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

(C) 133 个 P， 134 个 Q， 133 个 R

(A) 132 个 P， 134 个 Q， 134 个 R

(B) 133 个 P， 133 个 Q， 134 个 R

(D) 134 个 P， 132 个 Q， 134 个 R

(E) 134 个 P， 133 个 Q， 133 个 R

2023 AMC 8 真题与在线限时测验

计时

40:00

1.

下列表达式的值是多少？ $(8 \times 4 + 2) - (8 + 4 \times 2)$

What is the value of $(8 \times 4 + 2) - (8 + 4 \times 2)?$

$0$

$6$

$10$

$18$

$24$

答案：D

知识点：运算顺序

难度评级：370

视频讲解：

Click to load, then click again to play

文字解答：

按运算顺序计算。

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

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?

答案：E

知识点：折纸对称性

难度评级：660

视频讲解：

Click to load, then click again to play

文字解答：

展开纸张时，剪痕会关于两条折线分别对称出现，得到如下图形。

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，下列哪一项最接近近似风寒温度？

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: $W=T-0.7\cdot S$ Where $W$ 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?

$18$

$23$

$28$

$32$

$35$

答案：B

知识点：换元法小数估算

难度评级：450

视频讲解：

Click to load, then click again to play

文字解答：

代入公式得 $\begin{align*} 36 - 0.7 \cdot 18 &= 36 - 12.6 \\ &=23.4. \end{align*}$ 最接近的是二十三。

所以正确答案是 B。

Using the formula, we can calculate the wind chill to be $\begin{align*} 36 - 0.7 \cdot 18 &= 36 - 12.6 \\ &=23.4. \end{align*}$

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?

$0$

$1$

$2$

$3$

$4$

答案：D

知识点：质数找规律

难度评级：960

视频讲解：

Click to load, then click again to play

文字解答：

完成螺旋后得到下图。

阴影方格中的质数是 $19, 23$ 和 $47$ ，共有三个。

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?

$1250$

$1500$

$1750$

$1800$

$2000$

答案：B

知识点：比与比例

难度评级：720

视频讲解：

Click to load, then click again to play

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?

$0$

$8$

$9$

$16$

$18$

答案：C

知识点：指数最优化

难度评级：900

视频讲解：

Click to load, then click again to play

文字解答：

不要把 $0$ 放在底数位置，因为那会让这个因子等于 $0$ 。

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

为了让这项尽量小，应该使用 $2^0$ 。

剩下另一项可以是 $2^3$ 或 $3^2$ 。显然 $3^2$ 更大。

于是最大值为 $2^0 \times 3^2 = 1 \times 9 = 9.$

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 $1.$

Therefore, we should minimize this number, so we can put $2^0.$

Now the other term is $2^3$ or $3^2.$ Clearly, $3^2$ is larger, so we should use that.

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?

$0$

$1$

$2$

$3$

$4$

答案：B

知识点：坐标几何一次方程

难度评级：1070

视频讲解：

Click to load, then click again to play

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.

Lola、Lolo、Tiya 和 Tiyo 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，Lola 赢了五场，输了第四场。Tiyo 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?

$000101$

$001001$

$010000$

$010101$

$011000$

答案：A

知识点：逻辑推理奇偶性

难度评级：1020

视频讲解：

Click to load, then click again to play

文字解答：

每一轮有两场比赛，因此每列记录中有 $2$ 个胜者和 $2$ 个负者。

也就是说，每列应有 $2$ 个一和 $2$ 个零。

逐列补足这个条件，可得 Tiyo 的胜负记录为 $000101$ 。

所以正确答案是 A。

Note that for each round, there are going to $2$ winners and $2$ losers (two matches each with a winner and loser).

This means that for each match, there should be $2$ ones and $2$ zeros.

Using this fact, we can deduce that Tiyo's win-loss record is $000101.$

Thus, A is the correct answer.

9.

Malaika 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?

$6$

$8$

$10$

$12$

$14$

答案：B

知识点：数据与图表解读

难度评级：1020

视频讲解：

Click to load, then click again to play

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。

总时间为 $2 + 4 + 2 = 8$ 秒。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 answer is therefore $2 + 4 + 2 = 8$ Thus, B is the correct answer.

