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2025 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.

下图所示的八角星是一种流行的拼布图案。这个星形覆盖了整个 $4$ 乘 $4$ 网格的百分之多少？

The eight-pointed star, shown in the figure below, is a popular quilting pattern. What percent of the entire $4$ -by- $4$ grid is covered by the star?

$40$

$50$

$60$

$75$

$80$

知识点：面积对称性

难度评级：720

小提示：

利用对称性，只看网格的四分之一

Use symmetry to focus on one quarter of the grid

大提示：

比较该四分之一区域内的阴影和非阴影面积

Compare the shaded and unshaded areas inside that quarter

视频讲解：

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文字解答：

图形具有对称性，所以整个正方形中被星形覆盖的比例，等于任意一个四分之一区域中被覆盖的比例。

只看左上四分之一区域。设想把其中一个三角形平移一下：题目中的阴影面积与下图中的阴影面积完全相同（下图中左上四分之一用粗线标出）：

这样就能一眼看出左上四分之一区域恰好有一半被涂上阴影，所以答案是 $50\%$ ，即选项 B。

There is symmetry, and so whatever fraction of the top-left quarter is shaded, is the same as the fraction of the entire square that is shaded.

Focus on the top-left quarter. Consider moving a single triangle. The shaded area in the problem is exactly the same as the shaded area in this diagram (and the top-left quarter is outlined in bold):

But then it is obvious that exactly half of the top-left quarter is shaded, and so the answer is $50\%,$ which is choice B.

2.

下表显示了古埃及象形文字中表示不同数字的符号。

例如，数字 $32$ 表示为：

下面这组象形文字表示什么数字？

The table below shows the ancient Egyptian hieroglyphs that were used to represent different numbers.

For example, the number $32$ was represented by:

What number was represented by the following combination of hieroglyphs?

$1{,}423$

$10{,}423$

$14{,}023$

$14{,}203$

$14{,}230$

知识点：位值

难度评级：450

小提示：

分别按位值使用这些符号

Use place values separately

大提示：

数出每种符号的个数，再乘以它的数值并相加

Count each kind of glyph, then multiply by its value and add

视频讲解：

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文字解答：

只需数出每种符号各有多少个。

有 $1$ 个表示 $10{,}000$ 的符号。

有 $4$ 个表示 $100$ 的符号。

有 $2$ 个表示 $10$ 的符号。

有 $3$ 个表示 $1$ 的符号。

所以答案是 $10{,}423$ ，即 B。

We just need to count how many of each type of glyph there are.

There is $1$ glyph worth $10{,}000.$

There are $4$ glyphs worth $100$ each.

There are $2$ glyphs worth $10$ each.

There are $3$ glyphs worth $1$ each.

So, the answer is $10{,}423,$ which is choice B.

3.

Buffalo Shuffle-o 是一种纸牌游戏，游戏开始时所有牌平均分给所有玩家。Annika 和她的 $3$ 个朋友玩时，每位玩家分到 $15$ 张牌。假设下一局又有 $2$ 个朋友加入，每位玩家会分到多少张牌？

Buffalo Shuffle-o is a card game in which all the cards are distributed evenly among all players at the start of the game. When Annika and $3$ of her friends play Buffalo Shuffle-o, each player is dealt $15$ cards. Suppose $2$ more friends join the next game. How many cards will be dealt to each player?

$8$

$9$

$10$

$11$

$12$

知识点：比与比例

难度评级：660

小提示：

先求整副牌的总张数

First find the total number of cards in the deck

大提示：

第一局有 $4$ 位玩家，第二局有 $6$ 位玩家

The first game has $4$ players and the second game has $6$

视频讲解：

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文字解答：

起初共有 $4$ 人（Annika 和 $3$ 个朋友），每人 $15$ 张牌，所以总牌数为 $4 \times 15 = 60$ 。如果又有 $2$ 个朋友加入，共有 $6$ 人，因此每人分到 $60 \div 6 = 10$ 张牌，选 C。

At the start, there are $4$ total people (Annika plus $3$ friends), each with $15$ cards, so there are $4 \times 15 = 60$ cards in total. If $2$ more friends join, there will be $6$ people in total, and so each should get $60 \div 6 = 10$ cards, which is choice C.

4.

Lucius 正在每次减 $7$ 倒着数。他的前三个数是 $100$ 、 $93$ 、 $86$ 。他的第 $10$ 个数是多少？

Lucius is counting backward by $7$ s. His first three numbers are $100,$ $93,$ and $86.$ What is his $10$ th number?

$30$

$37$

$42$

$44$

$47$

知识点：等差数列

难度评级：720

小提示：

第 $10$ 个数比第 $1$ 个数晚 $9$ 步

The $10$ th number is $9$ steps after the $1$ st

大提示：

从第一项连续减去九次七

Subtract nine 7s from the first term

视频讲解：

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文字解答：

从他的第一个数 $100$ 到第 $10$ 个数之间有 $10 - 1 = 9$ 个间隔，每个间隔的大小都是 $7$ 。

因此总共减去 $9 \times 7 = 63$ ，从 $100$ 剩下 $100 - 63 = 37$ ，选 B。

There will be $10 - 1 = 9$ gaps between his first number $100$ and his $10$ th number. Each gap has size $7.$

So, he will subtract a total of $9 \times 7 = 63$ from $100,$ leaving an answer of $100 - 63 = 37,$ which is choice B.

5.

Betty 开卡车在一个街区送包裹，街道地图如下。Betty 从工厂（标为 $F$ ）出发，依次开到地点 $A$ 、 $B$ 、 $C$ ，然后回到 $F$ 。完成这条路线的最短距离是多少个街区？

Betty drives a truck to deliver packages in a neighborhood whose street map is shown below. Betty starts at the factory (labeled $F$ ) and drives to location $A,$ then $B,$ then $C,$ before returning to $F.$ What is the shortest distance, in blocks, she can drive to complete the route?

$20$

$22$

$24$

$26$

$28$

知识点：坐标几何最优化

难度评级：960

小提示：

把路线分成四段最短的网格路径

Break the route into four shortest grid trips

大提示：

在街道网格上，最短距离等于水平街区数加竖直街区数

On a street grid, count horizontal blocks plus vertical blocks

视频讲解：

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文字解答：

关键是：若在街道网格中从坐标 $(x_1, y_1)$ 开到 $(x_2, y_2)$ ，最短距离为 $|x_2 - x_1| + |y_2 - y_1|$ 。这通常称为曼哈顿距离。它也等于两地之间的水平街区数加上竖直街区数。

从 $F$ 到 $A$ 的最短距离是 $1 + 2 = 3$ 。

从 $A$ 到 $B$ 的最短距离是 $7 + 3 = 10$ 。

从 $B$ 到 $C$ 的最短距离是 $2 + 4 = 6$ 。

从 $C$ 到 $F$ 的最短距离是 $4 + 1 = 5$ 。

相加得到 $3 + 10 + 6 + 5 = 24$ ，选 C。

一个小捷径是注意，从 $B$ 到 $C$ 再到 $F$ 时，经过 $C$ 正好位于从 $B$ 到 $F$ 的某条最短路径上，所以甚至可以先忽略必须停在 $C$ 的要求。

The key idea is that if driving from coordinates $(x_1, y_1)$ to $(x_2, y_2),$ then the shortest distance is $|x_2 - x_1| + |y_2 - y_1|.$ This is often called the Manhattan distance. It is also equal to the number of horizontal blocks between the locations, plus the number of vertical blocks between the locations.

