2018 年 AMC 8 真题（P版）

1 答案 A

复制模型的高度是 \(\dfrac{289}{20}=14.45\) 英尺，四舍五入得到 \(14\) 英尺。

所以正确答案是 A。

English

The replica is \(\dfrac{289}{20}=14.45\) feet tall, which rounds to \(14\) feet.

Thus, the correct answer is A.

2 答案 D

先注意，形如 \(1 + \frac{1}{n}\) 的因子可以改写为 \[\frac{n}{n} + \frac{1}{n} = \frac{n+1}{n}\text{。}\]因此题中的乘积可以写成 \[\begin{align*} &\left(1+\frac{1}{1}\right)\cdot\left(1+\frac{1}{2}\right)\\ &\quad{}\cdot\left(1+\frac{1}{3}\right)\cdot\left(1+\frac{1}{4}\right)\\ &\quad{}\cdot\left(1+\frac{1}{5}\right)\cdot\left(1+\frac{1}{6}\right)\\ &=\dfrac{2}{1} \cdot \dfrac{3}{2} \cdot \dfrac{4}{3} \cdot\dfrac{5}{4} \cdot\dfrac{6}{5} \cdot\dfrac{7}{6}\\ &=7 \end{align*}\] 所以正确答案是 D。

English

Let’s first note that if we are given an expression of the form \(1 + \frac{1}{n},\) we can rewrite this as \[\frac{n}{n} + \frac{1}{n} = \frac{n+1}{n}.\] With that in mind, we can rewrite the expression given to us in the problem, as shown below: \[\begin{align*} &\left(1+\frac{1}{1}\right)\cdot\left(1+\frac{1}{2}\right)\\ &\quad{}\cdot\left(1+\frac{1}{3}\right)\cdot\left(1+\frac{1}{4}\right)\\ &\quad{}\cdot\left(1+\frac{1}{5}\right)\cdot\left(1+\frac{1}{6}\right)\\ &=\dfrac{2}{1} \cdot \dfrac{3}{2} \cdot \dfrac{4}{3} \cdot\dfrac{5}{4} \cdot\dfrac{6}{5} \cdot\dfrac{7}{6}\\ &=7 \end{align*}\] Thus, the correct answer is D.

3 答案 D

最先会让人离开的五个数是 \(7,14,17,21,27\)。只跟踪这些轮次即可。

Arn 报到 \(7\)，Cyd 报到 \(14\)，Fon 报到 \(17\)，Bob 报到 \(21\)，Eve 报到 \(27\)。

Dan 是唯一留在圆圈里的学生，所以正确答案是 D。

English

The first five removal numbers are \(7,14,17,21,27.\) Tracking only those turns gives:

Arn says \(7\), Cyd says \(14\), Fon says \(17\), Bob says \(21\), and Eve says \(27\).

Dan is the only student remaining, so D is the correct answer.

4 答案 C

如图所示，可以把图形分解：

现在可以看出，中间是一个 \(3 \times 3\) 正方形，周围有 \(4\) 个较小的阴影三角形。

正方形面积为 \(3 \cdot 3 = 9\)。每个小三角形的底为 \(2\)，高为 \(1\)，所以面积为 \[\dfrac{bh}{2} = \dfrac{2\cdot 1}{2} =1\text{。}\]这样的三角形有 \(4\) 个，所以它们的总面积为 \(1\cdot 4 = 4\)。

因此总面积是 \(9+4 = 13\)。

所以正确答案是 C。

English

To solve for the area of the figure, we separate the compound shape into parts that are easier to work with, as such:

As is now clear, there is the center \(3 \times 3\) square, with \(4 \) smaller shaded triangles surrounding it.

The area of the square is \(3 \cdot 3 = 9.\) The other triangles each have a base of \(2\) and a height of \(1,\) so their area is equal to \[\dfrac{bh}{2} = \dfrac{2\cdot 1}{2} =1 .\] There are \(4\) of these triangles, so their total area is \(1\cdot 4 = 4.\)

Therefore, the total area is \(9+4 = 13.\)

Thus, the correct answer is C.

5 答案 E

重新分组可得 \[\begin{align*}&1 + (3-2) + (5-4) + \cdots +\\ &(2017-2016) + (2019-2018) \end{align*}\text{。}\]每一项都等于 \(1\)，而这样的项共有 \(\frac{2019-1}2+1 = 1010\) 项，所以总和为 \(1010\cdot1 = 1010\)。

所以正确答案是 E。

English

Rearranging the terms, notice that the expression in the question is equal to: \[\begin{align*}&1 + (3-2) + (5-4) + \cdots +\\ &(2017-2016) + (2019-2018). \end{align*}\] Each term is equal to \(1,\) and there are \(\frac{2019-1}2+1 = 1010\) terms, so the total sum is \(1010\cdot1 = 1010.\)

Thus, E is the correct answer.

