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AMC10 2005 B

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AMC10 · 2005 (B)

Q1
A scout troop buys 1000 candy bars at a price of five for \$2. They sell all the candy bars at a price of two for \$1. What was their profit, in dollars?
一个童子军部队以5条2美元的价格购买了1000条糖果。他们以2条1美元的价格卖出了所有糖果。他们的利润是多少美元？
Correct Answer: A
The scouts bought $1000/5 = 200$ groups of 5 candy bars at a total cost of $200 \cdot 2 = 400$ dollars. They sold $1000/2 = 500$ groups of 2 candy bars for a total of $500 \cdot 1 = 500$ dollars. Their profit was $500 - 400 = 100$.
童子军买了$1000/5 = 200$组5条糖果，总成本$200 \cdot 2 = 400$美元。他们卖了$1000/2 = 500$组2条糖果，总收入$500 \cdot 1 = 500$美元。利润是$500 - 400 = 100$美元。
Q2
A positive number $x$ has the property that $x\%$ of $x$ is 4. What is $x$?
一个正数$x$具有这样的性质：$x$的$x\%$是4。$x$是多少？
Correct Answer: D
$\frac{x}{100} \cdot x = 4$, so $x^2 = 400$. Because $x > 0$, it follows that $x = 20$.
$\frac{x}{100} \cdot x = 4$，所以$x^2 = 400$。因为$x > 0$，所以$x = 20$。
Q3
A gallon of paint is used to paint a room. One third of the paint is used on the first day. On the second day, one third of the remaining paint is used. What fraction of the original amount of paint is available to use on the third day?
一加仑油漆用于粉刷一个房间。第一天用了三分之一的油漆。第二天，用了剩余油漆的三分之一。第三天可用的油漆占原始油漆的比例是多少？
Correct Answer: D
(D) After the first day, $1-\frac{1}{3}=\frac{2}{3}$ of the paint remains. On the second day, $\frac{1}{3}\cdot\frac{2}{3}=\frac{2}{9}$ of the paint is used. So for the third day $1-\frac{1}{3}-\frac{2}{9}=\frac{4}{9}$ of the original gallon of paint is available.
（D）第一天之后，剩余的油漆为 $1-\frac{1}{3}=\frac{2}{3}$。第二天，使用的油漆为 $\frac{1}{3}\cdot\frac{2}{3}=\frac{2}{9}$。因此到第三天，可用的原来一加仑油漆为 $1-\frac{1}{3}-\frac{2}{9}=\frac{4}{9}$。
Q4
For real numbers $a$ and $b$, define $a \diamond b = \sqrt{a^2 + b^2}$. What is the value of $(5 \diamond 12) \diamond ((-12) \diamond (-5))$?
对于实数$a$和$b$，定义$a \diamond b = \sqrt{a^2 + b^2}$。求$(5 \diamond 12) \diamond ((-12) \diamond (-5))$的值。
Correct Answer: D
(D) It follows from the definition that $$(5\diamond 12)\diamond((-12)\diamond(-5))=\sqrt{5^2+12^2}\diamond\sqrt{(-12)^2+(-5)^2}$$ $$=13\diamond 13=\sqrt{13^2+13^2}=13\sqrt{2}.$$
（D）由定义可知 $$(5\diamond 12)\diamond((-12)\diamond(-5))=\sqrt{5^2+12^2}\diamond\sqrt{(-12)^2+(-5)^2}$$ $$=13\diamond 13=\sqrt{13^2+13^2}=13\sqrt{2}。$$
Q5
Brianna is using part of the money she earned on her weekend job to buy several equally-priced CDs. She used one fifth of her money to buy one third of the CDs. What fraction of her money will she have left after she buys all the CDs?
Brianna用她周末工作赚的部分钱买几张同样价格的CD。她用五分之一的钱买了三分之一的CD。她买完所有CD后，还剩多少比例的钱？
Correct Answer: C
The number of CDs that Brianna will finally buy is three times the number she has already bought. The fraction of her money that will be required for all the purchases is $(3)(\frac{1}{5}) = \frac{3}{5}$. The fraction she will have left is $1 - \frac{3}{5} = \frac{2}{5}$.
Brianna最终买的CD数量是她已买数量的三倍。所有购买所需钱的比例是$(3)(\frac{1}{5}) = \frac{3}{5}$。剩余比例是$1 - \frac{3}{5} = \frac{2}{5}$。
Q6
At the beginning of the school year, Lisa's goal was to earn an A on at least 80% of her 50 quizzes for the year. She earned an A on 22 of the first 30 quizzes. If she is to achieve her goal, on at most how many of the remaining quizzes can she earn a grade lower than an A?
在学年初，Lisa的目标是在全年50次小测验中至少获得80%的A。她在前30次小测验中获得了22次A。如果她要实现目标，在剩余的小测验中最多能有多少次获得低于A的成绩？
Correct Answer: B
To earn an A on at least 80% of her quizzes, Lisa needs to receive an A on at least $(0.8)(50) = 40$ quizzes. Thus she must earn an A on at least $40 - 22 = 18$ of the remaining 20. So she can earn a grade lower than an A on at most 2 of the remaining quizzes.
