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AMC10 2000 A

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AMC10 · 2000 (A)

Q1
In the year 2001, the United States will host the International Mathematical Olympiad. Let $I$, $M$, and $O$ be distinct positive integers such that the product $I \cdot M \cdot O = 2001$. What is the largest possible value of the sum $I + M + O$?
在2001年，美国将举办国际数学奥林匹克竞赛。设$I$、$M$和$O$是互不相同的正整数，使得乘积$I \cdot M \cdot O = 2001$。$I + M + O$的最大可能值为多少？
Correct Answer: E
Answer (E): Factor 2001 into primes to get 2001 = 3 · 23 · 29. The largest possible sum of three distinct factors whose product is the one which combines the two largest prime factors, namely $I = 23 \cdot 29 = 667$, $M = 3$, and $O = 1$, so the largest possible sum is $1 + 3 + 667 = 671$.
答案（E）：将 2001 质因数分解得 $2001 = 3 \cdot 23 \cdot 29$。三个互不相同且乘积为 2001 的因数之和要尽可能大，应将最大的两个质因数相乘作为一个因数，即 $I = 23 \cdot 29 = 667$，另外两个因数为 $M = 3$ 和 $O = 1$，因此最大可能的和为 $1 + 3 + 667 = 671$。
Q2
$2000(2000^{2000}) =$
$2000(2000^{2000}) = $
Correct Answer: A
Answer (A): $2000(2000^{2000}) = (2000^1)(2000^{2000}) = 2000^{1+2000} = 2000^{2001}$. All the other options are greater than $2000^{2001}$.
答案（A）：$2000(2000^{2000}) = (2000^1)(2000^{2000}) = 2000^{1+2000} = 2000^{2001}$。 其他所有选项都大于 $2000^{2001}$。
Q3
Each day, Jenny ate 20% of the jellybeans that were in her jar at the beginning of that day. At the end of second day, 32 remained. How many jellybeans were in the jar originally?
每天，Jenny吃掉她罐子开始那天20%的果冻豆。到第二天结束时，还剩32颗。最初罐子里有多少果冻豆？
Correct Answer: B
Answer (B): Since Jenny ate $20\%$ of the jellybeans remaining each day, $80\%$ of the jellybeans are left at the end of each day. If $x$ is the number of jellybeans in the jar originally, then $(0.8)^2x=32$. Thus $x=50$.
答案（B）：因为珍妮每天吃掉剩余果冻豆的 $20\%$，所以每天结束时还剩下 $80\%$ 的果冻豆。若 $x$ 是罐子里最初的果冻豆数量，则 $(0.8)^2x=32$。因此 $x=50$。
Q4
Chandra pays an on-line service provider a fixed monthly fee plus an hourly charge for connect time. Her December bill was $12.48, but in January her bill was $17.54 because she used twice as much connect time as in December. What is the fixed monthly fee?
Chandra向在线服务提供商支付固定的月费加上按小时计费的连接时间费。她的12月账单是$12.48，但1月账单是$17.54，因为她1月的连接时间是12月的两倍。固定的月费是多少？
Correct Answer: D
Answer (D): Since Chandra paid extra \$5.06 in January, her December connect time must have cost her \$5.06. Therefore, her monthly fee is \$12.48 – \$5.06 = \$7.42.
答案（D）：由于 Chandra 在一月份多付了 \$5.06，她十二月份的连接时长费用一定是 \$5.06。因此，她的月费为 \$12.48 – \$5.06 = \$7.42。
Q5
Points $M$ and $N$ are the midpoints of sides $PA$ and $PB$ of $\triangle PAB$. As $P$ moves along a line that is parallel to side $AB$, how many of the four quantities listed below change? (a) the length of the segment $MN$ (b) the perimeter of $\triangle PAB$ (c) the area of $\triangle PAB$ (d) the area of trapezoid $ABNM$
点$M$和$N$分别是$ riangle PAB$的边$PA$和$PB$的中点。当$P$沿平行于边$AB$的直线移动时，下列四个量中有多少个会变化？(a) 线段$MN$的长度 (b) $\triangle PAB$的周长 (c) $\triangle PAB$的面积 (d) 梯形$ABNM$的面积
stem
Correct Answer: B
Answer (B): Since $\triangle ABP$ is similar to $\triangle MNP$ and $PM=\frac{1}{2}\cdot AP$, it follows that $MN=\frac{1}{2}\cdot AB$. Since the base $AB$ and the altitude to $AB$ of $\triangle ABP$ do not change, the area does not change. The altitude of the trapezoid is half that of the triangle, and the bases do not change as $P$ changes, so the area of the trapezoid does not change. Only the perimeter changes (reaching a minimum when $\triangle ABP$ is isosceles).