6 答案 C

Anh 在海边通道上开了 \(10\) 英里，用了 \(30\) 分钟。因此海边通道速度（记为 \(v_c\)）是 \[\begin{align*}v_c&=\dfrac{10}{30}\\ &=\dfrac13\end{align*}\text{，}\]也就是每分钟 \(\dfrac13\) 英里。因为他在高速公路上的速度是 \(3\) 倍，也就是 \(v_h=3v_c\)，所以高速公路速度为 \[3\cdot \dfrac13 = 1\] 英里每分钟。他在海边通道上开了 \(30\) 分钟，在高速公路上开了 \(50\) 英里，速度为每分钟 \(1\) 英里，所以这 \(50\) 英里需要 \(50\) 分钟。

因此总行程时间为 \(50+30=80\) 分钟。

所以正确答案是 C。

English

Anh drove \(10\) miles on the coastal road in \(30\) minutes. Therefore, his speed on the coastal road (notated as \(v_c\)) is \[\begin{align*}v_c&=\dfrac{10}{30}\\ &=\dfrac13.\end{align*}\] This is \(\dfrac13\) mile per minute. Since he drives \(3\) times as fast on the highway (i.e. \(v_h=3v_c\)), his highway speed is \[3\cdot \dfrac13 = 1\] mile per minute. Armed with these two facts, we know that Anh drove for \(30\) minutes on the coastal road, and he drove \(50\) miles at \(1\) mile per minute. This means it takes \(50\) minutes to drive the \(50\) miles on the highway.

As such, the total travel time is \(50+30=80\) minutes.

Thus, the correct answer is C.

7 答案 B

注意，一个数能被 \(9\) 整除，当且仅当它的各位数字之和也能被 \(9\) 整除。

题中这个 \(5\) 位数的数字和为 \[2+0+1+8+U= 11+U\text{。}\]因为 \(\underline{2}\) \(\underline{0}\) \(\underline{1}\) \(\underline{8}\) \(\underline{U}\) 能被 \(9\) 整除，\(11+U\) 也必须被 \(9\) 整除。又因为 \(U\) 是一位数字，\(0\le U\le 9\)，所以 \(U\) 只能是 \(7\)。

现在可知这个 \(5\) 位数是 \(20187\)，而我们要求 \(20187\) 除以 \(8\) 的余数。长除法给出 \(20187=2523\cdot 8 + 3\)，所以余数是 \(3\)。

所以正确答案是 B。

English

Notice that a number is divisible by \(9\) if and only if the sum of its digits is also divisible by \(9.\)

The sum of the digits of the \(5\)-digit number in the problem is: \[2+0+1+8+U= 11+U.\] As \(\underline{2}\) \(\underline{0}\) \(\underline{1}\) \(\underline{8}\) \(\underline{U}\) is divisible by \(9,\) \(11+U\) must also be divisible by \(9.\) Also, as \(U\) is a digit, we know that \(0\le U\le 9.\) This means that \(U\) can only be \(7.\)

Now we know that the \(5\)-digit number in question is \(20187,\) and we want to find the remainder when we divide \(20187\) by \(8.\) To solve this, simply use long division to see that \(20187=2523\cdot 8 + 3.\) Therefore, the remainder is \(3.\)

Thus, the correct answer is B.

8 答案 C

从条形图可知，锻炼 \(1,2,3,4,5,6,7\) 天的学生人数依次为 \(1,3,2,6,8,3,2\)，学生总数为 \(25\)。

学生报告的锻炼天数总和为 \[\begin{aligned} &1\cdot1+2\cdot3+3\cdot2+4\cdot6\\ &\qquad+5\cdot8+6\cdot3+7\cdot2=109 \end{aligned}\text{。}\]

平均数为 \(\frac{109}{25}=4.36\) 天。

所以正确答案是 C。

English

The bar heights for \(1,2,3,4,5,6,7\) days are \(1,3,2,6,8,3,2\), for a total of \(25\) students.

The total number of reported exercise days is \[\begin{aligned} &1\cdot1+2\cdot3+3\cdot2+4\cdot6\\ &\qquad+5\cdot8+6\cdot3+7\cdot2=109. \end{aligned}\]

The mean is \(\frac{109}{25}=4.36\) days.