要获得至少80%的A，Lisa需要至少获得$(0.8)(50) = 40$次A。因此，她必须在剩余20次中至少获得$40 - 22 = 18$次A。所以，她最多能在剩余小测验中获得2次低于A的成绩。
Q7
A circle is inscribed in a square, then a square is inscribed in this circle, and finally, a circle is inscribed in this square. What is the ratio of the area of the smaller circle to the area of the larger square?
一个圆内接于一个正方形，然后一个正方形内接于这个圆，最后一个圆内接于这个正方形。小圆的面积与大正方形的面积之比是多少？
stem
Correct Answer: B
(B) Let the radius of the smaller circle be \(r\). Then the side length of the smaller square is \(2r\). The radius of the larger circle is half the length of the diagonal of the smaller square, so it is \(\sqrt{2}r\). Hence the larger square has sides of length \(2\sqrt{2}r\). The ratio of the area of the smaller circle to the area of the larger square is therefore \[ \frac{\pi r^2}{(2\sqrt{2}r)^2}=\frac{\pi}{8}. \]
（B）设较小圆的半径为 \(r\)。则较小正方形的边长为 \(2r\)。较大圆的半径等于较小正方形对角线长度的一半，因此为 \(\sqrt{2}r\)。因此较大正方形的边长为 \(2\sqrt{2}r\)。所以较小圆面积与较大正方形面积之比为 \[ \frac{\pi r^2}{(2\sqrt{2}r)^2}=\frac{\pi}{8}. \]
Q8
An 8-foot by 10-foot floor is tiled with square tiles of size 1 foot by 1 foot. Each tile has a pattern consisting of four white quarter circles of radius 1/2 foot centered at each corner of the tile. The remaining portion of the tile is shaded. How many square feet of the floor are shaded?
一个8英尺乘10英尺的地板用1英尺乘1英尺的正方形瓷砖铺成。每块瓷砖上有四个以瓷砖每个角为中心、半径1/2英尺的白色四分之一圆。瓷砖的其余部分是遮荫的。地板的遮荫面积有多少平方英尺？
stem
Correct Answer: A
(A) The four white quarter circles in each tile have the same area as a whole circle of radius \$1/2\$, that is, \$\pi(1/2)^2=\pi/4\$ square feet. So the area of the shaded portion of each tile is \$1-\pi/4\$ square feet. Since there are \$8\cdot10=80\$ tiles in the entire floor, the area of the total shaded region in square feet is \[ 80\left(1-\frac{\pi}{4}\right)=80-20\pi. \]
（A）每块瓷砖上的四个白色四分之一圆的总面积等于半径为 \$1/2\$ 的整圆面积，即 \$\pi(1/2)^2=\pi/4\$ 平方英尺。因此，每块瓷砖阴影部分的面积为 \$1-\pi/4\$ 平方英尺。由于整个地面共有 \$8\cdot10=80\$ 块瓷砖，所以阴影总面积（单位：平方英尺）为 \[ 80\left(1-\frac{\pi}{4}\right)=80-20\pi. \]
Q9
One fair die has faces 1, 1, 2, 2, 3, 3 and another has faces 4, 4, 5, 5, 6, 6. The dice are rolled and the numbers on the top faces are added. What is the probability that the sum will be odd?
一个公平骰子有面1, 1, 2, 2, 3, 3，另一个有面4, 4, 5, 5, 6, 6。掷这两个骰子，将上面面的数字相加。和为奇数的概率是多少？
Correct Answer: D
An odd sum requires either that the first die is even and the second is odd or that the first die is odd and the second is even. The probability is $\frac{1}{3} \cdot \frac{1}{3} + \frac{2}{3} \cdot \frac{2}{3} = \frac{1}{9} + \frac{4}{9} = \frac{5}{9}$.
和为奇数需要第一骰子为偶数第二为奇数，或第一为奇数第二为偶数。概率为$\frac{1}{3} \cdot \frac{1}{3} + \frac{2}{3} \cdot \frac{2}{3} = \frac{1}{9} + \frac{4}{9} = \frac{5}{9}$。
Q10
In $\triangle ABC$, we have $AC = BC = 7$ and $AB = 2$. Suppose that $D$ is a point on line $AB$ such that $B$ lies between $A$ and $D$ and $CD = 8$. What is $BD$?
在$\triangle ABC$中，$AC = BC = 7$且$AB = 2$。设$D$是线$AB$上的一点，使得$B$在$A$和$D$之间，且$CD = 8$。$BD$是多少？
Correct Answer: A
Let $\overline{CH}$ be an altitude of $\triangle ABC$. Applying the Pythagorean Theorem to $\triangle CHB$ and to $\triangle CHD$ produces $8^2 - (BD + 1)^2 = CH^2 = 7^2 - 1^2 = 48$, so $(BD + 1)^2 = 16$. Thus $BD = 3$.
