/

AMC8 2018

You are not logged in. After submit, your report may not be available on other devices. Login

AMC8 · 2018

Q1
An amusement park has a collection of scale models, with ratio 1:20, of buildings and other sights from around the country. The height of the United States Capitol is 289 feet. What is the height in feet of its replica at this park, rounded to the nearest whole number?
一个游乐园有一系列比例为 1:20 的建筑物和其他景点的比例模型。美国国会大厦的高度是 289 英尺。这个公园里的复制品高度是多少英尺，四舍五入到最近的整数？
Correct Answer: A
Answer (A): The height of the replica is $\frac{289}{20}=14.45$ feet, so the answer is 14 feet when rounded to the nearest whole number.
答案（A）：该复制品的高度为$\frac{289}{20}=14.45$英尺，因此四舍五入到最接近的整数时，答案是14英尺。
Q2
What is the value of the product $(1 + \frac{1}{1}) \cdot (1 + \frac{1}{2}) \cdot (1 + \frac{1}{3}) \cdot (1 + \frac{1}{4}) \cdot (1 + \frac{1}{5}) \cdot (1 + \frac{1}{6})$?
计算乘积的值 $(1 + \frac{1}{1}) \cdot (1 + \frac{1}{2}) \cdot (1 + \frac{1}{3}) \cdot (1 + \frac{1}{4}) \cdot (1 + \frac{1}{5}) \cdot (1 + \frac{1}{6})$？
Correct Answer: D
Answer (D): The product may be written as $2\cdot\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot\frac{6}{5}\cdot\frac{7}{6}=7$.
答案（D）：这个乘积可以写成 $2\cdot\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot\frac{6}{5}\cdot\frac{7}{6}=7$。
Q3
Students Arn, Bob, Cyd, Dan, Eve, and Fon are arranged in that order in a circle. They start counting: Arn first, then Bob, and so forth. When the number contains a 7 as a digit (such as 47) or is a multiple of 7 that person leaves the circle and the counting continues. Who is the last one present in the circle?
学生 Arn、Bob、Cyd、Dan、Eve 和 Fon 按此顺序围成一圈。他们开始计数：Arn 先，然后 Bob，依此类推。当数字包含数字 7（如 47）或 是 7 的倍数时，那个人离开圈子，计数继续。谁是圈子里最后剩下的一个人？
Correct Answer: D
Answer (D): Dan is the last one present because Arn is out on $7$, Cyd on $14$, Fon on $17$, Bob on $21$ and Eve on $27$.
答案（D）：Dan 是最后一个在场的人，因为 Arn 在 $7$ 号离开，Cyd 在 $14$ 号离开，Fon 在 $17$ 号离开，Bob 在 $21$ 号离开，而 Eve 在 $27$ 号离开。
Q4
The twelve-sided figure shown has been drawn on 1 cm × 1 cm graph paper. What is the area of the figure in cm²?
图中所示的十二边形是在 1 cm × 1 cm 方格纸上绘制的。该图形的面积是多少平方厘米？
stem
Correct Answer: C
Answer (C): In the figure below, the square RSTU has side length 3, so its area is $3^2=9$, and $\triangle REF$ has area $1$. The other three triangles are congruent to $\triangle REF$, so the total area is $9+4\cdot1=13$.
解答（C）：如图所示，正方形 RSTU 的边长为 3，因此其面积为 $3^2=9$，且三角形 $\triangle REF$ 的面积为 $1$。其余三个三角形与 $\triangle REF$ 全等，所以总面积为 $9+4\cdot1=13$。
Q5
What is the value of $1+3+5+\dots+2017+2019-2-4-6-\dots-2016-2018$?
计算 $1+3+5+\dots+2017+2019-2-4-6-\dots-2016-2018$ 的值？
Correct Answer: E
Answer (E): The expression can be written as $1+(3-2)+(5-4)+(7-6)+\cdots+(2019-2018)=1+\frac{2018}{2}=1010$.
答案（E）：该表达式可以写成 $1+(3-2)+(5-4)+(7-6)+\cdots+(2019-2018)=1+\frac{2018}{2}=1010$。
Q6
On a trip to the beach, Anh traveled 50 miles on the highway and 10 miles on a coastal access road. He drove three times as fast on the highway as on the coastal road. If Anh spent 30 minutes driving on the coastal road, how many minutes did his entire trip take?