Thus, C is the correct answer.

9 答案 B

注意边框每平方英尺需要一块地砖，所以四边相加会得到 \(16+12+16+12=56\) 块。不过这样四个角各重复计算了一次，所以要减去 \(4\)。因此边框共需要 \(56-4=52\) 块 \(1\times 1\) 正方形地砖。

由于每边都去掉了一英尺宽的边框，剩下的内部矩形为 \(10\) 英尺乘 \(14\) 英尺。它要完全铺上 \(2 \times 2\) 的地砖，所以这部分需要 \[ \dfrac{10\cdot14}{2\cdot2} = 35 \] 块地砖。

边框需要 \(52\) 块 \(1\times 1\) 地砖，内部需要 \(35\) 块 \(2\times 2\) 地砖，因此一共需要 \(52+35=87\) 块地砖。

所以正确答案是 B。

English

Note that each square foot of the border would require one tile, meaning that the border will take \(16+12+16+12=56\) tiles. However, notice that this will cause overlapping tiles in each of the four corners, so to fix this, we subtract \(4.\) Therefore, the border will take \(56-4=52\) \(1\times 1\) square tiles to completely tile.

Since we have removed one foot from each side due to the border, the remaining rectangle is \(10\) feet by \(14\) feet. This must be tiled completely by \(2 \times 2 \) tiles, so it will take \[ \dfrac{10\cdot14}{2\cdot2} = 35 \] tiles in total to tile this area.

As it takes \(52\) \(1\times 1\) square tiles to tile the border, and \(35\) \(2\times 2\) square tiles to tile the remaining area, it will take \(52+35=87\) tiles in total to fill in Tyler’s entire living room floor.

Thus, the correct answer is B.

10 答案 C

\(1\)、\(2\) 和 \(4\) 的倒数分别是 \(\dfrac11\)、\(\dfrac12\) 和 \(\dfrac14\)。它们的平均数为 \[\begin{align*}\dfrac{\left(1 + \dfrac12 + \dfrac14\right)}{3} &= \dfrac{\left(\dfrac{7}{4}\right)}{3} \\&= \dfrac{7}{12} \end{align*}\text{。}\]

调和平均数是这些倒数的平均数的倒数；刚才算出的平均数是 \(\dfrac{7}{12}\)，所以调和平均数为 \(\dfrac{12}{7}\)。

正确答案是 C。

English

The reciprocals of \(1\), \(2\), and \(4\) are \(\dfrac11\), \(\dfrac12\), and \(\dfrac14\), respectively. The average of these reciprocals is \[\begin{align*}\dfrac{\left(1 + \dfrac12 + \dfrac14\right)}{3} &= \dfrac{\left(\dfrac{7}{4}\right)}{3} \\&= \dfrac{7}{12}. \end{align*}\]

As the harmonic mean is the reciprocal of the average of the reciprocals of the numbers (which we just calculated to be \(\dfrac{7}{12}\)), we conclude that the harmonic mean is \(\dfrac{12}{7}.\)

Thus, the correct answer is C.

11 答案 C

把情况分成两类：第 \(1\) 类是 Abby 坐在中间列的两个座位之一，第 \(2\) 类是她坐在外侧的 \(4\) 个座位之一。

Abby 坐在中间列的概率是 \(\dfrac26 = \dfrac13\)，这就是第 \(1\) 类。此时 Bridget 要与 Abby 相邻，可以坐在 Abby 左右两个座位中的一个，或者坐在同一列的座位，共有 \(3\) 个可选位置；剩下共有 \(5\) 个空位，所以条件概率是 \(\frac35\)。因此这一类的概率为 \[ \dfrac13 \cdot \dfrac35 = \dfrac{3}{15}\text{。}\]

Abby 坐在外侧的概率是 \(\dfrac46 = \dfrac23\)，这就是第 \(2\) 类。此时 Bridget 要与 Abby 相邻，只能坐在同一行相邻的那个座位，或同一列的座位，共有 \(2\) 个可选位置；剩下仍有 \(5\) 个空位，所以条件概率是 \(\frac25\)。因此这一类的概率为 \[ \frac23 \cdot \frac25 = \frac{4}{15}\text{。}\]

因此任一类发生的总概率为 \(\dfrac{3}{15} + \dfrac{4}{15} = \dfrac{7}{15} \)。

所以正确答案是 C。

English

We can split the problem into two cases. In case \(1,\) Abby is in one of the middle two seats, and in case \(2,\) she is in one of the outer \(4\) seats.