The shortest distance from $F$ to $A$ is then $1 + 2 = 3.$

The shortest distance from $A$ to $B$ is $7 + 3 = 10.$

The shortest distance from $B$ to $C$ is $2 + 4 = 6.$

The shortest distance from $C$ to $F$ is $4 + 1 = 5.$

Adding up all of these numbers, we get $3 + 10 + 6 + 5 = 24,$ which is choice C.

One small possible shortcut for the solution is to notice that when going from $B$ to $C$ to $F,$ the visit to $C$ is conveniently along a shortest path from $B$ to $F$ anyway, so we can even remove the requirement to stop at $C$ from the problem.

6.

Sekou 写下 $15$ 、 $16$ 、 $17$ 、 $18$ 、 $19$ 。擦掉其中一个数后，剩下四个数的和是 $4$ 的倍数。他擦掉了哪个数？

Sekou writes the numbers $15,$ $16,$ $17,$ $18,$ $19.$ After he erases one of the numbers, the sum of the remaining four numbers is a multiple of $4.$ Which number did he erase?

$15$

$16$

$17$

$18$

$19$

知识点：模运算整除性

难度评级：900

小提示：

比较擦掉前后的总和模 $4$ 的余数

Compare the total sum modulo $4$ before and after erasing

大提示：

被擦掉的数必须与原总和有相同的模四余数

The erased number must have the same remainder as the original total

视频讲解：

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文字解答：

这五个数除以 $4$ 的余数分别为 $3$ 、 $0$ 、 $1$ 、 $2$ 、 $3$ 。因此五个数的总和模 $4$ 与 $3 + 0 + 1 + 2 + 3 = 9$ 相同，余数为 $1$ （模 $4$ ）。要擦掉一个数后使总和余 $0$ （模 $4$ ），必须擦掉一个余数为 $1$ （模 $4$ ）的数，也就是 $17$ 。所以答案是 C。

The remainders of the five numbers after dividing by $4$ are: $3,$ $0,$ $1,$ $2,$ and $3.$ So, the sum of all five numbers modulo $4$ is the same as $3 + 0 + 1 + 2 + 3 = 9$ which has remainder $1$ modulo $4.$ Therefore, in order to erase a single number and get a sum that is $0$ modulo $4,$ we must erase the number which was $1$ modulo $4,$ which was $17.$ Therefore, the answer is C.

7.

在 Xochi 教授班最近的一次考试中：

• $5$ 名学生得分至少为 $95\%$ ，

• $13$ 名学生得分至少为 $90\%$ ，

• $27$ 名学生得分至少为 $85\%$ ，

• $50$ 名学生得分至少为 $80\%$ 。

有多少名学生得分至少为 $80\%$ 且低于 $90\%$ ？

On the most recent exam in Prof. Xochi’s class,

• $5$ students earned a score of at least $95\%,$

• $13$ students earned a score of at least $90\%,$

• $27$ students earned a score of at least $85\%,$ and

• $50$ students earned a score of at least $80\%.$

How many students earned a score of at least $80\%$ and less than $90\%?$

$8$

$14$

$22$

$37$

$45$

知识点：数据与图表解读

难度评级：770

小提示：

这些分数组是嵌套的

The score groups are nested

大提示：

从至少 $80\%$ 的人数中减去至少 $90\%$ 的人数

Subtract the students with scores at least $90\%$ from those at least $80\%$

视频讲解：

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文字解答：

注意各档人数逐步增加，这是因为条件变得越来越宽。

要包括所有至少 $80\%$ 的 $50$ 名学生，但排除所有至少 $90\%$ 的 $13$ 名学生。因此答案是 $50 - 13 = 37$ ，选 D。

Notice that the numbers of students in each category keep increasing, which makes sense because the categories are getting broader.

We want to include all $50$ of the students who earned a score of at least $80\%$ but exclude all $13$ of the students who earned a score of at least $90\%.$ So, the answer is $50 - 13 = 37,$ which is choice D.

8.

Isaiah 沿一些边剪开一个纸板立方体，形成下图所示的平面图形，其面积为 $18$ 平方厘米。原立方体的体积是多少立方厘米？

Isaiah cuts open a cardboard cube along some of its edges to form the flat shape shown, which has an area of $18$ square centimeters. What was the volume of the cube in cubic centimeters?

$3\sqrt{3}$

$6$

$9$

$6\sqrt{3}$

$9\sqrt{3}$

知识点：展开图（立体几何）表面积体积

难度评级：960

小提示：

立方体展开图有 $6$ 个全等正方形面

A cube net has $6$ congruent square faces

大提示：

用展开图总面积除以 $6$ ，得到一个面的面积

Divide the net area by $6$ to get one face area

视频讲解：

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文字解答：

展开图由 $6$ 个正方形面组成，所以每个面的面积是 $18 \div 6 = 3$ 。立方体边长为 $\sqrt{3}$ 。体积为 $\sqrt{3} \times \sqrt{3} \times \sqrt{3} = 3\sqrt{3}$ 。正确答案是 A。

There are $6$ squares, so each has area $18 \div 6 = 3.$ Then the side length of the cube is $\sqrt{3}.$ The volume is $\sqrt{3} \times \sqrt{3} \times \sqrt{3} = 3\sqrt{3},$ which is choice A.

9.

Ningli 看时钟上 $6$ 对正对着的数字。她求每一对数字的平均数。所得 $6$ 个数的平均数是多少？

Ningli looks at the $6$ pairs of numbers directly across from each other on a clock. She takes the average of each pair of numbers. What is the average of the resulting $6$ numbers?