答案（B）：由于 $\triangle ABP$ 与 $\triangle MNP$ 相似，且 $PM=\frac{1}{2}\cdot AP$，因此 $MN=\frac{1}{2}\cdot AB$。由于 $\triangle ABP$ 的底边 $AB$ 以及对应于 $AB$ 的高不变，所以面积不变。梯形的高是三角形高的一半，并且当点 $P$ 变化时两条底边不变，因此梯形的面积也不变。只有周长会变化（当 $\triangle ABP$ 为等腰三角形时周长达到最小值）。
Q6
The Fibonacci sequence $1, 1, 2, 3, 5, 8, 13, 21, \dots$ starts with two 1s, and each term afterwards is the sum of its two predecessors. Which one of the ten digits is the last to appear in the units position of a number in the Fibonacci sequence?
斐波那契数列 $1, 1, 2, 3, 5, 8, 13, 21, \dots$ 以两个 $1$ 开头，此后每一项是前两项之和。十个数字中，哪个数字是最晚出现在斐波那契数列中某个数的个位上的？
Correct Answer: C
Answer (C): The sequence of units digits is 1, 1, 2, 3, 5, 8, 3, 1, 4, 5, 9, 4, 3, 7, 0, 7, 7, 4, 1, 5, 6, ... The digit 6 is the last of the ten digits to appear.
答案（C）：个位数字的序列为 1, 1, 2, 3, 5, 8, 3, 1, 4, 5, 9, 4, 3, 7, 0, 7, 7, 4, 1, 5, 6, … 数字 6 是十个数字中最后出现的一个。
Q7
In rectangle ABCD, AD = 1, P is on AB, and DB and DP trisect $\angle ADC$. What is the perimeter of $\Delta BDP$?
在矩形 ABCD 中，AD = 1，P 在 AB 上，DB 和 DP 三等分 $\angle ADC$。求 $\Delta BDP$ 的周长？
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Correct Answer: B
Answer (B): Both triangles $APD$ and $CBD$ are $30-60-90$ triangles. Thus $DP=\frac{2\sqrt{3}}{3}$ and $DB=2$. Since $\angle BDP=\angle PDB$, it follows that $PB=PD=\frac{2\sqrt{3}}{3}$. Hence the perimeter of $\triangle BDP$ is $\frac{2\sqrt{3}}{3}+\frac{2\sqrt{3}}{3}+2=2+\frac{4\sqrt{3}}{3}$.
答案（B）：三角形 $APD$ 和 $CBD$ 都是 $30-60-90$ 三角形。因此 $DP=\frac{2\sqrt{3}}{3}$，且 $DB=2$。由于 $\angle BDP=\angle PDB$，可得 $PB=PD=\frac{2\sqrt{3}}{3}$。因此 $\triangle BDP$ 的周长为 $\frac{2\sqrt{3}}{3}+\frac{2\sqrt{3}}{3}+2=2+\frac{4\sqrt{3}}{3}$。
Q8
At Olympic High School, 2/5 of the freshmen and 4/5 of the sophomores took the AMC 10. Given that the number of freshmen and sophomore contestants was the same, which of the following must be true?
在奥林匹克高中，$2/5$ 的新生和 $4/5$ 的二年级生参加了 AMC 10。已知新生和二年级参赛人数相同，下列哪项一定正确？
Correct Answer: D
Answer (D): Let $f$ and $s$ represent the numbers of freshmen and sophomores at the school, respectively. According to the given condition, $(2/5)f=(4/5)s$. Thus, $f=2s$. That is, there are twice as many freshmen as sophomores.