Firstly notice that there is a \( \dfrac26 = \dfrac13 \) probability of case \(1\) being true (i.e. Abby is in the middle two seats). For Bridget to be adjacent to Abby in this case, she must be in either of the two seats beside Abby in the same row, or she is in the same column as her. There are \(3\) ways to make this happen out of a possible \(5\) open seats, so there is a \( \frac35 \) chance of this happening. Therefore, the total probability of this case is \[ \dfrac13 \cdot \dfrac35 = \dfrac{3}{15} .\]

Next, notice that there is a \( \dfrac46 = \dfrac23 \) probability of case \(2\) being true (i.e. Abby is in the outer four seats). For Bridget to be adjacent to Abby in this case, she must either be in the single seat next to Abby in the same row, or she is in the same column as Abby. There are \(2\) ways to make this happen out of a possible \(5\) open seats, so there is a \( \frac25 \) chance of this happening. Therefore, the total probability of this case is \[ \frac23 \cdot \frac25 = \frac{4}{15} .\]

Therefore, the final probability of either of these cases happening is \( \dfrac{3}{15} + \dfrac{4}{15} = \dfrac{7}{15} .\)

Thus, C is the correct answer.

12 答案 B

从 \(12:00\) 中午开始，真实经过 \(30\) 分钟时，车钟走快到经过了 \(35\) 分钟。

因此车钟每走一分钟，真实时间经过 \(\frac {30}{35} = \frac 67\) 分钟。从 \(12:00\) 到 \(7:00\)，车钟走了 \(7\cdot 60 = 420\) 分钟，所以真实时间经过了 \(420 \cdot \frac67 = 360\) 分钟，也就是 \(6\) 小时。从 \(12:00\) 开始经过 \(6\) 小时，实际时间是 \(6:00 \)。

所以正确答案是 B。

English

Starting from \(12:00\) noon, after \(30\) minutes of time elapsed, the car clock went \(35\) minutes ahead.

Therefore, for every minute the car clock goes ahead, \( \frac {30}{35} = \frac 67 \) minutes of actual time pass by. From the time \(12:00 \) to \(7:00,\) the car clock goes ahead \(7\cdot 60 = 420 \) minutes, and therefore, \(420 \cdot \frac67 = 360 \) minutes, or \(6\) hours, of actual time have passed by. If we start at \(12:00 \) and \(6\) hours pass by, the time is \(6:00 .\)

Thus, B is the correct answer.

13 答案 A

因为五次测试平均分是 \(82\)，所以五次总分为 \(5\cdot82 = 410 \)。

设 \(f\) 为前 \(4\) 次测试的分数，\(l\) 为最后一次测试的分数。

我们知道 \[f < l \leq 100\] 且 \[4f + l = 410\text{。}\]又因为 \(410 = 4f + l < 5l \)，所以 \(\frac{410}{5} = 82 < l \)。

另外，由于 \(4f + l = 410 \)，而 \(410\) 除以 \(4\) 的余数是 \(2\)，所以 \(l\) 除以 \(4\) 也必须余 \(2\)，因为 \(4f\) 除以 \(4\) 没有余数。等价地，\[l \equiv 2 \mod 4\text{。}\]因为 \(82 < l \leq 100\) 且 \(l \equiv 2 \mod 4\)，所以 \(l\) 只能是 \(86,90,94,98\)。这给出四组不同的解：\[\begin{align*} (f,l) =& (81,86);\\ &(80,90);\\ &(79,94);\\ &(78,98) \end{align*}\] 因此共有 \(4\) 种可能，正确答案是 A。

English

Since the average score on the five tests is \(82,\) the total score of those five tests must be \(5\cdot82 = 410 .\)

Now, let \(f\) be the score on the first \(4\) tests and let \(l\) be the score for the last test.

We know that \[f < l \leq 100\] and \[4f + l = 410. \] And as \(410 = 4f + l < 5l ,\) we know \(\frac{410}{5} = 82 < l .\)

Also, since \(4f + l = 410 ,\) and dividing \(410\) by \(4\) gives us a remainder of \(2,\) we know that dividing \(l\) by \(4\) must leave a remainder of \(2\) as \(4f\) will leave no remainder when divided by \(4.\) Equivalently: \[l \equiv 2 \mod 4 .\] Since \(82 < l \leq 100 \) and \(l \equiv 2 \mod 4, \) the only options for \(l\) are \(86,90,94,98.\) This yields four distinct solutions as follows: \[\begin{align*} (f,l) =& (81,86);\\ &(80,90);\\ &(79,94);\\ &(78,98) \end{align*}\] Therefore, there are \(4\) solutions, and A is the correct answer.