在去海滩的旅行中，Anh 在高速公路上行驶了50英里，在沿海通道上行驶了10英里。他在高速公路上的速度是沿海道路速度的三倍。如果Anh在沿海道路上行驶了30分钟，整个旅行总共用了多少分钟？
Correct Answer: C
Answer (C): Anh’s speed on the coastal road was $20$ miles per hour because he drove $10$ miles in half an hour. This implies that Anh traveled $60$ miles per hour on the highway, and so it took Anh $50$ minutes to drive the $50$ miles. Anh’s entire trip took $30+50=80$ minutes.
答案（C）：Anh 在海岸公路上的速度是每小时 $20$ 英里，因为他半小时行驶了 $10$ 英里。这意味着 Anh 在高速公路上的速度是每小时 $60$ 英里，因此行驶那 $50$ 英里用了 $50$ 分钟。Anh 的整个行程用了 $30+50=80$ 分钟。
Q7
The 5-digit number $2\ 0\ 1\ 8\ 7$ is divisible by 9. What is the remainder when this number is divided by 8?
五位数 $2\ 0\ 1\ 8\ 7$ 能被9整除。这个数除以8的余数是多少？
Correct Answer: B
Answer (B): To be divisible by $9$, the sum of the digits must be divisible by $9$, so $2+0+1+8+U=11+U$ is divisible by $9$. Thus $U=7$, and the $5$-digit number is $20187$. Then because $20187=8\cdot2523+3$, the remainder is $3$.
答案（B）：要能被 $9$ 整除，各位数字之和必须能被 $9$ 整除，因此 $2+0+1+8+U=11+U$ 必须能被 $9$ 整除。于是 $U=7$，这个 $5$ 位数是 $20187$。又因为 $20187=8\cdot2523+3$，所以余数是 $3$。
Q8
Mr. Garcia asked the members of his health class how many days last week they exercised for at least 30 minutes. The results are summarized in the following bar graph, where the heights of the bars represent the number of students. What was the mean number of days of exercise last week, rounded to the nearest hundredth, reported by the students in Mr. Garcia's class?
Garcia先生询问他的健康课成员上周锻炼至少30分钟的天数。结果总结在下面的条形图中，条的高度表示学生人数。学生们报告的上周锻炼天数的平均数，四舍五入到最近的百分位，是多少？
stem
Correct Answer: C
Answer (C): The number of students surveyed was $1+3+2+6+8+3+2=25$, and the number of days of exercise reported was $1\cdot1+3\cdot2+2\cdot3+6\cdot4+8\cdot5+3\cdot6+2\cdot7=109$. So the mean number of days of exercise was $109\div25=4.36$.
答案（C）：被调查的学生人数为 $1+3+2+6+8+3+2=25$，报告的运动天数总和为 $1\cdot1+3\cdot2+2\cdot3+6\cdot4+8\cdot5+3\cdot6+2\cdot7=109$。因此，平均每人运动的天数为 $109\div25=4.36$。
Q9
Tyler is tiling the floor of his 12 foot by 16 foot living room. He plans to place one-foot by one-foot square tiles to form a border along the edges of the room and to fill in the rest of the floor with two-foot by two-foot square tiles. How many tiles will he use?
Tyler正在为他12英尺乘16英尺的客厅铺瓷砖。他计划沿房间边缘放置一英尺乘一英尺的方形瓷砖作为边框，并用两英尺乘两英尺的方形瓷砖填充其余地板。他将使用多少块瓷砖？
Correct Answer: B
Answer (B): The perimeter of the room is $2(12+16)=56$ feet. The $4$ corner tiles each occupy $2$ feet of the perimeter, so Tyler needs those $4$ plus another $56-4\cdot2=48$ one-foot square tiles to make the border. The remaining space in the room is a $10$ foot by $14$ foot rectangle, so Tyler needs $\frac{10}{2}\cdot\frac{14}{2}=35$ two-foot square tiles to cover it. Therefore he needs a total of $4+48+35=87$ tiles.