答案（D）：设 $f$ 和 $s$ 分别表示学校中新生和二年级学生的人数。根据给定条件，$(2/5)f=(4/5)s$。因此，$f=2s$。也就是说，新生人数是二年级学生人数的两倍。
Q9
If $|x - 2| = p$, where $x < 2$, then $x - p =$
若 $|x - 2| = p$，其中 $x < 2$，则 $x - p =$
Correct Answer: C
Answer (C): Since $x<2$, it follows that $\left|x-2\right|=2-x$. If $2-x=p$, then $x=2-p$. Thus $x-p=2-2p$.
答案（C）：由于 $x<2$，所以 $\left|x-2\right|=2-x$。若 $2-x=p$，则 $x=2-p$。因此 $x-p=2-2p$。
Q10
The sides of a triangle with positive area have lengths 4, 6, and x. The sides of a second triangle with positive area have lengths 4, 6, and y. What is the smallest positive number that is not a possible value of |x - y|?
一个具有正面积的三角形的边长为 4、6 和 $x$。第二个具有正面积的三角形的边长为 4、6 和 $y$。最小的不可能是 $|x - y|$ 值的正数是多少？
Correct Answer: D
Answer (D): By the Triangle Inequality, each of $x$ and $y$ can be any number strictly between 2 and 10, so $0 \le |x-y| < 8$. Therefore, the smallest positive number that is not a possible value of $|x-y|$ is $10-2=8$.
答案（D）：根据三角不等式，$x$ 和 $y$ 都可以是严格介于 2 和 10 之间的任意数，因此 $0 \le |x-y| < 8$。所以，不可能作为 $|x-y|$ 的取值的最小正数是 $10-2=8$。
Q11
Two different prime numbers between 4 and 18 are chosen. When their sum is subtracted from their product, which of the following number could be obtained?
从4和18之间选择两个不同的质数。当它们的和从它们的积中减去时，下列哪个数可能得到？
Correct Answer: C
Answer (C): There are five prime numbers between 4 and 18: 5, 7, 11, 13, and 17. Hence the product of any two of these is odd and the sum is even. Because $xy-(x+y)=(x-1)(y-1)-1$ increases as either $x$ or $y$ increases (since both $x$ and $y$ are bigger than 1), the answer must be an odd number that is no smaller than $23=5\cdot 7-(5+7)$ and no larger than $191=13\cdot 17-(13+17)$. The only possibility among the options is 119, and indeed $119=11\cdot 13-(11+13)$.
答案（C）：在 4 和 18 之间有五个质数：5、7、11、13 和 17。因此，其中任意两个的乘积为奇数，而它们的和为偶数。因为 $xy-(x+y)=(x-1)(y-1)-1$ 会随着 $x$ 或 $y$ 的增大而增大（由于 $x$ 和 $y$ 都大于 1），所以答案必须是一个不小于 $23=5\cdot 7-(5+7)$ 且不大于 $191=13\cdot 17-(13+17)$ 的奇数。选项中唯一可能的是 119，且确实有 $119=11\cdot 13-(11+13)$。
Q12
Figure 0, 1, 2, and 3 consist of 1, 5, 13, and 25 nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?
图0、1、2和3分别由1、5、13和25个不重叠的单位正方形组成。如果图案继续，如何多不重叠的单位正方形在图100中？
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Correct Answer: C
Answer (C): Calculating the number of squares in the first few figures uncovers a pattern. Figure 0 has $2(0)+1=2(0^2)+1$ squares, figure 1 has $2(1)+3=2(1^2)+3$ squares, figure 2 has $2(1+3)+5=2(2^2)+5$ squares, and figure 3 has $2(1+3+5)+7=2(3^2)+7$ squares. In general, the number of unit squares in figure $n$ is $$2(1+3+5+\cdots+(2n-1))+2n+1=2(n^2)+2n+1.$$ Therefore, the figure 100 has $2(100^2)+2\cdot 100+1=20201$.