14 答案 D

为了使 \(5\) 位数最大，必须先让第一位，也就是万位，尽可能大。

小于 \(10\) 且能整除 \(120\) 的最大数字是 \(8\)，所以第一位必须是 \(8\)。剩下各位数字的乘积为 \(15\)。

同样地，现在要让第二位尽可能大。

小于 \(10\) 且能整除 \(15\) 的最大数字是 \(5\)，所以第二位是 \(5\)。剩下各位数字的乘积为 \(3\)。

接着让第三位尽可能大。

小于 \(10\) 且能整除 \(3\) 的最大数字是 \(3\)，所以第三位是 \(3\)。剩下的乘积是 \(1\)，这说明第 \(4\) 位和第 \(5\) 位都是 \(1\)。

于是 \(N = 85311\)，各位数字之和为 \[8+5+3+1+1=18\]

所以正确答案是 D。

English

To make the largest possible \(5\) digit number, we must maximize the first digit (the digit in the ten-thousands place).

The largest number that is strictly less than \(10\) and divides \(120\) is \(8,\) so the first digit must be \(8.\) Therefore, the product of the remaining number is \(15.\)

Similarly, we must now maximize the second digit.

The largest number that is less than \(10\) and divides \(15\) is \(5,\) so the second digit is \(5.\) Therefore, the product of the remaining number is \(3.\)

We must then maximize the third digit.

The largest number that is less than \(10\) and divides \(3\) is \(3,\) so the third digit is \(3.\) Therefore, the product of the remaining number is \(1.\) This means the \(4\)th and \(5\)th digits are \(1.\)

This makes \(N = 85311,\) so the sum of the digits is \[8+5+3+1+1=18\]

Thus, D is the correct answer.

15 答案 D

设大圆面积为 \(A\)。

每个小圆的直径等于大圆半径，所以每个小圆的半径是大圆半径的一半。

用符号表示，若大圆半径为 \(r_l\)，每个小圆半径为 \(r_s\)，则 \[r_s = \dfrac12 r_l\] 因为大圆面积为 \(A=\pi r_l^2\)，所以每个小圆面积为 \[\begin{align*}\pi r_s^2 &= \pi \left(\dfrac12 r_l\right)^2 \\&= \dfrac14 (\pi r_l^2)\\&=\dfrac14 A\end{align*}\text{。}\]两个小圆总面积为 \(1\) 平方单位，于是 \(2\cdot \dfrac14 A=1\) 平方单位，所以 \(A=2\) 平方单位。

阴影面积等于大圆面积 \((A)\) 减去两个小圆的总面积 \((1)\)，所以阴影面积为 \(A - 1=2-1=1\) 平方单位。

所以正确答案是 D。

English

Let \(A\) be the area of the large circle.

Since the diameter of each of the two smaller circles is itself the radius of the larger circle, the radius of each smaller circle is half that of the larger circle.

Symbolically, if we allow \(r_l\) to be the radius of the large circle and \(r_s\) to be the radius of each of the smaller circles: \[r_s = \dfrac12 r_l\] As the area of the larger circle is equal to \(A=\pi r_l^2,\) the area of the smaller circles are equal to \[\begin{align*}\pi r_s^2 &= \pi \left(\dfrac12 r_l\right)^2 \\&= \dfrac14 (\pi r_l^2)\\&=\dfrac14 A.\end{align*}\] As the area of two of these smaller circles combined is equal to \(1\) square unit, then it follows that \(2\cdot \dfrac14 A=1\) square unit, implying that \(A=2\) square units.

As the area of the shaded region is equal to the area of the larger circle \((A)\) minus the combined area of the two smaller circles \((1),\) the area of the shaded region is \( A - 1=2-1=1\) square unit.

Thus, the correct answer is D

16 答案 C

把两本阿拉伯语书作为一个块，四本西班牙语书作为一个块。

此时共有 \(5\) 个对象：三本德语书、一个阿拉伯语书块、一个西班牙语书块。可用 \(5!\) 种方式排列这 \(5\) 个对象。阿拉伯语书块内部有 \(2!\) 种排列，西班牙语书块内部有 \(4!\) 种排列，所以总排列数为 \[\begin{align*}5! \cdot 4! \cdot 2! &= 120 \cdot 24 \cdot 2 \\&= 5760 \end{align*}\]

所以正确答案是 C。

English

Since we are keeping the Arabic books together and the Spanish books together, we can look at each group as a single block.