答案（B）：房间的周长是 $2(12+16)=56$ 英尺。四个角上的瓷砖每块占据周长中的 $2$ 英尺，所以 Tyler 需要这 $4$ 块角砖，再加上另外 $56-4\cdot2=48$ 块边框用的 1 英尺见方瓷砖来铺边框。房间剩余的空间是一个 $10$ 英尺乘 $14$ 英尺的长方形，因此 Tyler 需要 $\frac{10}{2}\cdot\frac{14}{2}=35$ 块边长为 2 英尺的正方形瓷砖来覆盖。于是他总共需要 $4+48+35=87$ 块瓷砖。
Q10
The harmonic mean of a set of non-zero numbers is the reciprocal of the average of the reciprocals of the numbers. What is the harmonic mean of 1, 2, and 4?
一组非零数的调和平均数是这些数的倒数的平均数的倒数。1、2和4的调和平均数是多少？
Correct Answer: C
Answer (C): The reciprocals of $1$, $2$, and $4$ are $1$, $\frac{1}{2}$, and $\frac{1}{4}$, and their average is $\frac{1+\frac{1}{2}+\frac{1}{4}}{3}=\frac{7}{12}$. So the harmonic mean of $1$, $2$, and $4$ is the reciprocal of $\frac{7}{12}$, which is $\frac{12}{7}$.
答案（C）：$1$、$2$、$4$ 的倒数分别是 $1$、$\frac{1}{2}$、$\frac{1}{4}$，它们的平均值为 $\frac{1+\frac{1}{2}+\frac{1}{4}}{3}=\frac{7}{12}$。因此，$1$、$2$、$4$ 的调和平均数是 $\frac{7}{12}$ 的倒数，即 $\frac{12}{7}$。
Q11
Abby, Bridget, and four of their classmates will be seated in two rows of three for a group picture, as shown. X X X X X X If the seating positions are assigned randomly, what is the probability that Abby and Bridget are adjacent to each other in the same row or the same column?
Abby、Bridget 和他们的四位同学将坐在两排三列的位置上拍团体照，如图所示。 X X X X X X 如果座位位置是随机分配的，那么Abby和Bridget在同一行或同一列相邻的概率是多少？
Correct Answer: C
Answer (C): There are $6$ possible positions for Abby, and this leaves $5$ possible positions for Bridget. Because their order doesn’t matter, the two girls can be placed in any of $\frac{6\cdot5}{2}=15$ pairs of positions. There are $2$ pairs of positions that are adjacent in the top row, $2$ pairs that are adjacent in the bottom row, and $3$ pairs that are adjacent in the same column. So the probability that they occupy adjacent positions is $\frac{2+2+3}{15}=\frac{7}{15}$.
答案（C）：Abby 有 $6$ 个可能的位置，这样 Bridget 剩下 $5$ 个可能的位置。由于两人的先后顺序不重要，这两个女孩可以放在 $\frac{6\cdot5}{2}=15$ 对位置中的任意一对。上排有 $2$ 对相邻位置，下排有 $2$ 对相邻位置，同一列有 $3$ 对相邻位置。因此，她们占据相邻位置的概率是 $\frac{2+2+3}{15}=\frac{7}{15}$。
Q12
The clock in Sri’s car, which is not accurate, gains time at a constant rate. One day as he begins shopping he notes that his car clock and his watch (which is accurate) both say 12:00 noon. When he is done shopping, his watch says 12:30 and his car clock says 12:35. Later that day, Sri loses his watch. He looks at his car clock and it says 7:00. What is the actual time?
Sri车上的钟不准确，以恒定速率快走。有一天他开始购物时，注意到他的车钟和他的手表（准确的）都显示12:00中午。购物结束后，他的手表显示12:30，车钟显示12:35。那天晚些时候，Sri丢了手表。他看车钟显示7:00，实际时间是几点？
Correct Answer: B
Answer (B): The ratio of time as measured by Sri’s car clock to actual time is $\frac{35}{30}=\frac{7}{6}$. Therefore when the clock shows that $7$ hours have passed since noon, only $6$ hours have actually passed, so the actual time is $6$:00.