10.

Harold 做了一个李子派去野餐。他只吃了 $\frac{1}{4}$ 个派，把剩下的留给朋友。一只驼鹿经过，吃掉了 Harold 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?

$\dfrac{1}{12}$

$\dfrac{1}{6}$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{5}{12}$

答案：D

知识点：分数

难度评级：900

视频讲解：

Click to load, then click again to play

文字解答：

Harold 后剩下 $1 - \frac{1}{4} = \frac{3}{4}$ 。

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 Perseverance 探测车于 $2020$ 年七月 $30$ 日发射。它行进 $292,526,838$ 英里后，约 $6.5$ 个月后降落在火星的 Jezero 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?

$6,000$

$12,000$

$60,000$

$120,000$

$600,000$

答案：C

知识点：路程、速度与时间估算单位换算

难度评级：1100

视频讲解：

Click to load, then click again to play

文字解答：

把距离近似为 $300,000,000$ 英里。把 $6.5$ 个月近似为 $30 \times 6.5 = 195 \approx 200$ 天，因此每天约行进 $300,000,000 \div 200 = 1,500,000$ 英里。再除以 $24$ 小时，约为 $62,500$ 英里每小时，最接近 $60,000$ 。

所以正确答案是 C。

We can round the distance up to $300,000,000$ miles. We can also approximate $6.5$ months as $30 \times 6.5 = 195 \approx 200$ days. This means that the Rover traveled $300,000,000 \div 200 = 1,500,000$ miles per day. Dividing this by $24$ to get the speed in miles per hour yields $62,500,$ which is close 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?

$\dfrac{1}{4}$

$\dfrac{11}{36}$

$\dfrac{1}{3}$

$\dfrac{19}{36}$

$\dfrac{5}{9}$

答案：B

知识点：圆面积面积比

难度评级：1370

视频讲解：

Click to load, then click again to play

文字解答：

不妨设每个小正方形边长为 $2$ 个单位。

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 $\begin{align*} 4^2 \pi - 2 \cdot 2^2 \pi &= 16 \pi - 8 \pi \\ &= 8 \pi. \end{align*}$

合计阴影面积为 $8\pi + 3\pi = 11\pi.$ 大未涂色圆面积为 $6^2\pi = 36\pi.$ 所以所求分数为 $\dfrac{11\pi}{36\pi} = \dfrac{11}{36}.$

所以正确答案是 B。

WLOG, assume that each square has a side length of $2$ units.

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 $\begin{align*} 4^2 \pi - 2 \cdot 2^2 \pi &= 16 \pi - 8 \pi \\ &= 8 \pi. \end{align*}$

Adding these values together yields $8\pi + 3\pi = 11\pi.$ 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?

$8$

$16$

$24$

$48$

$96$

答案：D

知识点：分数比与比例

难度评级：1240

视频讲解：

Click to load, then click again to play

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。两者相差 $\dfrac{3}{8} - \dfrac{1}{3} = \dfrac{1}{24}$ 个全程。这段距离是 $2$ 英里，所以全程为 $24 \cdot 2 = 48$ 英里。

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 two is $\dfrac{3}{8} - \dfrac{1}{3} = \dfrac{1}{24}$ the length of the race. We know that this equals $2,$ which means the race is $24 \cdot 2 = 48$ miles long.

Thus, D is the correct answer.

14.

Nicolas 计划给朋友 Anton 寄一个包裹，Anton 是集邮爱好者。为了支付邮资，Nicolas 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。Nicolas 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.)