As such, there are \(5\) objects on the bookshelf: three German books, one collection of Arabic books, and one collection of Spanish books. There are \(5!\) ways to order the \(5\) objects. As we already have the books together, there are \(2!\) ways of ordering the Arabic books and \(4!\) ways of ordering the Spanish books. Therefore, the total ways to order the books is \[\begin{align*}5! \cdot 4! \cdot 2! &= 120 \cdot 24 \cdot 2 \\&= 5760 \end{align*}\]

Thus, the correct answer is C.

17 答案 A

Ella 每骑 \(5\) 英尺，Bella 走一英尺，所以相遇时 Bella 走了两家距离的 \(\frac16\)。因此 Bella 走了 \[\dfrac16 \cdot 10560=1760\] 英尺。因为她每步走 \(2.5\) 英尺，到相遇时她走了 \[\dfrac{1760}{2.5} = 704 \] 步。

所以正确答案是 A。

English

Since for every foot Bella walks, Ella rides \(5\) feet, we know that Bella will walk \(\frac16\) of the distance between the two houses, and so she walks \[\dfrac16 \cdot 10560=1760\] feet. Since she walks \(2.5\) feet per step, she takes \[\dfrac{1760}{2.5} = 704 \] steps by the time she meets Ella.

Thus, A is the correct answer.

18 答案 E

质因数分解为 \(23{,}232=2^6\cdot3\cdot11^2\)。构造正因数时，指数可以从 \(0\) 到 \(6\)（对应质因数 \(2\)）、从 \(0\) 到 \(1\)（对应 \(3\)）、从 \(0\) 到 \(2\)（对应 \(11\)）中选择。因此正因数的个数为 \((6+1)(1+1)(2+1)=42\)。

所以正确答案是 E。

English

The prime factorization is \(23{,}232=2^6\cdot3\cdot11^2.\) A divisor may use any exponent from \(0\) through \(6\) on \(2\), from \(0\) through \(1\) on \(3\), and from \(0\) through \(2\) on \(11\). Therefore, the number of positive divisors is \((6+1)(1+1)(2+1)=42.\)

Thus, E is the correct answer.

19 答案 C

设有两个相邻的下方格子和它们上方的格子。如果已知左下格和上方格，就总能确定右下格：

如果上方格是 \(+\)，则右下格必须与左下格相同；如果上方格是 \(-\)，则右下格必须与左下格相反。

现在假设已知某一行。只要选择这一行最左格左下方的那个格子的符号，就能依次确定下一整行。

因此，顶行已知为 \(+\) 时，下一行有 \(2\) 种选择；同理，第三行有 \(2\) 种选择，第四行也有 \(2\) 种选择。底行共有 \(2\cdot2\cdot2=8\) 种可能。

所以正确答案是 C。

English

Suppose we have two cells and the cell above them. If we are given the bottom left cell and the top cell, we can always find the bottom right cell as follows:

If the top cell is \(+,\) then the bottom right cell must be the same as the bottom left cell, and if the top cell is \(-,\) the bottom right cell must be the opposite of the bottom left cell.

Now, suppose we are given a row. Then, suppose we choose a value for the cell below and to the left of the leftmost cell in our given row. We then can inductively determine the entire row below our given by first finding the bottom-right cell of the leftmost cell in our row, and using that newly found cell as the bottom-left reference for the second to the left cell in the given row to find its bottom-right counterpart. The process continues on until the row below the given row is fully solved.

Therefore, since we know that the top row has a cell labelled \(+,\) we have \(2\) choices for the row below, depending on our choice of the bottom-left cell. Similarly, we have \(2\) choices for the third row, and thus \(2\) choices for the fourth row. This makes \(2\cdot2\cdot2=8\) total choices for the bottom row of the sign pyramid.

Thus, the correct answer is C.