答案（B）：Sri 的车钟所测量的时间与实际时间之比为 $\frac{35}{30}=\frac{7}{6}$。因此，当车钟显示从中午起已经过去 $7$ 小时时，实际上只过去了 $6$ 小时，所以实际时间是 $6$:00。
Q13
Laila took five math tests, each worth a maximum of 100 points. Laila’s score on each test was an integer between 0 and 100, inclusive. Laila received the same score on the first four tests, and she received a higher score on the last test. Her average score on the five tests was 82. How many values are possible for Laila’s score on the last test?
Laila参加了五次数学测验，每次满分100分。Laila每次测验的得分是0到100之间的整数。前四次测验得分相同，最后一次得分更高。五次测验的平均分是82。最后一次测验的得分有多少种可能值？
Correct Answer: A
Answer (A): Because Laila’s score on the last test was higher than her other test scores and all of her scores were integers, for each point below $82$ that she scored on each of the first four tests, she had to score four points above $82$ on the last test. Therefore, her possible scores on the last test were $86$, $90$, $94$, and $98$ (with scores on the first four tests of $81$, $80$, $79$, and $78$, respectively). Thus there are four possible values for her score on the last test.
答案（A）：因为 Laila 的最后一次测验分数高于她其他测验的分数，并且她的所有分数都是整数，所以在前四次测验中，她每比 $82$ 低 $1$ 分，最后一次测验就必须比 $82$ 高 $4$ 分。于是，她最后一次测验可能的分数是 $86$、$90$、$94$、$98$（分别对应前四次测验分数为 $81$、$80$、$79$、$78$）。因此，她最后一次测验的分数共有 4 种可能。
Q14
Let $N$ be the greatest five-digit number whose digits have a product of 120. What is the sum of the digits of $N$?
设 $N$ 是各位数字乘积为120的最大五位数。$N$ 的各位数字之和是多少？
Correct Answer: D
Answer (D): The leftmost digit of $N$ must be the greatest single-digit factor of $120=2^3\cdot3\cdot5$. Note that $2^3=8$ is a factor of $120$, but $9$ is not, hence the leftmost digit of $N$ must be $8$. Because $5\cdot3=15$ cannot be a digit, it follows that $5$ must be immediately to the right of the $8$, and $3$ must be immediately to the right of the $5$. The remaining rightmost two digits of $N$ must be $1$, so that $N=85311$. Thus the sum of the digits of $N$ is $18$.
答案（D）：$N$ 的最左边一位必须是 $120=2^3\cdot3\cdot5$ 的最大一位数因子。注意 $2^3=8$ 是 $120$ 的因子，但 $9$ 不是，因此 $N$ 的最左一位必须是 $8$。由于 $5\cdot3=15$ 不能作为一位数字，所以 $5$ 必须紧挨在 $8$ 的右边，而 $3$ 必须紧挨在 $5$ 的右边。剩下最右边两位必须是 $1$，因此 $N=85311$。所以 $N$ 的各位数字之和为 $18$。
Q15
In the diagram below, a diameter of each of the two smaller circles is a radius of the larger circle. If the two smaller circles have a combined area of 1 square unit, then what is the area of the shaded region, in square units?
下图中，两个小圆的直径均为大圆的半径。如果两个小圆的总面积为1平方单位，则阴影区域的面积是多少平方单位？
stem
Correct Answer: D
Answer (D): If one of the smaller circles has radius $r$, then its area is $\pi r^2$, and the area of the larger circle is $\pi(2r)^2=4\pi r^2$. Thus the area of the large circle is $4$ times the area of one of the small circles. So the area of the shaded region is equal to the combined area of the two smaller circles, which is $1$ square unit.
答案（D）：如果较小的一个圆的半径为 $r$，那么它的面积是 $\pi r^2$，而较大圆的面积是 $\pi(2r)^2=4\pi r^2$。因此，大圆的面积是小圆面积的 $4$ 倍。所以阴影部分的面积等于两个小圆面积之和，也就是 $1$ 平方单位。
Q16
Professor Chang has nine different language books lined up on a bookshelf: two Arabic, three German, and four Spanish. How many ways are there to arrange the nine books on the shelf keeping the Arabic books together and keeping the Spanish books together?