$5$

$6.5$

$8$

$9.5$

$12$

知识点：平均数等差数列

难度评级：1070

小提示：

这些成对平均数的平均数等于全部 $12$ 个数字的平均数

The average of pair-averages equals the average of all $12$ numbers

大提示：

$1$ 到 $12$ 的平均数是两个端点的中点

The average of $1$ through $12$ is halfway between the endpoints

视频讲解：

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文字解答：

答案等于全部 $12$ 个数字的平均数。一般地，把一组数两两配对，先求每对平均，再求这些平均数的平均，结果仍是所有原数的平均。

例如看较小的 $6$ 个数 $a$ 、 $b$ 、 $c$ 、 $d$ 、 $e$ 、 $f$ 。这些成对平均数的平均为 $\frac{1}{3} \left( \frac{a+b}{2} + \frac{c+d}{2} + \frac{e+f}{2} \right)\text{，}$ 化简后等于 $\frac{a+b+c+d+e+f}{6}\text{。}$ 对 $12$ 个数也同理。

由于这 $12$ 个数字等差排列，平均数等于首尾两项的平均，即 $\frac{1+12}{2} = \frac{13}{2} = 6.5\text{，}$ 所以选 B。

The answer is the same as the average of all $12$ numbers. To understand why this is the case, it is actually generally true that if a whole bunch of numbers is split up into pairs, and each pair is averaged, and then all those pair-averages are averaged, the answer is the average of all the numbers.

To see why this is true, consider a smaller example of $6$ numbers $a,$ $b,$ $c,$ $d,$ $e,$ and $f.$ The average of the pair-averages is: $\frac{1}{3} \left( \frac{a+b}{2} + \frac{c+d}{2} + \frac{e+f}{2} \right)$ and that is equal to $\frac{a+b+c+d+e+f}{6}.$ The same type of simplification happens with $12$ numbers.

Since the $12$ numbers are equally spaced (in an arithmetic progression), the answer is the same as the average of the first and last number, which is $\frac{1+12}{2} = \frac{13}{2} = 6.5,$ or choice B.

10.

下图中， $ABCD$ 是一个矩形，边长 $AB = 5$ 英寸， $AD = 3$ 英寸。矩形 $ABCD$ 顺时针旋转 $90^\circ$ ，旋转中心是边 $DC$ 的中点，得到第二个矩形。两个重叠矩形覆盖的总面积是多少平方英寸？

In the figure below, $ABCD$ is a rectangle with sides of length $AB = 5$ inches and $AD = 3$ inches. Rectangle $ABCD$ is rotated $90^\circ$ clockwise around the midpoint of side $DC$ to give a second rectangle. What is the total area, in square inches, covered by the two overlapping rectangles?

$21$

$22.25$

$23$

$23.75$

$25$

知识点：变换面积容斥原理

难度评级：1220

小提示：

把两个矩形面积相加，再减去重叠面积

Add the two rectangle areas and subtract the overlap

大提示：

重叠部分是一个边长为 $5$ 英寸边一半的正方形

The overlap is a square whose side is half of the $5$ -inch side

视频讲解：

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文字解答：

最直接的方法是用容斥：把两个矩形的面积相加，再减去重叠的正方形面积。

每个矩形面积为 $5 \times 3 = 15$ 。

重叠部分是边长 $2.5$ 的正方形，面积为 $2.5^2 = 6.25$ 。

因此总覆盖面积为 $15 + 15 - 6.25 = 23.75$ ，选 D。

The easiest way to solve this problem is using the Inclusion-Exclusion formula. That is: add the areas of the two rectangles, and then subtract the overlapping (square) area.

Each rectangle has area $5 \times 3 = 15.$

Their overlap is a square that has side length $2.5,$ and so its area is $2.5^2 = 6.25.$

Therefore, the total area is $15 + 15 - 6.25 = 23.75,$ which is choice D.

11.

一个四格骨牌由四个沿边相连的正方形组成。可能的五种形状 $\mathrm{I}$ 、 $\mathrm{O}$ 、 $\mathrm{L}$ 、 $\mathrm{T}$ 、 $\mathrm{S}$ 如下图所示，可以旋转或翻转。用三个四格骨牌完全覆盖一个 $3 \times 4$ 矩形。至少有一块是 $\mathrm{S}$ 形。另外两块是什么？

A tetromino consists of four squares connected along their edges. There are five possible tetromino shapes, $\mathrm{I}$ , $\mathrm{O}$ , $\mathrm{L}$ , $\mathrm{T}$ , and $\mathrm{S}$ , shown below, which can be rotated or flipped over. Three tetrominoes are used to completely cover a $3 \times 4$ rectangle. At least one of the tiles is an $\mathrm{S}$ tile. What are the other two tiles?

$\mathrm{I}$ 和 $\mathrm{L}$

$\mathrm{I}$ and $\mathrm{L}$

$\mathrm{I}$ 和 $\mathrm{T}$

$\mathrm{I}$ and $\mathrm{T}$

$\mathrm{L}$ 和 $\mathrm{L}$

$\mathrm{L}$ and $\mathrm{L}$

$\mathrm{L}$ 和 $\mathrm{S}$

$\mathrm{L}$ and $\mathrm{S}$

$\mathrm{O}$ 和 $\mathrm{T}$

$\mathrm{O}$ and $\mathrm{T}$

知识点：铺砖分类讨论

难度评级：1140

小提示：

先尝试放置 $\mathrm{S}$ 形，使剩余区域容易分开

Try placing the $\mathrm{S}$ tile so the leftover region separates cleanly

大提示：

放下一块 $\mathrm{S}$ 后，检查剩余部分可由哪一对拼块覆盖

After one $\mathrm{S}$ tile is placed, check which pair can tile the rest

视频讲解：

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文字解答：

稍加尝试可以想到，把 $\mathrm{S}$ 形四格骨牌放在不会挡住太多地方的位置，例如这样：

剩余区域正好可以分成两个 $\mathrm{L}$ 形四格骨牌，所以正确答案是 C。

A little trial-and-error suggests placing the $\mathrm{S}$ tetromino in a way that does not block too much, like this:

It is then easy to see that the remainder can be partitioned into two $\mathrm{L}$ tetrominoes, and so the answer is C.

12.

下图区域由 $24$ 个边长为 $1$ 厘米的正方形组成。能放入该区域内部、可以接触边界的最大圆的面积是多少平方厘米？

The region shown below consists of $24$ squares, each with side length $1$ centimeter. What is the area, in square centimeters, of the largest circle that can fit inside the region, possibly touching the boundaries?

$3\pi$

$4\pi$

$5\pi$

$6\pi$

$8\pi$

知识点：圆面积勾股定理最优化

难度评级：1280

小提示：

最大圆的圆心在该区域中心

The largest circle is centered at the center of the region

大提示：

求中心到最近内凹角点的距离

Find the distance from the center to the nearest concave corners

视频讲解：

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文字解答：

离区域中心最近的边界角点，是图中位于这个圆上的 $8$ 个点：

由勾股定理，每个点到中心的距离为 $\sqrt{1^2 + 2^2} = \sqrt{5}\text{。}$ 所以圆面积为 $\pi (\sqrt{5})^2 = 5\pi\text{，}$ 选 C。

The corners of the region which are closest to the center are the $8$ points which lie on this circle:

By the Pythagorean Theorem, each of these points has this distance from the center: $\sqrt{1^2 + 2^2} = \sqrt{5}.$ The area of the circle is then $\pi (\sqrt{5})^2 = 5\pi,$ which is choice C.

