference. Using the largest possible value $10$ for $a_1$ and the smallest possible value $241$ for $a_{15},$ we have $d \geq \dfrac{241 - 10}{14} = 16.5.$ Since all the numbers are integers, $d$ must be at least $17.$

Using the smallest possible value $1$ for $a_1$ and the largest possible value $250$ for $a_{15},$ we have $d \leq \dfrac{250 - 1}{14} \approx 17.8.$ Since all the numbers are integers, $d$ is at most $17.$ Therefore, $d=17.$

Note that $17 \cdot 14 = 238.$ Since $a_{15}$ is at least $241,$ $a_1$ must be at least $3.$

On the other hand, if $a_1$ were greater than $3,$ then $a_2=a_1+17$ would be greater than $20,$ which is not allowed.

Now we know that $a_1 = 3$ and $d = 17.$ This tells us that $a_{14} = 3 + 13 \cdot 17 = 224.$

Therefore, sum of the digits is $2 + 2 + 4 = 8.$

Thus, A is the correct answer.