张教授有九本不同的语言书籍排放在书架上：两本阿拉伯语，三本德语，四本西班牙语。保持阿拉伯语书籍在一起且西班牙语书籍在一起，将这九本书排放在书架上的方法有多少种？
Correct Answer: C
Answer (C): There are two ways to arrange the two Arabic books among themselves and there are $4\cdot3\cdot2\cdot1=24$ ways to arrange the four Spanish books among themselves. If the two Arabic books are considered as a unit and the four Spanish books are considered as a unit, then, including the three German books, there are five objects to arrange on the shelf. These five objects can be arranged on the shelf in $5\cdot4\cdot3\cdot2\cdot1=120$ ways. So altogether the books can be arranged in $2\cdot24\cdot120=5760$ ways.
答案（C）：两本阿拉伯书在内部可以有 2 种排列方式，四本西班牙书在内部有 $4\cdot3\cdot2\cdot1=24$ 种排列方式。若把两本阿拉伯书看作一个整体、把四本西班牙书看作一个整体，再加上三本德语书，则书架上相当于有 5 个“对象”需要排列。这 5 个对象在书架上的排列方式有 $5\cdot4\cdot3\cdot2\cdot1=120$ 种。因此总的排列方式为 $2\cdot24\cdot120=5760$ 种。
Q17
Bella begins to walk from her house toward her friend Ella's house. At the same time, Ella begins to ride her bicycle toward Bella's house. They each maintain a constant speed, and Ella rides 5 times as fast as Bella walks. The distance between their houses is 2 miles, which is 10,560 feet, and Bella covers $2\frac{1}{2}$ feet with each step. How many steps will Bella take by the time she meets Ella?
贝拉从她家开始向朋友艾拉的家步行。与此同时，艾拉开始骑自行车向贝拉家骑行。她们各自保持恒定速度，艾拉骑行的速度是贝拉步行速度的5倍。两家之间的距离是2英里，即10,560英尺，贝拉每步走 $2\frac{1}{2}$ 英尺。到她与艾拉相遇时，贝拉会走多少步？
Correct Answer: A
Answer (A): By the time that the two girls meet, Ella will ride $5$ times the distance that Bella walks, so Ella will ride $\frac{5}{6}$ of the total distance, and Bella will walk $\frac{1}{6}$ of the total distance, or $\frac{1}{6}(10,560)=1760$ feet. Each of her steps covers $2\frac{1}{2}=\frac{5}{2}$ feet, so the number of steps she will take is $1760\div\frac{5}{2}=1760\cdot\frac{2}{5}=704$.
答案（A）：当两个女孩相遇时，Ella 骑行的距离是 Bella 步行距离的 $5$ 倍，因此 Ella 骑行总路程的 $\frac{5}{6}$，Bella 步行总路程的 $\frac{1}{6}$，即 $\frac{1}{6}(10,560)=1760$ 英尺。Bella 每一步走 $2\frac{1}{2}=\frac{5}{2}$ 英尺，所以她将走的步数为 $1760\div\frac{5}{2}=1760\cdot\frac{2}{5}=704$。
Q18
How many positive factors does 23,232 have?
23,232 有多少个正因数？
Correct Answer: E
Answer (E): The prime factorization of $23{,}232$ is $2^6\cdot3\cdot11^2$. Each factor of $23{,}232$ must be of the form $2^a\cdot3^b\cdot11^c$, where $a=0,1,2,3,4,5,$ or $6$, $b=0$ or $1$, and $c=0,1,$ or $2$. Therefore, the number of factors of $23{,}232$ is $7\cdot2\cdot3=42$.
答案（E）：$23{,}232$ 的质因数分解为 $2^6\cdot3\cdot11^2$。$23{,}232$ 的每个因数都必须形如 $2^a\cdot3^b\cdot11^c$，其中 $a=0,1,2,3,4,5$ 或 $6$，$b=0$ 或 $1$，$c=0,1$ 或 $2$。因此，$23{,}232$ 的因数个数为 $7\cdot2\cdot3=42$。
Q19
In a sign pyramid a cell gets a “+” if the two cells below it have the same sign, and it gets a “-” if the two cells below it have different signs. The diagram below illustrates a sign pyramid with four levels. How many possible ways are there to fill the four cells in the bottom row to produce a “+” at the top of the pyramid?