答案（C）：计算前几个图形中的小正方形数量可以发现规律。图形0有 $2(0)+1=2(0^2)+1$ 个正方形，图形1有 $2(1)+3=2(1^2)+3$ 个正方形，图形2有 $2(1+3)+5=2(2^2)+5$ 个正方形，图形3有 $2(1+3+5)+7=2(3^2)+7$ 个正方形。一般地，第 $n$ 个图形中的单位正方形数为 $$2(1+3+5+\cdots+(2n-1))+2n+1=2(n^2)+2n+1.$$ 因此，第100个图形有 $2(100^2)+2\cdot 100+1=20201$ 个正方形。
Q13
There are 5 yellow pegs, 4 red pegs, 3 green pegs, 2 blue pegs, and 1 orange peg to be placed on a triangular peg board. In how many ways can the pegs be placed so that no (horizontal) row or (vertical) column contains two pegs of the same color?
有一个三角形钉板，要放置5个黄色钉、4个红色钉、3个绿色钉、2个蓝色钉和1个橙色钉。有多少种方法可以放置这些钉，使得没有（水平）行或（垂直）列包含相同颜色的两个钉？
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Correct Answer: B
Answer (B): To avoid having two yellow pegs in the same row or column, there must be exactly one yellow peg in each row and in each column. Hence, starting at the top of the array, the peg in the first row must be yellow, the second peg of the second row must be yellow, the third peg of the third row must be yellow, etc. To avoid having two red pegs in some row, there must be a red peg in each of rows 2, 3, 4, and 5. The red pegs must be in the first position of the second row, the second position of the third row, etc. Continuation yields exactly one ordering that meets the requirements, as shown.
答案（B）：为避免在同一行或同一列出现两个黄色棋子，每一行和每一列必须恰好有一个黄色棋子。因此，从阵列顶部开始，第一行中的棋子必须是黄色，第二行的第二个棋子必须是黄色，第三行的第三个棋子必须是黄色，依此类推。为避免在某一行中出现两个红色棋子，第2、3、4、5行中每一行都必须有一个红色棋子。红色棋子必须位于第二行的第一个位置、第三行的第二个位置，等等。继续推下去可得到恰好一种满足要求的排列，如图所示。
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Q14
Mrs. Walter gave an exam in a mathematics class of five students. She entered the scores in random order into a spreadsheet, which recalculated the class average after each score was entered. Mrs. Walter noticed that after each score was entered, the average was always an integer. The scores (listed in ascending order) were 71, 76, 80, 82, and 91. What was the last score Mrs. Walter entered?
沃尔特夫人给一个五名学生的数学班考试。她随机顺序将分数输入电子表格，每次输入一个分数后电子表格重新计算班级平均分。沃尔特夫人注意到，每次输入分数后，平均分总是整数。分数（按升序排列）是71、76、80、82和91。沃尔特夫人最后输入的分数是什么？
Correct Answer: C
Answer (C): Note that the integer average condition means that the sum of the scores of the first $n$ students is a multiple of $n$. The scores of the first two students must be both even or both odd, and the sum of the scores of the first three students must be divisible by $3$. The remainders when $71$, $76$, $80$, $82$, and $91$ are divided by $3$ are $2$, $1$, $2$, $1$, and $1$, respectively. Thus the only sum of three scores divisible by $3$ is $76+82+91=249$, so the first two scores entered are $76$ and $82$ (in some order), and the third score is $91$. Since $249$ is $1$ larger than a multiple of $4$, the fourth score must be $3$ larger than a multiple of $4$, and the only possible is $71$, leaving $80$ as the score of the fifth student.