13.

将所有偶数 $2$ 、 $4$ 、 $6$ 、 $\dots$ 、 $50$ 分别除以 $7$ 。记录余数。哪一个直方图显示了每个余数出现的次数？

Each of the even numbers $2,$ $4,$ $6,$ $\dots,$ $50$ is divided by $7.$ The remainders are recorded. Which histogram displays the number of times each remainder occurs?

知识点：模运算数据与图表解读

难度评级：1220

小提示：

偶数模 $7$ 的余数按每次加 $2$ 循环

The remainders of even numbers modulo $7$ cycle by adding $2$

大提示：

找出哪四个余数会多出现一次

Track which four remainders appear one extra time

视频讲解：

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文字解答：

余数依次为 $2$ 、 $4$ 、 $6$ 、 $1$ 、 $3$ 、 $5$ 、 $0$ ，然后循环。

所有选项都有 $3$ 根高度为 $3$ 的柱和 $4$ 根高度为 $4$ 的柱。

因此应选择较高的柱落在最先出现的余数上的直方图，也就是 $2$ 、 $4$ 、 $6$ 、 $1$ 。这是选项 A。

The remainders go in order $2,$ $4,$ $6,$ $1,$ $3,$ $5,$ $0,$ and then repeat.

All of the answer choices have $3$ bars of height $3$ and $4$ bars of height $4.$

So, we should pick the answer choice where the taller bars are on the first remainders to appear in our order, which are $2,$ $4,$ $6,$ and $1.$ That is option A.

14.

一个数 $N$ 被插入列表 $2$ 、 $6$ 、 $7$ 、 $7$ 、 $28$ 中。现在平均数是中位数的两倍。 $N$ 是多少？

A number $N$ is inserted into the list $2,$ $6,$ $7,$ $7,$ $28.$ The mean is now twice as great as the median. What is $N?$

$7$

$14$

$20$

$28$

$34$

知识点：平均数中位数（数据）

难度评级：1170

小提示：

因为每个选项都至少为 $7$ ，中位数保持不变

Since every choice is at least $7,$ the median stays fixed

大提示：

令总和等于 $6$ 乘以中位数的两倍

Make the total sum equal to $6$ times twice the median

视频讲解：

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文字解答：

插入 $N$ 后共有 $6$ 个数。个数为偶数，所以中位数是中间两个数的平均。

所有选项都至少为 $7$ ，所以插入后中间两个数仍是 $7$ 和 $7$ 。方便的是，无论选哪个选项，中位数都是 $7$ 。

平均数是中位数的两倍，即 $2 \times 7 = 14$ 。

要让 $6$ 个数的平均数为 $14$ ，总和必须为 $6 \times 14 = 84$ 。

原来 $5$ 个数的和为 $2 + 6 + 7 + 7 + 28 = 50\text{，}$ 所以答案是 $84 - 50 = 34$ ，选 E。

After inserting $N$ into the list, there will be $6$ total numbers. That is even, so the median will be the average of the middle two numbers.

All of the answer choices are at least $7,$ so when they are inserted into the list, the middle two numbers will be $7$ and $7.$ It is convenient that the median will always be $7,$ no matter which answer choice is picked.

The mean becomes twice the median, which is $2 \times 7 = 14.$

To have a total of $6$ numbers with mean $14,$ their sum must become $6 \times 14 = 84.$

The sum of the original $5$ numbers is $2 + 6 + 7 + 7 + 28 = 50,$ so the answer is $84 - 50 = 34,$ which is choice E.

15.

凯画了一个 $6$ 乘 $6$ 的网格。他把 $13$ 个单位正方形涂成银色，其余涂成金色。然后凯沿竖直方向把网格对折，形成重叠的单位正方形对。设 $m$ 和 $M$ 分别为金色叠金色的对数的最小值和最大值。求 $m+M$ 。

Kei draws a $6$ -by- $6$ grid. He colors $13$ of the unit squares silver and the remaining squares gold. Kei then folds the grid in half vertically, forming pairs of overlapping unit squares. Let $m$ and $M$ equal the least and greatest possible number of gold-on-gold pairs, respectively. What is the value of $m+M?$

$12$

$14$

$16$

$18$

$20$

知识点：配对与分组极端原理

难度评级：1480

小提示：

折叠后形成 $18$ 对重叠方格

The fold makes $18$ overlapping pairs

大提示：

要最小化双金色对，先把金色方格尽量分散到不同对中

To minimize double-gold pairs, spread gold squares among pairs first

视频讲解：

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文字解答：

金色方格数为 $6 \times 6 - 13 = 36 - 13 = 23\text{。}$ 折叠后， $36$ 个方格形成 $18$ 对重叠方格。

要让双金色对数最小，先把金色方格分散到所有对中。这用掉 $18$ 个金色方格，剩下 $23 - 18 = 5$ 个必须与已有金色方格重叠，所以 $m = 5$ 。

要让双金色对数最大，就尽量把 $23$ 个金色方格两两配对。最多可形成 $M = 11$ 对，并剩下 $1$ 个金色方格，因为 $23 \div 2$ 商 $11$ 余 $1$ 。

答案是 $m + M = 5 + 11 = 16$ ，选 C。

The number of gold squares is $6 \times 6 - 13 = 36 - 13 = 23.$ The $36$ total squares overlap as $18$ pairs.

To minimize the number of pairs with two gold squares, the gold squares should first be spread out across all pairs. That uses up $18$ of them. The remaining $23 - 18 = 5$ gold squares double-up and create a total of $m = 5$ gold-on-gold pairs.

To maximize the number of pairs with two gold squares, the $23$ gold squares should first be paired up as much as possible. That can be done to create $M = 11$ pairs, with $1$ gold square left over, because $23 \div 2$ is $11$ with a remainder of $1.$

The answer is $m + M = 5 + 11 = 16,$ which is choice C.

16.

从 $1$ 到 $10$ 中选出五个不同整数，并从 $11$ 到 $20$ 中选出五个不同整数。没有两个选出的数相差正好 $10$ 。这十个选出的数字之和是多少？

Five distinct integers from $1$ to $10$ are chosen, and five distinct integers from $11$ to $20$ are chosen. No two numbers differ by exactly $10.$ What is the sum of the ten chosen numbers?