在一个符号金字塔中，如果下方两个单元格符号相同，则上方的单元格为“+”，如果不同则为“-”。下图展示了一个四层的符号金字塔。填充底行四个单元格有多少种可能的方式，使得金字塔顶端为“+”？
stem
Correct Answer: C
Answer (C): Think of the $+$ sign as $+1$, and the $-$ sign as $-1$. Let $a$, $b$, $c$, and $d$ denote the values of the four cells at the bottom of the pyramid, in that order. Then the cells in the second row from the bottom have values $a\cdot b$, $b\cdot c$ and $c\cdot d$, and the cells in the row above this are $a\cdot b\cdot b\cdot c=a\cdot c$ and $b\cdot c\cdot c\cdot d=b\cdot d$ (because both $1$ and $-1$ squared are $1$). Finally, the top cell has value $a\cdot b\cdot c\cdot d$. This value is $+1$ if all four variables are $+1$ or all four are $-1$, giving two ways; or, if two of the variables are $+1$ and two are $-1$, giving $6$ additional ways ($++--$, $+-+-$, $+--+$, $-++-$, $-+-+$, and $--++$). Thus there are a total of $8$ ways to fill the fourth row.
答案（C）：把 $+$ 号看作 $+1$，把 $-$ 号看作 $-1$。设金字塔底部一行从左到右四个格子的值分别为 $a,b,c,d$。那么倒数第二行三个格子的值分别为 $a\cdot b$、$b\cdot c$、$c\cdot d$；再上一行两个格子的值为 $a\cdot b\cdot b\cdot c=a\cdot c$ 和 $b\cdot c\cdot c\cdot d=b\cdot d$（因为 $1$ 和 $-1$ 的平方都等于 $1$）。最后，顶端格子的值为 $a\cdot b\cdot c\cdot d$。当四个变量全为 $+1$ 或全为 $-1$ 时，该值为 $+1$，共有 2 种；或者当其中两个为 $+1$、两个为 $-1$ 时，该值也为 $+1$，共有另外 $6$ 种（$++--$、$+-+-$、$+--+$、$-++-$、$-+-+$、$--++$）。因此，填充第四行共有 $8$ 种方法。
Q20
In $\triangle ABC$, point $E$ is on $\overline{AB}$ with $AE = 1$ and $EB = 2$. Point $D$ is on $\overline{AC}$ so that $\overline{DE} \parallel \overline{BC}$ and point $F$ is on $\overline{BC}$ so that $\overline{EF} \parallel \overline{AC}$. What is the ratio of the area of $\triangle CDEF$ to the area of $\triangle ABC$?
在 $\triangle ABC$ 中，点 $E$ 在 $\overline{AB}$ 上，$AE = 1$，$EB = 2$。点 $D$ 在 $\overline{AC}$ 上，使得 $\overline{DE} \parallel \overline{BC}$，点 $F$ 在 $\overline{BC}$ 上，使得 $\overline{EF} \parallel \overline{AC}$。$ riangle CDEF$ 的面积与 $\triangle ABC$ 的面积之比是多少？
stem
Correct Answer: A
Answer (A): Triangles $AED$, $EBF$, and $ABC$ are similar with their sides in the ratio of $1:2:3$. Therefore their areas are in the ratio of $1:4:9$. The combined areas of $\triangle AED$ and $\triangle EBF$ constitute $\frac{5}{9}$ of the area of $\triangle ABC$, so the area of $CDEF$ is $\frac{4}{9}$ of the area of $\triangle ABC$.
答案（A）：三角形 $AED$、$EBF$ 和 $ABC$ 相似，它们的边长比为 $1:2:3$，因此面积比为 $1:4:9$。$\triangle AED$ 与 $\triangle EBF$ 的面积之和占 $\triangle ABC$ 面积的 $\frac{5}{9}$，所以 $CDEF$ 的面积占 $\triangle ABC$ 面积的 $\frac{4}{9}$。
Q21
How many positive three-digit integers have a remainder of 2 when divided by 6, a remainder of 5 when divided by 9, and a remainder of 7 when divided by 11?