$45$

$46$

$51$

$54$

$55$

答案：E

知识点：最优化模运算

难度评级：1480

视频讲解：

Click to load, then click again to play

文字解答：

目标金额是 $7 \cdot 100 + 10 = 710$ 分。先尝试尽量用完所有 $5$ 分和 $10$ 分邮票。

若用完全部五分和十分邮票，它们共值 $20(5 + 10) = 20 \cdot 15 = 300$ 分。还需 $710 - 300 = 410$ 分，不能只用 $25$ 分邮票凑出。

不过， $425$ 分可以用 $25$ 分邮票凑出。若 $5$ 分和 $10$ 分邮票各用 $19$ 张，它们共值 $300 - 5 - 10 = 285$ 分。此时还需 $710 - 285 = 425$ 分，也就是 $425 \div 25 = 17$ 张 $25$ 分邮票。

总张数为 $\begin{align*} 2 \cdot 19 + 17 &= 38 + 17 \\ &= 55 \end{align*}$

所以正确答案是 E。

Note that we want to get $7 \cdot 100 + 10 = 710$ cents. Let us try to see if it is possible to use up all of the $5$ -cent and $10$ -cent stamps.

All of these two types of stamps combined would be worth $20(5 + 10) = 20 \cdot 15 = 300$ cents. We then would need $710 - 300 = 410$ cents, which cannot be created with just $25$ -cent stamps.

We can, however, make $425$ cents with $25$ -cent stamps. Using only $19$ each of $5$ and $10$ -cent stamps would total $300 - 5 - 10 = 285$ cents. This means we would then need $710 - 285 = 425$ cents. This can be achieved with $425 \div 25 = 17$ $25$ -cent stamps.

This lets us use $\begin{align*} 2 \cdot 19 + 17 &= 38 + 17 \\ &= 55 \end{align*}$ stamps.

Thus, E is the correct answer.

15.

Visvam 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，Visvam 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？

Visvam 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, Visvam 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 Visvam walk in order to arrive at school at his usual time?

$4$

$4.2$

$4.5$

$4.8$

$5$

答案：B

知识点：路程、速度与时间单位换算

难度评级：1410

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，Visvam 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。

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, Visvam has to walk $7 \cdot \dfrac{1}{20} = \dfrac{7}{20}$ miles.

Normally, from this spot Visvam 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 $\begin{align*} \dfrac{\frac{7}{20}}{\frac{1}{12}} &= 12 \cdot \dfrac{7}{20} \\&= \dfrac{21}{5} \\ &= 4.2 \end{align*}$ mph.

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 Ps, Qs, and Rs will appear in the completed table?

$132$ 个 P， $134$ 个 Q， $134$ 个 R

$132$ Ps, $134$ Qs, $134$ Rs

$133$ 个 P， $133$ 个 Q， $134$ 个 R

$133$ Ps, $133$ Qs, $134$ Rs

$133$ 个 P， $134$ 个 Q， $133$ 个 R

$133$ Ps, $134$ Qs, $133$ Rs

$134$ 个 P， $132$ 个 Q， $134$ 个 R

$134$ Ps, $132$ Qs, $134$ Rs

$134$ 个 P， $133$ 个 Q， $133$ 个 R

$134$ Ps, $133$ Qs, $133$ Rs

答案：C

知识点：找规律模运算

难度评级：1540

视频讲解：

Click to load, then click again to play

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。

同样分析可知， $R$ 和 $P$ 在 $7$ 列中处于第三个位置，而 $Q$ 只在这些列中出现 $6$ 次。

因此在 $20 - 7 = 13$ 列中， $P$ 和 $R$ 各出现 $7$ 次，共 $13 \cdot 7 = 91$ 次。

它们还会在另外 $7$ 列中各出现 $6$ 次，增加 $6 \cdot 7 = 42$ 次。因此 $P$ 和 $R$ 各出现 $91 + 42 = 133$ 次。

$Q$ 的出现次数为 $\begin{align*} 20 \cdot 20 - 2 \cdot 133 &=400 - 266 \\ &= 134 \end{align*}$

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.

Using the same analysis, we can see that $R$ and $P$ appear in the third row $7$ times, whereas $Q$ only appears $6$ times.

Therefore, in $20 - 7 = 13$ columns, $P$ and $R$ appear $7$ times, for a total of $13 \cdot 7 = 91$ times.