答案（C）：注意，整数平均数条件意味着前 $n$ 个学生的分数之和是 $n$ 的倍数。前两名学生的分数必须同为偶数或同为奇数，并且前三名学生分数之和必须能被 $3$ 整除。$71$、$76$、$80$、$82$ 和 $91$ 分别除以 $3$ 的余数是 $2$、$1$、$2$、$1$ 和 $1$。因此，三个分数之和能被 $3$ 整除的唯一组合是 $76+82+91=249$，所以前两次输入的分数是 $76$ 和 $82$（顺序不定），第三个分数是 $91$。由于 $249$ 比某个 $4$ 的倍数大 $1$，第四个分数必须比某个 $4$ 的倍数大 $3$，唯一可能是 $71$，于是第五名学生的分数为 $80$。
Q15
Two non-zero real numbers, a and b, satisfy $ab = a - b$. Find a possible value of $\frac{a}{b} + \frac{b}{a} - ab$.
两个非零实数$a$和$b$满足$ab=a-b$。求$\frac{a}{b}+\frac{b}{a}-ab$的一个可能值。
Correct Answer: E
Answer (E): Find the common denominator and replace the $ab$ in the numerator with $a-b$ to get $\frac{a}{b}+\frac{b}{a}-ab=\frac{a^2+b^2-(ab)^2}{ab}$ $=\frac{a^2+b^2-(a-b)^2}{ab}$ $=\frac{a^2+b^2-(a^2-2ab+b^2)}{ab}$ $=\frac{2ab}{ab}=2.$
答案（E）：求出公分母，并将分子中的 $ab$ 用 $a-b$ 替换，得到 $\frac{a}{b}+\frac{b}{a}-ab=\frac{a^2+b^2-(ab)^2}{ab}$ $=\frac{a^2+b^2-(a-b)^2}{ab}$ $=\frac{a^2+b^2-(a^2-2ab+b^2)}{ab}$ $=\frac{2ab}{ab}=2.$
Q16
The diagram shows 28 lattice points, each one unit from its nearest neighbors. Segment AB meets segment CD at E. Find the length of segment AE.
图中显示了28个格点，每个点与其最近邻点相距一单位。线段AB与线段CD在点E相交。求线段AE的长度。
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Correct Answer: B
Answer (B): Extend $\overline{DC}$ to $F$. Triangle $FAE$ and $DBE$ are similar with ratio $5:4$. Thus $AE = 5\cdot AB/9$, $AB = \sqrt{3^2 + 6^2} = \sqrt{45} = 3\sqrt{5}$, and $AE = 5(3\sqrt{5})/9 = 5\sqrt{5}/3$.
答案（B）：将$\overline{DC}$延长至点$F$。三角形$FAE$与$DBE$相似，相似比为$5:4$。因此$AE = 5\cdot AB/9$，$AB = \sqrt{3^2 + 6^2} = \sqrt{45} = 3\sqrt{5}$，且$AE = 5(3\sqrt{5})/9 = 5\sqrt{5}/3$。
Q17
Boris has an incredible coin changing machine. When he puts in a quarter, it returns five nickels; when he puts in a nickel, it returns five pennies; and when he puts in a penny, it returns five quarters. Boris starts with just one penny. Which of the following amounts could Boris have after using the machine repeatedly?
鲍里斯有一个神奇的找零机。当他投入一个25美分硬币时，机器返回五个5美分硬币；投入一个5美分硬币时，返回五个1美分硬币；投入一个1美分硬币时，返回五个25美分硬币。鲍里斯开始只有一个1美分硬币。经过反复使用机器后，他可能拥有下列哪种金额？
Correct Answer: D
Answer (D): Neither of the exchanges quarter → five nickels nor nickel → five pennies changes the total value of Boris’s coins. The exchange penny → five quarters increase the total value of Boris’s coins by \$1.24. Hence, Boris must have \$0.01 + \$1.24$n$ after $n$ uses of the last exchange. Only option D is of this form: $745 = 1 + 124 \cdot 6$. In cents, option A is 115 more than a multiple of 124, B is 17 more than a multiple of 124, C is 10 more than a multiple of 124, and E is 39 more than a multiple of 124.