20 答案 A

设 \(\triangle ABC\) 的面积为 \(t\)。由于 \(DE \parallel BC\) 且 \(FE \parallel CA \)，可得 \[ADE \sim ABC\] 以及 \[EFB \sim ABC\text{。}\]因为 \(AE = \dfrac{AB}3\)，所以 \(ADE\) 的面积为 \(\left(\dfrac13\right)^2 t = \dfrac{t}{9} \)。因为 \(EB = \dfrac{2AB}3\)，所以 \(EFB\) 的面积为 \(\left(\dfrac23\right)^2 t = \dfrac{4}{9}t \)。最后，要找 \(CDEF\) 的面积，就从 \(ABC =t\) 中减去 \(ADE\) 和 \(EFB\) 的面积，即 \[t- \frac{t}{9} - \frac{4t}{9} = \frac{4t}{9}\text{。}\]因此 \(CDEF\) 与 \(ABC\) 的面积比为 \(\dfrac{\left(\dfrac{4t}{9}\right)}{t} = \dfrac{4}{9} \)。

所以正确答案是 A。

English

Let the area of \(\triangle ABC\) be equal to \(t.\) Since \(DE \parallel BC\) and \(FE \parallel CA ,\) we can deduce that \[ADE \sim ABC\] and \[EFB \sim ABC. \] Since \(AE = \dfrac{AB}3,\) the area of \(ADE\) is equal to \(\left(\dfrac13\right)^2 t = \dfrac{t}{9} .\) Since \(EB = \dfrac{2AB}3,\) the area of \(EFB\) is equal to \(\left(\dfrac23\right)^2 t = \dfrac{4}{9}t .\) Finally, to find the area of \(CDEF,\) we take the area of \(ABC =t\) and subtract the areas of \(ADE \) and \(EFB.\) This is equivalent to the expression \[t- \frac{t}{9} - \frac{4t}{9} = \frac{4t}{9} .\] Therefore, the ratio of the area of \(CDEF\) and \(ABC \) is \(\dfrac{\left(\dfrac{4t}{9}\right)}{t} = \dfrac{4}{9} .\)

Thus, A is the correct answer.

21 答案 E

每个所给余数都比相应除数小 \(4\)，所以 \(x+4\) 必须同时被 \(6,9\) 和 \(11\) 整除。因此 \(x+4\) 是 \(\operatorname{lcm}(6,9,11)=198\) 的倍数。

因为 \(x\) 是三位数，所以 \(104\le x+4\le1003\)。这个区间内 \(198\) 的倍数是 \(198,396,594,792,990\)，因此共有 \(5\) 个符合条件的整数。

正确答案是 E。

English

Each required remainder is \(4\) less than its divisor, so \(x+4\) must be divisible by \(6,9,\) and \(11\). Hence \(x+4\) is a multiple of \(\operatorname{lcm}(6,9,11)=198.\)

For a three-digit \(x\), we have \(104\le x+4\le1003\). The multiples of \(198\) in this interval are \(198,396,594,792,990\), giving \(5\) possible integers.

Thus, E is the correct answer.

22 答案 B

设正方形边长为 \(s\)，并设 \(H\) 是从 \(F\) 向 \(\overline{BC}\) 作垂线的垂足。直角三角形 \(CAB\) 与 \(CFH\) 相似，所以 \(CH=FH\)。另外，三角形 \(BFH\) 与 \(BEC\) 相似，因此 \[ \frac{FH}{EC}=\frac{BH}{BC}\text{。}\]因为 \(EC=\frac{s}{2}\)、\(BC=s\)，且 \(BH=s-CH=s-FH\)，所以 \(\frac{2FH}{s}=1-\frac{FH}{s}\)，从而 \(FH=CH=\frac{s}{3}\)。

因此，三角形 \(EFC\) 的底为 \(EC=\frac{s}{2}\)，高为 \(CH=\frac{s}{3}\)，所以面积为 \(\frac{s^2}{12}\)。于是 \[\begin{aligned} [AFED]&=[\triangle ACD]-[\triangle EFC]\\ &=\dfrac{s^2}{2}-\dfrac{s^2}{12}\\ &=\dfrac{5s^2}{12} \end{aligned}\text{。}\]因为这个面积是 \(45\)，所以 \(s^2=108\)，这就是正方形的面积。

English

Let the square have side length \(s,\) and let \(H\) be the foot of the perpendicular from \(F\) to \(\overline{BC}.\) The right triangles \(CAB\) and \(CFH\) are similar, so \(CH=FH.\) Also, triangles \(BFH\) and \(BEC\) are similar, giving \[ \frac{FH}{EC}=\frac{BH}{BC}. \] Since \(EC=\frac{s}{2},\) \(BC=s,\) and \(BH=s-CH=s-FH,\) this becomes \(\frac{2FH}{s}=1-\frac{FH}{s},\) so \(FH=CH=\frac{s}{3}.\)

Thus triangle \(EFC\) has base \(EC=\frac{s}{2}\) and height \(CH=\frac{s}{3},\) so its area is \(\frac{s^2}{12}.\) Therefore \[\begin{aligned} [AFED]&=[\triangle ACD]-[\triangle EFC]\\ &=\dfrac{s^2}{2}-\dfrac{s^2}{12}\\ &=\dfrac{5s^2}{12}. \end{aligned}\] Since this area is \(45\), we get \(s^2=108\), the area of the square.