They also appear $6$ times in $7$ columns, adding $6 \cdot 7 = 42$ appearances to the total. This gives us a total of $91 + 42 = 133$ appearances for $P$ and $R.$

$Q$ 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。Jun 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 $Q?$

$1$

$2$

$3$

$4$

$5$

答案：A

知识点：展开图（立体几何）多面体

难度评级：1580

视频讲解：

Click to load, then click again to play

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，因此只需在剩余面中考虑。

候选面是 $1$ 、 $6$ 和 $7$ 。从网格关系可看出， $6$ 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 sides 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 number 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 Greta 坐在池塘中一长排睡莲叶上。从任意一片睡莲叶出发，Greta 可以向右跳 $5$ 片，或向左跳 $3$ 片。Greta 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?

$405$

$407$

$409$

$411$

$413$

答案：D

知识点：丢番图方程模运算最优化

难度评级：1740

视频讲解：

Click to load, then click again to play

文字解答：

设向右跳 $r$ 次，向左跳 $l$ 次，则 $5r-3l=2023.$ 模 $5$ 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 $5r-3l=2023.$ Modulo $5$ , this 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.

$1 : 3$

$3 : 8$

$5 : 12$

$7 : 16$

$4 : 9$

答案：C

知识点：面积比相似

难度评级：1410

视频讲解：

Click to load, then click again to play

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 $\dfrac{\frac{5}{27}}{\frac{4}{9}} = \dfrac{5}{27} \cdot \dfrac{9}{4} = \dfrac{5}{12}.$

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?

$56$

$57$

$58$

$60$

$61$

答案：D

知识点：极差中位数（数据）众数最优化

难度评级：1600

视频讲解：

Click to load, then click again to play

文字解答：

原极差为 $28-3=25$ ，新极差必须为 $50$ 。为保持七个数的中位数仍为 $8$ ，新增的两个数应一个小于 $8$ ，一个大于 $8$ 。

若较小的新增数至少为 $3$ ，则最小值仍为 $3$ 。为了达到这个极差，最大值必须为 $3+50=53$ 。较小的新增数不能是 $3$ ，否则众数改变；它最大可取 $7$ 。

于是新增数为 $7$ 和 $53$ ，和为 $60$ 。

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 must be below $8$ and the other above $8$ .

If the smaller added number is at least $3$ , then the minimum stays $3$ , so the maximum must be $3+50=53$ . The smaller added number cannot be $3$ , because that would change the mode, so its largest possible value is $7$ .

This gives added numbers $7$ and $53$ , with sum $60$ .

Thus, D is the correct answer.

21.

Alina 把数字 $1,2,\cdots,9$ 分别写在卡片上，每张卡片一个数。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?

$0$

$1$

$2$

$3$

$4$

答案：C

知识点：分拆与有序分拆分类讨论

难度评级：1690

视频讲解：

Click to load, then click again to play

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 $1, 5, 9$ 。

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 $15$ ，所以这种情况贡献一种分法。

情况 $2$ ：一组是 $2, 4, 9$ 。

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 $15.$ 这给出另一种分法。

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 $1, 5, 9$

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 $2, 4, 9$

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?

$1$

$2$

$4$

$5$

$10$

答案：D

知识点：递推质因数分解

难度评级：1790

视频讲解：

Click to load, then click again to play

文字解答：

设第一项为 $x$ ，第二项为 $y$ 。

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

分解 $4000$ ： $4000 = 2^5 \cdot 5^3.$ 因为 $x$ 和 $y$ 是正整数， $4000$ 中可能作为五次方的因子只有 $1$ 或 $32$ 。

若 $y = 1$ ，则 $x^3 = 4000$ ，但 $4000$ 不是完全立方数。因此 $y^5 = 32$ $y = 2.$

这迫使 $x$ 等于 $5$ 。

正确答案是 D。

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

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

Factoring $4000,$ we get $4000 = 2^5 \cdot 5^3.$ We need $x$ and $y$ to be integers. The only fifth powers that divides $4000$ are $1$ and $32.$

If $y = 1,$ then $x^3 = 4000,$ which doesn't work since $4000$ is not a perfect cube. Therefore, $y^5 = 32$ $y = 2.$

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.