$95$

$100$

$105$

$110$

$115$

知识点：双射配对与分组

难度评级：1650

小提示：

把 $1$ 到 $10$ 中每个数与大 $10$ 的数配对

Pair each number from $1$ to $10$ with the number $10$ larger

大提示：

被选中的高位区间数字对应于低位区间中未被选中的数字

The chosen upper numbers correspond to the unchosen lower numbers

视频讲解：

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文字解答：

把 $1$ 到 $10$ 的整数称为低位区间，把 $11$ 到 $20$ 的整数称为高位区间。

从低位区间选出的 $5$ 个不同数字，会排除高位区间中比它们正好大 $10$ 的数。高位区间只有 $10$ 个数，所以还剩 $10 - 5 = 5$ 个没有被排除。

我们必须从高位区间选 $5$ 个不同数字，所以选中的高位区间数字正好就是未被排除的那些。它们各自比低位区间中一个未选数字大 $10$ 。

因此，选中的 $5$ 个高位区间数字之和，比低位区间中 $5$ 个未选数字之和正好多 $5 \times 10 = 50$ 。

所以这 $10$ 个选中数字的总和等于 $50$ ，再加上低位区间中选中和未选数字的总和 $1 + 2 + \dots + 10$ 。

从 $1$ 到 $10$ 的和是 $\frac{10 (10+1)}{2} = 55$ ，所以答案是 $50 + 55 = 105$ ，选 C。

Call the integers from $1$ to $10$ inclusive the lower range, and call the integers from $11$ to $20$ inclusive the higher range.

Each of the $5$ distinct numbers chosen from the lower range blocks out the number in the higher range that is exactly $10$ more than itself. There are only $10$ numbers in the higher range, so there are only $10 - 5 = 5$ numbers not yet blocked.

We need to choose $5$ distinct numbers from the higher range, so the numbers chosen from the higher range are precisely those which are not yet blocked. They are each exactly $10$ more than a not-chosen number in the lower range.

So, the sum of the $5$ distinct numbers chosen from the higher range is exactly $5 \times 10 = 50$ more than the sum of the $5$ not-chosen numbers in the lower range.

The sum of all $10$ chosen numbers is therefore equal to $50$ plus the sum of all chosen and not-chosen numbers in the lower range $1 + 2 + \dots + 10.$

The sum of the numbers from $1$ to $10$ is $\frac{10 (10+1)}{2} = 55,$ so the answer is $50 + 55 = 105,$ or choice C.

17.

在马尔科维亚，有三座城市： $A$ 、 $B$ 、 $C$ 。住在 $A$ 的有 $100$ 人，住在 $B$ 的有 $120$ 人，住在 $C$ 的有 $160$ 人。每个人都在三座城市之一工作，也可以在自己居住的城市工作。下图中，从一座城市指向另一座城市的箭头标有居住在第一座城市且在第二座城市工作的人所占比例。（例如，住在 $A$ 的人中有 $\frac{1}{4}$ 在 $B$ 工作。）有多少人在 $A$ 工作？

In the land of Markovia, there are three cities: $A,$ $B,$ and $C.$ There are $100$ people who live in $A,$ $120$ who live in $B,$ and $160$ who live in $C.$ Everyone works in one of the three cities, and a person may work in the same city where they live. In the figure below, an arrow pointing from one city to another is labeled with the fraction of people living in the first city who work in the second city. (For example, $\frac{1}{4}$ of the people who live in $A$ work in $B.$ ) How many people work in $A?$

$55$

$60$

$85$

$115$

$160$

知识点：分数

难度评级：1340

小提示：

把来自每个居住城市、在 $A$ 工作的人数相加

Add the people working in $A$ from each home city

大提示：

对住在 $A$ 的人，减去去 $B$ 和 $C$ 工作的比例

For city $A$ residents, subtract the fractions going to $B$ and $C$

视频讲解：

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文字解答：

住在 $A$ 且在 $A$ 工作的人数为 $100 - 100 \times \frac{1}{4} - 100 \times \frac{1}{5}$ 也就是 $100 - 25 - 20 = 55\text{。}$ 住在 $B$ 且在 $A$ 工作的人数为 $120 \times \frac{1}{3} = 40$ 住在 $C$ 且在 $A$ 工作的人数为 $160 \times \frac{1}{8} = 20$ 所以答案为 $55 + 40 + 20 = 115\text{，}$ 选 D。

The number of people who live in $A$ and work in $A$ is $100 - 100 \times \frac{1}{4} - 100 \times \frac{1}{5}$ which is $100 - 25 - 20 = 55.$ The number of people who live in $B$ and work in $A$ is $120 \times \frac{1}{3} = 40$ The number of people who live in $C$ and work in $A$ is $160 \times \frac{1}{8} = 20$ So, the answer is $55 + 40 + 20 = 115,$ which is choice D.

18.

左图中的圆半径为 $1$ 个单位。圆和内接正方形之间的区域被涂阴影。右图中的圆里，圆和内接正方形之间区域的四分之一被涂阴影。两圆中的阴影区域面积相同。右图圆的半径 $R$ 是多少？

The circle shown below on the left has a radius of $1$ unit. The region between the circle and the inscribed square is shaded. In the circle shown on the right, one quarter of the region between the circle and the inscribed square is shaded. The shaded regions in the two circles have the same area. What is the radius $R,$ in units, of the circle on the right?

$\sqrt{2}$

$2$

$2\sqrt{2}$

$4$

$4\sqrt{2}$

知识点：相似长度、面积与体积的缩放关系

难度评级：1310

小提示：

用缩放来比较阴影环形区域面积

Match the shaded gap areas by thinking about scaling

大提示：

面积按长度缩放因子的平方变化

Areas scale as the square of the length scale factor

视频讲解：

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文字解答：

右图中，圆与内接正方形之间的区域只有四分之一被涂上阴影。要使这块阴影面积等于左图中整块区域的面积，右图中圆与正方形之间的整块区域面积必须是左图对应区域的 $4$ 倍。

两个由圆和内接正方形组成的图形是相似的，它们的面积按半径的平方缩放。因此 $R^2 = 4 \cdot 1^2\text{，}$ 所以 $R = 2$ ，即选项 B。

Only one quarter of the gap between the circle and square is shaded on the right. For that shaded area to equal the whole gap on the left, the entire gap on the right must have $4$ times the area of the gap on the left.

The two circle-and-square figures are similar, and their areas scale as the square of their radii. Thus $R^2 = 4 \cdot 1^2,$ so $R = 2,$ which is choice B.

19.

城镇 $A$ 和 $B$ 由一条长 $15$ 英里的直路连接。从 $A$ 到 $B$ 行驶时，每 $5$ 英里限速变化一次：从 $25$ 到 $40$ 再到 $20$ 英里每小时。两辆车分别从 $A$ 和 $B$ 同时出发相向而行，并在每段路上都恰好按限速行驶。两车会在离城镇 $A$ 多少英里处相遇？

Two towns, $A$ and $B,$ are connected by a straight road, $15$ miles long. Traveling from town $A$ to town $B,$ the speed limit changes every $5$ miles: from $25$ to $40$ to $20$ miles per hour (mph). Two cars, one at town $A$ and one at town $B,$ start moving toward each other at the same time. They drive at exactly the speed limit in each portion of the road. How far from town $A,$ in miles, will the two cars meet?