答案（D）：无论是把 1 枚 25 美分硬币换成 5 枚 5 美分硬币，还是把 1 枚 5 美分硬币换成 5 枚 1 美分硬币，都不会改变 Boris 硬币的总价值。把 1 枚 1 美分硬币换成 5 枚 25 美分硬币，会使 Boris 硬币的总价值增加 \$1.24。因此，进行最后一种交换 $n$ 次后，Boris 的总金额应为 \$0.01 + \$1.24$n$。只有选项 D 具有这种形式：$745 = 1 + 124 \cdot 6$。按“美分”为单位看，选项 A 比某个 124 的倍数多 115，B 多 17，C 多 10，E 多 39。
Q18
Charlyn walks completely around the boundary of a square whose sides are each 5 km long. From any point on her path she can see exactly 1 km horizontally in all directions. What is the area of the region consisting of all points Charlyn can see during her walk, expressed in square kilometers and rounded to the nearest whole number?
查琳完全绕着一个边长各为5千米的正方形的边界走一圈。从她路径上的任意一点，她都能在所有方向水平看清1千米。求查琳在行走过程中能看到的区域的面积（以平方千米为单位，四舍五入到最接近的整数）。
Correct Answer: C
Answer (C): At any point on Charlyn’s walk, she can see all the points inside a circle of radius $1$ km. The portion of the viewable region inside the square consists of the interior of the square except for a smaller square with side length $3$ km. This portion of the viewable region has area $(25-9)\ \text{km}^2$. The portion of the viewable region outside the square consists of four rectangles, each $5$ km by $1$ km, and four quarter-circles, each with a radius of $1$ km. This portion of the viewable region has area $4\left(5+\frac{\pi}{4}\right)=(20+\pi)\ \text{km}^2$. The area of the entire viewable region is $36+\pi\approx 30\ \text{km}^2$.
答案（C）：在 Charlyn 行走路径上的任意一点，她都能看到以该点为圆心、半径为 $1$ km 的圆内所有点。正方形内部的可视区域等于正方形内部去掉一个边长为 $3$ km 的小正方形后剩余的部分。该部分面积为 $(25-9)\ \text{km}^2$。正方形外部的可视区域由四个长方形（每个为 $5$ km 乘 $1$ km）以及四个四分之一圆（每个半径为 $1$ km）组成。该部分面积为 $4\left(5+\frac{\pi}{4}\right)=(20+\pi)\ \text{km}^2$。整个可视区域的面积为 $36+\pi\approx 30\ \text{km}^2$。
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Q19
Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the triangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is $m$ times the area of the square. The ratio of the area of the other small right triangle to the area of the square is
在一个直角三角形的斜边上一点，作与两条直角边平行的直线，将三角形分割成一个正方形和两个更小的直角三角形。其中一个小的直角三角形的面积是正方形面积的$m$倍。另一个小直角三角形面积与正方形面积的比值为
Correct Answer: D
Answer (D): Without loss of generality, let the side of the square have length $1$ unit and let the area of triangle $ADF$ be $m$. Let $AD=r$ and $EC=s$. Because triangles $ADF$ and $FEC$ are similar, $\frac{s}{1}=\frac{1}{r}$. Since $\frac{1}{2}r=m$, the area of triangle $FEC$ is $\frac{1}{2}s=\frac{1}{2r}=\frac{1}{4m}$.
答案（D）：不失一般性，设正方形的边长为 $1$ 个单位，且三角形 $ADF$ 的面积为 $m$。设 $AD=r$，$EC=s$。由于三角形 $ADF$ 与 $FEC$ 相似，所以 $\frac{s}{1}=\frac{1}{r}$。又因为 $\frac{1}{2}r=m$，所以三角形 $FEC$ 的面积为 $\frac{1}{2}s=\frac{1}{2r}=\frac{1}{4m}$。
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Q20
Let $A$, $M$, and $C$ be nonnegative integers such that $A + M + C = 10$. What is the maximum value of $A \cdot M \cdot C + A \cdot M + M \cdot C + C \cdot A$?