$\dfrac{1}{1024}$

$\dfrac{1}{256}$

$\dfrac{1}{64}$

$\dfrac{1}{16}$

$\dfrac{1}{4}$

答案：C

知识点：基本概率乘法原理

难度评级：1840

视频讲解：

Click to load, then click again to play

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。

选定一个 $2\times2$ 小网格后，其中四块瓷砖的方向被确定，其余 $5$ 个方格任意填，有 $4^5$ 种。这个 $2\times2$ 小网格中的大菱形也随之确定。

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 $\dfrac{4^6}{4^9} = \dfrac{1}{4^3} = \dfrac{1}{64}.$

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。在下图中，画出若干与 $\overline{AC}$ 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 $\triangle ABC?$

$14.6$

$14.8$

$15$

$15.2$

$15.4$

答案：A

知识点：相似面积比

难度评级：1930

视频讲解：

Click to load, then click again to play

文字解答：

设 $\triangle ABC$ 的面积为 $a$ 。

小三角形都与 $\triangle ABC$ 相似，因此面积比等于对应高的比例的平方。

由两图阴影面积相等，得到 $a - a \cdot \left(\dfrac{11}{h}\right)^2$ $= a \cdot \left(\dfrac{h - 5}{h}\right)^2.$

左边是左图阴影面积，右边是右图阴影三角形面积。

化简得 $h^2 - 121 = h^2 - 10h + 25.$ 再化简： $10h = 146$ $h = 14.6.$

所以正确答案是 A。

Let $a$ be the area of $\triangle ABC.$

Note that the smaller triangles are similar to $\triangle ABC.$ This means that the ratio of their areas is the ratio of their side lengths squared.

Then we get that $a - a \cdot \left(\dfrac{11}{h}\right)^2$ $= a \cdot \left(\dfrac{h - 5}{h}\right)^2.$ The left side is the area of the whole triangle minus the area of the unshaded region of the left triangle.

The right hand side is the area of the shaded triangle. We get this by finding the ratio of their side lengths (which is the same as the ratio of their heights) and squaring it.

Simplifying yields $h^2 - 121 = h^2 - 10h + 25.$ This simplifies to $10h = 146$ $h = 14.6.$

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。 $1 \leq a_1 \leq 10,$ $13 \leq a_2 \leq 20,$ $241 \leq a_{15} \leq 250.$

这些整数等距排列。 $a_{14}$ 的数字和是多少？

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 $a_{14}?$

$8$

$9$

$10$

$11$

$12$

答案：A

知识点：等差数列极限情形界定数字

难度评级：1950

视频讲解：

Click to load, then click again to play

文字解答：

设公差为 $d$ 。若取 $a_1 = 10$ 、 $a_{15} = 241$ ，则 $d \geq \dfrac{241 - 10}{14} = 16.5.$ 因此整数公差 $d$ 至少为 $17$ 。

另一方面，若取 $a_1 = 1$ 、 $a_{15} = 250$ ，则 $d \leq \dfrac{250 - 1}{14} \approx 17.8.$ 因为公差是整数，所以 $d$ 至多为 $17$ 。因此 $d$ 是 $17$ 。

注意 $17 \cdot 14 = 238$ 。为了让 $a_{15}$ 落在给定范围内， $a_1$ 至少为 $3$ 。

如果 $a_1$ 大于 $3$ ，则 $a_2$ 会大于 $20$ ，不符合条件。

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

数字和为 $2 + 2 + 4 = 8.$

所以正确答案是 A。

Let $d$ be the common difference. If we let $a_1 = 10$ and $a_{15} = 241,$ we see that $d \geq \dfrac{241 - 10}{14} = 16.5.$ Since all the numbers are integers, $d$ must be at least $17.$

Also, if $a_1 = 1$ and $a_{15} = 250,$ we get that $d \leq \dfrac{250 - 1}{14} \approx 17.8.$ Once again since all the numbers are integers, $d$ is at most $17.$ This tells us that $d$ is $17.$

Note that $17 \cdot 14 = 238.$ This means that $a_1$ must be at least $3$ for $a_{15}$ to be within the desired range.

If $a_1$ is greater than $3,$ however, $a_2$ becomes greater than $20,$ which is not allowed.

Now we know that $a_1 = 3$ and $d = 17.$ This tell 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.