$7.75$

$8$

$8.25$

$8.5$

$8.75$

知识点：相对速度路程、速度与时间

难度评级：1650

小提示：

求每辆车何时到达中间 $5$ 英里路段

Find when each car reaches the middle $5$ -mile section

大提示：

两辆车都在中间路段后，它们速度相同

Once both cars are in the middle section, their speeds are equal

视频讲解：

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文字解答：

把道路看作左、中、右三段。每段长 $5$ 英里。

从 $A$ 出发的车到达中间路段需要 $\frac{5}{25} = \frac{1}{5}$ 小时。

从 $B$ 出发的车到达中间路段需要 $\frac{5}{20} = \frac{1}{4}$ 小时。到那时，从 $A$ 出发的车已经在中间路段行驶了 $\frac{1}{4} - \frac{1}{5} = \frac{1}{20}$ 小时。这辆从 $A$ 出发的车在中间路段已经行驶了 $40 \times \frac{1}{20} = 2$ 英里。

于是中间路段中两车之间还剩 $5 - 2 = 3$ 英里。

此时从 $A$ 出发的车距离 $A$ 为 $5 + 2 = 7$ 英里。

由于两车在中间路段都以每小时 $40$ 英里的速度行驶，它们再各行 $1.5$ 英里后相遇。因此从 $A$ 出发的车总共行驶 $7 + 1.5 = 8.5$ 英里，选 D。

Think of the road as having three sections: left, middle, and right. Each section is $5$ miles long.

The car from $A$ reaches the middle section in $\frac{5}{25} = \frac{1}{5}$ hours.

The car from $B$ reaches the middle section in $\frac{5}{20} = \frac{1}{4}$ hours. By that time, the car from $A$ has already driven in the middle section for $\frac{1}{4} - \frac{1}{5} = \frac{1}{20}$ hours. During this time, that car from $A$ has traveled $40 \times \frac{1}{20} = 2$ miles in the middle section.

That leaves $5 - 2 = 3$ miles between the two cars in the middle section.

At that moment, the car from $A$ is $5 + 2 = 7$ miles from $A.$

Since the cars drive at the same speed of $40$ mph in the middle section, they meet after each driving $1.5$ more miles. This takes the car from $A$ a total distance of $7 + 1.5 = 8.5$ miles, which is choice D.

20.

萨里卡、德夫和拉吉夫分享一大块奶酪。他们轮流切下剩余奶酪的一半并吃掉：先萨里卡吃掉一半，然后德夫吃掉剩余奶酪的一半，然后拉吉夫吃掉此时剩余奶酪的一半，接着又轮到萨里卡，以此类推。他们一直吃到奶酪小到看不见为止。萨里卡总共大约吃掉了原奶酪的几分之几？

Sarika, Dev, and Rajiv are sharing a large block of cheese. They take turns cutting off half of what remains and eating it: first Sarika eats half of the cheese, then Dev eats half of the remaining half, then Rajiv eats half of what remains, then back to Sarika, and so on. They stop when the cheese is too small to see. About what fraction of the original block of cheese does Sarika eat in total?

$\dfrac{4}{7}$

$\dfrac{3}{5}$

$\dfrac{2}{3}$

$\dfrac{3}{4}$

$\dfrac{7}{8}$

知识点：等比数列求和

难度评级：1580

小提示：

萨里卡吃的是连续二分序列中的每第三项

Sarika eats every third term of the halving sequence

大提示：

她吃到的份额构成公比为 $\frac{1}{8}$ 的等比级数

Her shares form a geometric series with ratio $\frac{1}{8}$

视频讲解：

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文字解答：

萨里卡第一步吃掉原奶酪的 $\frac{1}{2}$ 。

然后德夫吃掉原奶酪的 $\frac{1}{2^2}$ 。

接着拉吉夫吃掉原奶酪的 $\frac{1}{2^3}$ 。

萨里卡接下来又吃掉原奶酪的 $\frac{1}{2^4}$ 。

这个模式继续下去。最终萨里卡吃到原奶酪的比例为 $\frac{1}{2} + \frac{1}{2^4} + \frac{1}{2^7} + \dots$ 这是首项 $a = \frac{1}{2}$ 、公比 $r = \frac{1}{2^3}$ 的无穷等比级数。其和为 $\frac{a}{1-r} = \frac{\frac{1}{2}}{1 - \frac{1}{8}} = \frac{\frac{1}{2}}{\frac{7}{8}} = \frac{4}{7}$ 所以选 A。

Sarika gets $\frac{1}{2}$ of the cheese in the first step.

Then Dev gets $\frac{1}{2^2}$ of the cheese.

Then Rajiv gets $\frac{1}{2^3}$ of the cheese.

Sarika then gets another $\frac{1}{2^4}$ of the cheese.

This pattern continues. Ultimately, Sarika gets $\frac{1}{2} + \frac{1}{2^4} + \frac{1}{2^7} + \dots$ which is the sum of an infinite geometric series with first term $a = \frac{1}{2}$ and common ratio $r = \frac{1}{2^3}.$ That sum is $\frac{a}{1-r} = \frac{\frac{1}{2}}{1 - \frac{1}{8}} = \frac{\frac{1}{2}}{\frac{7}{8}} = \frac{4}{7}$ which is choice A.

21.

哥尼斯堡学校把 $1$ 到 $7$ 年级分别分配给 $A$ 到 $G$ 七个教学舱，每个教学舱一个年级。部分教学舱由步道连接，如下图所示。学校注意到每一对相连教学舱的年级差都至少为 $2$ 个年级。（例如， $1$ 年级和 $2$ 年级不会在由步道直接相连的教学舱中。）分配给教学舱 $C$ 、 $E$ 、 $F$ 的年级之和是多少？

The Konigsberg School has assigned grades $1$ through $7$ to pods $A$ through $G,$ one grade per pod. Some of the pods are connected by walkways, as shown in the figure below. The school noticed that each pair of connected pods has been assigned grades differing by $2$ or more grade levels. (For example, grades $1$ and $2$ will not be in pods directly connected by a walkway.) What is the sum of the grade levels assigned to pods $C,$ $E,$ and $F?$