设$A$、$M$、$C$是非负整数且$A+M+C=10$。求$A\cdot M\cdot C+A\cdot M+M\cdot C+C\cdot A$的最大值。
Correct Answer: C
Answer (C): Note that $AMC + AM + MC + CA = (A+1)(M+1)(C+1) - (A+M+C) - 1 = pqr - 11,$ where $p$, $q$, and $r$ are positive integers whose sum is $13$. A case-by-case analysis shows that $pqr$ is largest when two of the numbers $p$, $q$, $r$ are $4$ and the third is $5$. Thus the answer is $4\cdot 4\cdot 5 - 11 = 69$.
答案（C）：注意到 $AMC + AM + MC + CA = (A+1)(M+1)(C+1) - (A+M+C) - 1 = pqr - 11,$ 其中 $p$、$q$、$r$ 是和为 $13$ 的正整数。逐一分类分析可知，当 $p,q,r$ 中有两个为 $4$、另一个为 $5$ 时，$pqr$ 取得最大值。因此答案为 $4\cdot 4\cdot 5 - 11 = 69$。
Q21
If all alligators are ferocious creatures and some creepy crawlers are alligators, which statement(s) must be true? I. All alligators are creepy crawlers. II. Some ferocious creatures are creepy crawlers. III. Some alligators are not creepy crawlers.
如果所有短吻鳄都是凶猛的生物，并且一些诡异的爬行动物是短吻鳄，那么以下哪项（些）声明必须为真？I. 所有短吻鳄都是诡异的爬行动物。II. 一些凶猛的生物是诡异的爬行动物。III. 一些短吻鳄不是诡异的爬行动物。
Correct Answer: B
Answer (B): From the conditions we can conclude that some creepy crawlers are ferocious (since some are alligators). Hence, there are some ferocious creatures that are creepy crawlers, and thus II must be true. The diagram below shows that the only conclusion that can be drawn is existence of an animal in the region with the dot. Thus, neither I nor III follows from the given conditions.
答案（B）：根据条件我们可以推断，一些“爬行怪物”是凶猛的（因为其中一些是短吻鳄）。因此，存在一些既凶猛又属于“爬行怪物”的生物，所以 II 必然为真。下图表明，唯一能得出的结论是在带有点的区域中存在一种动物。因此，I 和 III 都不能由给定条件推出。
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Q22
One morning each member of Angela’s family drank an 8-ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?
一个早晨，Angela 全家每个人都喝了一杯 8 盎司的咖啡与牛奶混合物。每个杯子中咖啡和牛奶的量各不相同，但都不为零。Angela 喝了总牛奶量的四分之一和总咖啡量的六分之一。全家有多少人？
Correct Answer: C
Answer (C): Suppose that the whole family drank $x$ cups of milk and $y$ cups of coffee. Let $n$ denote the number of people in the family. The information given implies that $x/4 + y/6 = (x + y)/n$. This leads to $3x(n-4)=2y(6-n).$ Since $x$ and $y$ are positive, the only positive integer $n$ for which both sides have the same sign is $n=5.$
答案（C）：设全家一共喝了 $x$ 杯牛奶和 $y$ 杯咖啡。令 $n$ 表示家庭人数。由已知信息可得 $x/4 + y/6 = (x + y)/n$。于是推出 $3x(n-4)=2y(6-n)。$ 由于 $x$ 和 $y$ 均为正数，使得等式两边同号的唯一正整数 $n$ 是 $n=5。$
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Q23
When the mean, median, and mode of the list $10, 2, 5, 2, 4, 2, x$ are arranged in increasing order, they form a non-constant arithmetic progression. What is the sum of all possible real value of $x$?
当列表 $10, 2, 5, 2, 4, 2, x$ 的均值、中位数和众数按升序排列时，它们形成一个非恒等的等差数列。所有可能的实数 $x$ 的和是多少？
Correct Answer: E
Answer (E): If $x$ were less than or equal to $2$, then $2$ would be both the median and the mode of the list. Thus $x>2$. Consider the two cases $2<x<4$, and $x\ge 4$. Case 1: If $2<x<4$, then $2$ is the mode, $x$ is the median, and $\frac{25+x}{7}$ is the mean, which must equal $2-(x-2)$, $\frac{x+2}{2}$, or $x+(x-2)$, depending on the size of the mean relative to $2$ and $x$. These give $x=\frac{3}{8}$, $x=\frac{36}{5}$, and $x=3$, of which $x=3$ is the only value between $2$ and $4$. Case 2: If $x\ge 4$, then $4$ is the median, $2$ is the mode, and $\frac{25+x}{7}$ is the mean, which must be $0,3$, or $6$. Thus $x=-25$, $-4$, or $17$, of which $17$ is the only one of these values greater than or equal to $4$. Thus the $x$-value sum to $3+17=20$.