有有多少个正的三位数，除以6余2，除以9余5，除以11余7？
Correct Answer: E
Answer (E): Let $n$ be a positive three-digit integer that satisfies the conditions stated in the problem. Note that $n$ has a remainder of $2$ when divided by $6$ and $5$ when divided by $9$. Similarly, it has a remainder of $7$ when divided by $11$. Hence $n+4$ is divisible by the least common multiple of $6$, $9$, and $11$, which is $198$. Thus $n+4=198k$ where $k=1,2,3,4,$ or $5$, so $n=194,392,590,788, or\ 986$. The five numbers satisfy the conditions in the problem, so the answer is $5$.
解答 (E)：设 $n$ 为满足题意的三位正整数。注意到 $n$ 被 $6$ 除余 $2$、被 $9$ 除余 $5$，并且被 $11$ 除余 $7$。因此 $n+4$ 能被 $6,9,11$ 的最小公倍数 $198$ 整除。于是有 $n+4=198k$，其中 $k=1,2,3,4,5$，所以 $n=194,392,590,788,986$。这五个数都满足条件，所以答案为 $5$。
Q22
Point E is the midpoint of side CD in square ABCD, and BE meets diagonal AC at F. The area of quadrilateral AFED is 45. What is the area of ABCD?
在正方形ABCD中，点E是边CD的中点，BE与对角线AC相交于F。四边形AFED的面积是45。ABCD的面积是多少？
stem
Correct Answer: B
Answer (B): To see what fraction of the square is occupied by quadrilateral $AFED$, first note that $\triangle ADC$ occupies half of the area. Next note that because $AB$ and $CD$ are parallel, it follows that $\triangle ABF$ and $\triangle CEF$ are similar, and $CE=\frac{1}{2}AB$. Draw $PQ$ through $F$ such that $PF$ and $FQ$ are altitudes of $\triangle ABF$ and $\triangle CEF$, respectively. Then $FQ=\frac{1}{2}PF$, so $FQ=\frac{1}{3}PQ=\frac{1}{3}BC$. Therefore the area of $\triangle CEF$ is $\frac{1}{2}(CE)(FQ)=\frac{1}{2}\left(\frac{1}{2}AB\right)\left(\frac{1}{3}BC\right)=\frac{1}{12}(AB)(BC)$, which is $\frac{1}{12}$ of the area of the square. Because the area of quadrilateral $AFED$ equals the area of $\triangle ADC$ minus the area of $\triangle CEF$, the fraction of the square occupied by quadrilateral $AFED$ is $\frac{1}{2}-\frac{1}{12}=\frac{5}{12}$. Because the area of $AFED$ is $45$, the area of $ABCD$ is $\left(\frac{12}{5}\right)(45)=108$.
答案（B）：为了求四边形 $AFED$ 占正方形的面积比例，先注意到 $\triangle ADC$ 占正方形面积的一半。再注意到因为 $AB$ 与 $CD$ 平行，可知 $\triangle ABF$ 与 $\triangle CEF$ 相似，并且 $CE=\frac{1}{2}AB$。过点 $F$ 作直线 $PQ$，使得 $PF$ 与 $FQ$ 分别是 $\triangle ABF$ 与 $\triangle CEF$ 的高。则 $FQ=\frac{1}{2}PF$，所以 $FQ=\frac{1}{3}PQ=\frac{1}{3}BC$。因此 $\triangle CEF$ 的面积为 $\frac{1}{2}(CE)(FQ)=\frac{1}{2}\left(\frac{1}{2}AB\right)\left(\frac{1}{3}BC\right)=\frac{1}{12}(AB)(BC)$，这等于正方形面积的 $\frac{1}{12}$。由于四边形 $AFED$ 的面积等于 $\triangle ADC$ 的面积减去 $\triangle CEF$ 的面积，所以 $AFED$ 占正方形的比例为 $\frac{1}{2}-\frac{1}{12}=\frac{5}{12}$。已知 $AFED$ 的面积为 $45$，则正方形 $ABCD$ 的面积为 $\left(\frac{12}{5}\right)(45)=108$。
Q23
From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probability that at least one of the sides of the triangle is also a side of the octagon?