$12$

$13$

$14$

$15$

$16$

知识点：图论分类讨论逻辑推理

难度评级：1840

小提示：

四个两两相连的教学舱必须得到年级 $1$ 、 $3$ 、 $5$ 、 $7$

The four mutually connected pods must get grades $1,$ $3,$ $5,$ and $7$

大提示：

利用教学舱 $G$ 限制较小和较大的年级可以放在哪里

Use pod $G$ to force where small and large grades can go

视频讲解：

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文字解答：

教学舱 $A$ 、 $B$ 、 $C$ 、 $F$ 两两相连。 $1$ 到 $7$ 中四个两两至少相差 $2$ 的数只能是 $\{1,3,5,7\}$ 。因此 $D$ 、 $E$ 、 $G$ 使用偶数年级 $\{2,4,6\}$ 。

教学舱 $G$ 与 $A$ 和 $F$ 相连。若 $G=4$ ，则 $A$ 和 $F$ 必须是 $1$ 和 $7$ ，留下 $B$ 和 $C$ 为 $3$ 和 $5$ 。但 $E$ 与 $C$ 和 $F$ 相连，剩下的偶数年级都会与其中一个只差 $1$ ，所以 $G$ 不能是 $4$ 。

若 $G=2$ ，则 $A$ 和 $F$ 必须是 $5$ 和 $7$ 。若 $F=5$ ，则 $E$ 不能是 $4$ 或 $6$ ，所以 $F=7$ 。并且 $C$ 不能是 $3$ ，否则 $E$ 同样不能是 $4$ 或 $6$ 。因此 $C=1$ ， $E=4$ ，得到 $C+E+F=1+4+7=12$ 。

$G=6$ 的情况对称，得到 $C=7$ 、 $E=4$ 、 $F=1$ 。和同样是 $12$ 。

所以正确答案是 A。

Pods $A,$ $B,$ $C,$ and $F$ are all pairwise connected. Four numbers from $1$ through $7$ that are all at least $2$ apart must be $\{1,3,5,7\}$ . Thus $D,$ $E,$ and $G$ get the even grades $\{2,4,6\}$ .

Pod $G$ is connected to $A$ and $F$ . If $G=4$ , then $A$ and $F$ would have to be $1$ and $7$ , leaving $B$ and $C$ as $3$ and $5$ . But then $E$ , which is connected to $C$ and $F$ , cannot be either remaining even grade without being only $1$ away from one of them. So $G$ cannot be $4$ .

If $G=2$ , then $A$ and $F$ must be $5$ and $7$ . If $F=5$ , then $E$ cannot be $4$ or $6$ , so $F=7$ . Also, $C$ cannot be $3$ , since then $E$ again cannot be $4$ or $6$ . Therefore $C=1$ and $E=4$ , giving $C+E+F=1+4+7=12$ .

The case $G=6$ is symmetric, giving $C=7,$ $E=4,$ and $F=1$ . The same sum is $12$ .

Thus, A is the correct answer.

22.

一个教室有一排 $35$ 个挂衣钩。保利娜喜欢外套等距挂放，也就是说第一件外套前、最后一件外套后、以及每两件相邻外套之间的空钩数都相同。假设至少有 $1$ 件外套，且至少有 $1$ 个空钩。有多少种不同的外套数量能满足保利娜的排列方式？

A classroom has a row of $35$ coat hooks. Paulina likes coats to be equally spaced, so that there is the same number of empty hooks before the first coat, after the last coat, and between every coat and the next one. Suppose there is at least $1$ coat and at least $1$ empty hook. How many different numbers of coats can satisfy Paulina’s pattern?

$2$

$4$

$5$

$7$

$9$

知识点：因数个数双射

难度评级：1710

小提示：

想象在最后额外增加一个挂有外套的钩子

Add one imaginary final coat to make equal blocks

大提示：

块大小和块数的乘积必须为 $36$

The block size and number of blocks must multiply to $36$

视频讲解：

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文字解答：

想象在最后一件外套后再加一个挂着外套的钩子。现在共有 $36$ 个挂钩，并形成重复模式：每个块由若干空钩后接一件外套组成。设 $b$ 为每块中的位置数，设 $d$ 为块数。注意 $d$ 正好比原来的外套件数多一，因为末尾加了一件外套。

于是有 $bd = 36$ 。

对 $b$ 的限制是 $b \geq 2$ ，因为每块至少有一个空钩，并以一件外套结尾。

对 $d$ 的限制是 $d \geq 2$ ，因为原来至少有一件外套，末尾又额外加了一件。

所以只需数出有多少种方式把 $36$ 分解为两个至少为 $2$ 的整数之积。

把 $36$ 分解成两个正整数之积的方式数，正好等于 $36$ 的因数个数。因为 $36 = 2^2 \times 3^2$ ，所以 $36$ 的因数个数为 $(2+1)(2+1) = 9\text{。}$

其中正好两种不合格： $1 \times 36$ 和 $36 \times 1$ 。所以答案是 $9 - 2 = 7$ ，选 D。

Imagine adding an extra coat hook with a coat on it after the last coat. Now, there will be $36$ coat hooks, and a repeating pattern, where each block of the pattern has a bunch of empty hooks followed by a coat. Let $b$ be the number of items in each block. Let $d$ be the number of blocks. Note that $d$ is exactly one more than the number of coats, because we added an extra coat at the end.

We then have $bd = 36.$

The constraint on $b$ is that $b \geq 2$ because each block has at least one empty hook, and ends with a coat.

The constraint on $d$ is that $d \geq 2$ because there was at least one coat before, and we added one extra coat at the end.

So, we just need to find out how many ways there are to factorize $36$ into the product of two integers that are at least $2.$

The number of ways to factorize $36$ into the product of two positive integers is exactly equal to the number of factors of $36.$ There is a formula for that: since $36 = 2^2 \times 3^2$ , the number of factors of $36$ is $(2+1)(2+1) = 9.$

Out of these factorizations, exactly two are disqualified: $1 \times 36$ and $36 \times 1.$ So, the answer is $9 - 2 = 7,$ which is choice D.

23.

有多少个四位数同时满足以下三个性质？

(I) 十位数字和个位数字都是 $9$ 。

(II) 这个数比一个完全平方数小 $1$ 。

(III) 这个数是恰好两个质数的乘积。

How many four-digit numbers have all three of the following properties?

(I) The tens digit and ones digit are both $9.$

(II) The number is $1$ less than a perfect square.

(III) The number is the product of exactly two prime numbers.