23 答案 D

用补集计数：计算任意两个所选顶点都不相邻的三角形。从一个所选顶点出发，设三个间隔中未选顶点的正整数个数为 \(x,y,z\)。那么 \(x+y+z=5\)，共有 \(\binom42=6\) 个正整数解。

起点有 \(8\) 种选择，而每个三角形会从它的 \(3\) 个顶点各被计算一次，所以补集中有 \(\frac{8\cdot6}{3}=16\) 个三角形。全部三角形共有 \(\binom83=56\) 个，因此所求概率是 \(1-\frac{16}{56}=\frac{5}{7}\)。

所以正确答案是 D。

English

Count the complement: triangles with no two chosen vertices adjacent. Starting at a chosen vertex, let \(x,y,z\) be the positive numbers of unchosen vertices in the three gaps. Then \(x+y+z=5\), which has \(\binom42=6\) positive solutions.

There are \(8\) choices for the starting vertex, and each triangle is counted from each of its \(3\) vertices, so the complement contains \(\frac{8\cdot6}{3}=16\) triangles. Out of \(\binom83=56\) total triangles, the desired probability is \(1-\frac{16}{56}=\frac{5}{7}.\)

Thus, D is the correct answer.

24 答案 C

设立方体边长为 \(s\)。注意截面四边形每条边长度相同，所以 \(EJCI\) 是菱形。菱形面积等于两条对角线乘积的一半，因此它的面积为 \(\frac12 IJ\cdot CE\)。由勾股定理，\[IJ=FH=s\sqrt{2}\text{。}\]再次使用勾股定理可得 \[\begin{align*}CE&=\sqrt{AC^2+AE^2}\\&=\sqrt{(s\sqrt{2})^2+s^2}\\&=\sqrt{2s^2+s^2}\\&=s\sqrt{3}\end{align*}\] 因此 \[\begin{align*}R&=\dfrac{\frac12 IJ\cdot CE}{s^2}\\&=\dfrac{\frac12 s^2\sqrt{2}\sqrt{3}}{s^2}\\&=\sqrt{\dfrac32}\end{align*}\] 所以 \(R^2=\dfrac32\)，正确答案是 C。

English

Allow \(s\) to represent the length of an edge of the cube. Noting that each side of the cross section is equal in length, we conclude that \(EJCI\) is a rhombus. The area of this rhombus can be calculated as \(\frac12 IJ\cdot CE,\) as the area of a rhombus is equal to half the product of its diagonals. Using the Pythagorean Theorem: \[IJ=FH=s\sqrt{2}.\] Similarly, using the Pythagorean Theorem again lets us see that: \[\begin{align*}CE&=\sqrt{AC^2+AE^2}\\&=\sqrt{(s\sqrt{2})^2+s^2}\\&=\sqrt{2s^2+s^2}\\&=s\sqrt{3}\end{align*}\] Therefore, \[\begin{align*}R&=\dfrac{\frac12 IJ\cdot CE}{s^2}\\&=\dfrac{\frac12 s^2\sqrt{2}\sqrt{3}}{s^2}\\&=\sqrt{\dfrac32}\end{align*}\] Thus, \(R^2=\dfrac32,\) and the correct answer is C.

25 答案 E

因为 \(6^3=216<257\)，且 \(257<343=7^3\)，区间内最小的完全立方数是 \(7^3\)。又因为 \(64^3=2^{18}<2^{18}+1<65^3\)，最大的完全立方数是 \(64^3\)。

所以整数立方根为 \(7,8,\ldots,64\)，共有 \(64-7+1=58\) 个。

所以正确答案是 E。

English

Because \(6^3=216<257\) and \(257<343=7^3\), the smallest cube in the interval is \(7^3\). Also, \(64^3=2^{18}<2^{18}+1<65^3\), so the largest cube is \(64^3\).

The integer cube roots are therefore \(7,8,\ldots,64\), a total of \(64-7+1=58\).

Thus, E is the correct answer.