从一个正八边形中，通过连接随机选择的三个顶点形成一个三角形。三角形至少有一条边也是八边形的边的概率是多少？
stem
Correct Answer: D
Answer (D): For each side of the octagon, there are $6$ triangles containing that side. Because the $8$ triangles containing two adjacent sides of the octagon are counted twice, there are a total of $8\cdot6-8=40$ triangles sharing a side with the octagon. The total number of triangles that can be formed from the eight vertices is $\frac{8\cdot7\cdot6}{3!}=56$, so the probability is $\frac{40}{56}=\frac{5}{7}$.
答案（D）：对于八边形的每一条边，都有 $6$ 个包含该边的三角形。由于包含八边形两条相邻边的那 $8$ 个三角形被重复计算了两次，所以与八边形共享一条边的三角形总数为 $8\cdot6-8=40$。由这 8 个顶点能组成的三角形总数为 $\frac{8\cdot7\cdot6}{3!}=56$，因此所求概率为 $\frac{40}{56}=\frac{5}{7}$。
Q24
In the cube ABCDEFGH with opposite vertices C and E, J and I are the midpoints of edges $\overline{FB}$ and $\overline{HD}$, respectively. Let $R$ be the ratio of the area of the cross-section EJCI to the area of one of the faces of the cube. What is $R^2$?
在立方体ABCDEFGH中，相对顶点为C和E，J和I分别是边$\overline{FB}$和$\overline{HD}$的中点。设$R$为截面EJCI的面积与立方体一个面的面积之比。$R^2$是多少？
stem
Correct Answer: C
Answer (C): Let $s$ denote the length of an edge of the cube. Now $EJCI$ is a non-square rhombus whose area is $\frac{1}{2}EC\cdot JI$, because the area of a rhombus is half the product of the lengths of its diagonals. By the Pythagorean Theorem, $JI=FH=s\sqrt{2}$, and using the Pythagorean Theorem twice, $EC=s\sqrt{3}$. Thus $R=\frac{\frac{1}{2}EC\cdot JI}{s^2}=\frac{\frac{1}{2}(s\sqrt{3})(s\sqrt{2})}{s^2}=\frac{\sqrt{6}}{2}$ and $R^2=\frac{6}{4}=\frac{3}{2}$.
答案（C）：设 $s$ 表示立方体棱长。$EJCI$ 是一个非正方形的菱形，其面积为 $\frac{1}{2}EC\cdot JI$，因为菱形面积等于两条对角线长度乘积的一半。由勾股定理可得 $JI=FH=s\sqrt{2}$；再两次使用勾股定理可得 $EC=s\sqrt{3}$。因此 $R=\frac{\frac{1}{2}EC\cdot JI}{s^2}=\frac{\frac{1}{2}(s\sqrt{3})(s\sqrt{2})}{s^2}=\frac{\sqrt{6}}{2}$，并且 $R^2=\frac{6}{4}=\frac{3}{2}$。
Q25
How many perfect cubes lie between $2^8 + 1$ and $2^{18} + 1$, inclusive?
有多少个完全立方数位于$2^8 + 1$和$2^{18} + 1$之间（包含边界）？
Correct Answer: E
Answer (E): Note that $2^8+1=257$ and that $216=6^3<257<7^3=343$. Also note that $2^{18}=(2^6)^3=64^3$, so the perfect cube that is closest to and less than $2^{18}+1$ is $64^3$. Thus the numbers $7^3$, $8^3$, $\ldots$, $63^3$, $64^3$ are precisely the perfect cubes that lie between the two given numbers, and so there are $64-6=58$ of these perfect cubes.
答案（E）：注意到 $2^8+1=257$，并且 $216=6^3<257<7^3=343$。还注意到 $2^{18}=(2^6)^3=64^3$，因此小于且最接近 $2^{18}+1$ 的完全立方数是 $64^3$。所以 $7^3$、$8^3$、$\ldots$、$63^3$、$64^3$ 恰好是位于给定两个数之间的完全立方数，因此这些完全立方数共有 $64-6=58$ 个。