答案（E）：如果 $x\le 2$，那么 $2$ 将同时是该列表的中位数和众数。因此 $x>2$。考虑两种情况：$2<x<4$，以及 $x\ge 4$。 情况 1：若 $2<x<4$，则 $2$ 是众数，$x$ 是中位数，且 $\frac{25+x}{7}$ 是平均数。根据平均数相对于 $2$ 和 $x$ 的大小，它必须等于 $2-(x-2)$、$\frac{x+2}{2}$ 或 $x+(x-2)$。解得 $x=\frac{3}{8}$、$x=\frac{36}{5}$、$x=3$，其中只有 $x=3$ 落在 $2$ 与 $4$ 之间。 情况 2：若 $x\ge 4$，则 $4$ 是中位数，$2$ 是众数，且 $\frac{25+x}{7}$ 是平均数，它必须为 $0$、$3$ 或 $6$。因此 $x=-25$、$-4$ 或 $17$，其中只有 $17$ 大于或等于 $4$。 因此 $x$ 的取值之和为 $3+17=20$。
Q24
Let $f$ be a function for which $f(x/3) = x^2 + x + 1$. Find the sum of all values of $z$ for which $f(3z) = 7$.
设 $f$ 是一个函数，使得 $f(x/3) = x^2 + x + 1$。求所有满足 $f(3z) = 7$ 的 $z$ 的值之和。
Correct Answer: B
Answer (B): Let $x=9z$. Then $f(3z)=f(9z/3)=f(3z)=(9z)^2+9z+1=7$. Simplifying and solving the equation for $z$ yields $81z^2+9z-6=0$, so $3(3z+1)(9z-2)=0$. Thus $z=-1/3$ or $z=2/9$. The sum of these values is $-1/9$. Note. The answer can also be obtained by using the sum-of-roots formula on $81z^2+9z-6=0$. The sum of the roots is $-9/81=-1/9$.
答案（B）：令 $x=9z$。则 $f(3z)=f(9z/3)=f(3z)=(9z)^2+9z+1=7$。化简并解关于 $z$ 的方程得 $81z^2+9z-6=0$，因此 $3(3z+1)(9z-2)=0$。所以 $z=-1/3$ 或 $z=2/9$。这些值的和为 $-1/9$。 注：也可以对 $81z^2+9z-6=0$ 使用根和公式得到答案。两根之和为 $-9/81=-1/9$。
Q25
In year $N$, the 300th day of the year is a Tuesday. In year $N + 1$, the 200th day is also a Tuesday. On what day of the week did the 100th day of year $N - 1$ occur?
在年份 $N$，当年的第 300 天是星期二。在年份 $N + 1$，第 200 天也是星期二。年份 $N - 1$ 的第 100 天是星期几？
Correct Answer: A
Note that, if a Tuesday is d days after a Tuesday, then d is a multiple of 7... It follows that year $N −1$ is not a leap year. Therefore, the 100th day of year $N −1$ precedes the given Tuesday in year N by 365−100+300 = 565 days, and therefore is a Thursday, since 565 = 7 · 80 + 5 is 5 larger than a multiple of 7.
注意，如果一个星期二是另一个星期二后 $d$ 天，则 $d$ 是 7 的倍数……由此可知，年份 $N −1$ 不是闰年。因此，年份 $N −1$ 的第 100 天比年份 $N$ 的给定星期二早 $365−100+300 = 565$ 天，因此是星期四，因为 $565 = 7 · 80 + 5$ 比 7 的倍数大 5 天。