$0$

$1$

$2$

$3$

$4$

知识点：平方差质数

难度评级：1770

小提示：

比它大一的平方数必须以 $00$ 结尾

The square one larger must end in $00$

大提示：

检查哪些 $10$ 的倍数两边相邻数都是质数

Check multiples of $10$ whose two neighbors are prime

视频讲解：

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文字解答：

这个数形如 $XX99$ ，所以大 $1$ 的完全平方数以 $00$ 结尾，因此它是一个以 $0$ 结尾的数的平方。设这个平方为 $a^2$ 。

为了让 $a^2 - 1$ 是一个 $4$ 位数且末两位为 $99$ ， $a$ 只能是 $\{40, 50, 60, 70, 80, 90, 100\}$ 。

由于 $a^2 - 1 = (a-1)(a+1)$ ，还需要 $a-1$ 和 $a+1$ 都是质数。也就是说，要在 $\{40, 50, \dots, 100\}$ 附近寻找相邻的质数对。

逐一检查，只有 $59$ 和 $61$ 可行，因此这样的数恰有 $1$ 个。答案是 B。

The number has the form $XX99,$ and so the perfect square that is $1$ more ends in $00,$ and so it is the square of a number ending in $0.$ Suppose that square is $a^2.$

In order for $a^2 - 1$ to be a $4$ -digit number ending in $99$ , the only possibilities for $a$ are $\{40, 50, 60, 70, 80, 90, 100\}.$

Since $a^2 - 1 = (a-1)(a+1),$ we also need both $a-1$ and $a+1$ to be prime. We are then looking for pairs of prime numbers that are right around $\{40, 50, \dots, 100\}.$

Going through all the possibilities, the only ones that work are $59$ and $61$ , and so there is exactly $1$ way to do this. The answer is B.

24.

在梯形 $ABCD$ 中，角 $B$ 和角 $C$ 都是 $60^\circ$ ，且 $AB = DC$ 。所有边长都是正整数，梯形 $ABCD$ 的周长为 $30$ 个单位。有多少个互不全等的梯形满足这些条件？

In trapezoid $ABCD,$ angles $B$ and $C$ measure $60^\circ$ and $AB = DC.$ The side lengths are all positive integers and the perimeter of $ABCD$ is $30$ units. How many non-congruent trapezoids satisfy all of these conditions?

$0$

$1$

$2$

$3$

$4$

知识点：梯形等边三角形丢番图方程

难度评级：1840

小提示：

过 $A$ 作一条与 $CD$ 平行的线

Draw a line through $A$ parallel to $CD$

大提示：

周长可写成 $3x+2y$ ，其中 $x,y$ 为正整数

The perimeter becomes $3x+2y$ for positive integers $x,y$

视频讲解：

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文字解答：

过 $A$ 作一条平行于 $CD$ 的线，并观察到有些长度会自动相等：

这是因为 $\angle AEB = \angle DCB = 60^\circ$ ，所以三角形 $ABE$ 是等边三角形。所有标为 $x$ 的线段总是相等。

另外， $ADCE$ 是平行四边形（不一定是菱形），所以其余标为 $y$ 的线段彼此相等，但不一定等于 $x$ 。

周长为 $3x + 2y$ ，题目要求它等于 $30$ 。

问题就变成数方程 $3x + 2y = 30$ 有多少组正整数解。

考虑正整数 $x$ 时，只有偶数值会给出整数 $y$ ，因为 $30 - 3x$ 必须是偶数 $2y$ 。

而当 $x = 10$ 时，满足方程的 $y$ 为 $y = \frac{30 - 3 \times 10}{2} = 0$ ，不是正整数。

所以可行的 $x$ 是小于 $10$ 的正偶数，即 $\{2, 4, 6, 8\}$ 。共有 $4$ 种，答案是 E。

Draw a line through $A$ parallel to line $CD,$ and observe that some lengths are automatically equal to each other:

That is because $\angle AEB = \angle DCB = 60^\circ,$ and so triangle $ABE$ is equilateral. All lengths labeled $x$ are always equal.

Also, $ADCE$ is a parallelogram (not necessarily a rhombus), so the remaining lengths labeled $y$ are always equal to each other, but not necessarily equal to $x.$

The perimeter is $3x + 2y,$ but it is also supposed to be $30.$

The problem then amounts to counting how many positive integer solutions there are to the equation $3x + 2y = 30.$

As we consider positive integers for $x,$ observe that it is precisely the even integers which give an integer solution for $y,$ because $30 - 3x$ is the even number $2y.$

And, once we get to $x = 10,$ the $y$ that satisfies the equation is $y = \frac{30 - 3 \times 10}{2} = 0,$ which is not a positive integer.

So, the values that work for $x$ are the positive even integers less than $10,$ or $\{2, 4, 6, 8\}.$ There are $4$ options, which gives the answer of E.

25.

麦凯拉找出在一个 $5 \times 5$ 菱形网格中画路径的所有可能方式。每条路径从网格底部开始，到顶部结束，每一步都向东北或西北移动一个单位。她计算每条路径与网格右边界之间区域的面积。下图给出了两个例子。所有可能路径确定的这些面积之和是多少？

Makayla finds all the possible ways to draw a path in a $5 \times 5$ diamond-shaped grid. Each path starts at the bottom of the grid and ends at the top, always moving one unit northeast or northwest. She computes the area of the region between each path and the right side of the grid. Two examples are shown in the figures below. What is the sum of the areas determined by all possible paths?

$2520$

$3150$

$3840$

$4730$

$5050$

知识点：格路对称性组合

难度评级：1930

小提示：

把每条路径与它的镜像路径配对

Pair every path with its mirror image

大提示：

一条路径和它的镜像所对应的两个区域会填满整个菱形

For a path and its mirror, the two regions fill the whole diamond

视频讲解：

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文字解答：

设所求和为 $X$ 。

由对称性，如果改为求每条路径与左边界之间的面积和，结果也为 $X$ 。

但若把这个和与原来的答案相加，就等于对所有路径，把左侧面积与右侧面积之和全部相加。

对每一条路径，这两个面积之和正好是 $25$ 。

路径数等于重排 $LLLLLRRRRR$ 的方式数，其中 $L$ 表示向左， $R$ 表示向右。重排数是 $10$ 选 $5$ ，记为 $\binom{10}{5}$ 。

所以 $2X = 25 \times \binom{10}{5}$ 。两边除以 $2$ ，得到 $X$ 等于 $\frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} \times \frac{25}{2}\text{，}$ 也就是 $3150$ ，选 B。

Let $X$ be the answer.

By symmetry, if the question asked for the sum of areas between each path and the left side of the grid, then the answer would be exactly the same $X.$

But if that answer is added to the original answer, that is exactly the same as the sum over all paths, of the sum of areas to the left and to the right.

For each path, that sum of areas is exactly $25.$

The number of paths is equal to the number of ways to rearrange $LLLLLRRRRR,$ where $L$ stands for left and $R$ stands for right, as the path walks up. The number of rearrangements is $10$ choose $5,$ denoted $\binom{10}{5}.$

So, $2X = 25 \times \binom{10}{5}.$ Dividing by $2,$ we get that $X$ equals $\frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} \times \frac{25}{2},$ which is $3150,$ or choice B.