设$\overline{CH}$为$\triangle ABC$的高。对$\triangle CHB$和$\triangle CHD$应用勾股定理，得$8^2 - (BD + 1)^2 = CH^2 = 7^2 - 1^2 = 48$，所以$(BD + 1)^2 = 16$。因此$BD = 3$。
solution
Q11
The first term of a sequence is 2005. Each succeeding term is the sum of the cubes of the digits of the previous term. What is the 2005th term of the sequence?
一个数列的第一项是 2005。每项之后的项是前一项各位数字立方的和。这个数列的第 2005 项是多少？
Correct Answer: E
The sequence begins 2005, 133, 55, 250, 133, ... Thus after the initial term 2005, the sequence repeats the cycle 133, 55, 250. Because $2005 = 1 + 3 \cdot 668$, the 2005th term is the same as the last term of the repeating cycle, 250.
数列开始是 2005, 133, 55, 250, 133, ... 因此在初始项 2005 之后，数列重复循环 133, 55, 250。因为 $2005 = 1 + 3 \cdot 668$，第 2005 项与重复循环的最后一项相同，即 250。
Q12
Twelve fair dice are rolled. What is the probability that the product of the numbers on the top faces is prime?
掷 12 个公平的骰子。顶面数字的乘积是质数的概率是多少？
Correct Answer: E
Exactly one die must have a prime face on top, and the other eleven must have 1’s. The prime die can be any one of the twelve, and the prime can be 2, 3, or 5. Thus the probability of a prime face on any one die is 1/2, and the probability of a prime product is $12 \cdot (\frac{1}{2}) \cdot (\frac{1}{6})^{11} = (\frac{1}{6})^{10}$.
必须恰好有一个骰子顶面是质数，其他十一个必须是 1。质数骰子可以是 12 个中的任意一个，质数可以是 2, 3 或 5。因此单个骰子出现质数的概率是 $1/2$，质数乘积的概率是 $12 \cdot (\frac{1}{2}) \cdot (\frac{1}{6})^{11} = (\frac{1}{6})^{10}$。
Q13
How many numbers between 1 and 2005 are integer multiples of 3 or 4 but not 12?
1 到 2005 之间有多少个数是 3 或 4 的整数倍但不是 12 的整数倍？
Correct Answer: C
Between 1 and 2005, there are 668 multiples of 3, 501 multiples of 4, and 167 multiples of 12. So there are $(668 -167) + (501 -167) = 835$ numbers between 1 and 2005 that are integer multiples of 3 or of 4 but not of 12.
在 1 到 2005 之间，有 668 个 3 的倍数，501 个 4 的倍数，167 个 12 的倍数。因此有 $(668 -167) + (501 -167) = 835$ 个数是 1 到 2005 之间 3 或 4 的整数倍但不是 12 的整数倍。
Q14
Equilateral $\triangle ABC$ has side length 2, $M$ is the midpoint of $\overline{AC}$, and $C$ is the midpoint of $\overline{BD}$. What is the area of $\triangle CDM$?
等边 $\triangle ABC$ 边长为 2，$M$ 是 $\overline{AC}$ 的中点，$C$ 是 $\overline{BD}$ 的中点。求 $\triangle CDM$ 的面积。
stem
Correct Answer: C
Drop $MQ$ perpendicular to $BC$. Then $\triangle MQC$ is a 30–60–90 triangle, so $MQ = \sqrt{3}/2$, and the area of $\triangle CDM$ is $\frac{1}{2} (2 \cdot \sqrt{3}/2) = \sqrt{3}/2$.
从 $M$ 向 $BC$ 作垂线 $MQ$。则 $\triangle MQC$ 是 30–60–90 三角形，因此 $MQ = \sqrt{3}/2$，$\triangle CDM$ 的面积是 $\frac{1}{2} (2 \cdot \sqrt{3}/2) = \sqrt{3}/2$。
Q15
An envelope contains eight bills: 2 ones, 2 fives, 2 tens, and 2 twenties. Two bills are drawn at random without replacement. What is the probability that their sum is $20$ or more?
一个信封里有 8 张钞票：2 张 1 元，2 张 5 元，2 张 10 元，2 张 20 元。随机不放回抽取 2 张钞票。它们的和为 20 元或更多的概率是多少？
Correct Answer: D
(D) There are \[ \binom{8}{2}=\frac{8!}{6!\cdot 2!}=28 \] ways to choose the bills. A sum of at least \$20 is obtained by choosing both \$20 bills, one of the \$20 bills and one of the six smaller bills, or both \$10 bills. Hence the probability is \[ \frac{1+2\cdot 6+1}{28}=\frac{14}{28}=\frac{1}{2}. \]
(D) 共有 \[ \binom{8}{2}=\frac{8!}{6!\cdot 2!}=28 \] 种选取钞票的方法。至少得到 \$20 的总金额可以通过选取两张 \$20 钞票、选取一张 \$20 钞票并再选取六张较小面额钞票中的一张，或者选取两张 \$10 钞票来实现。因此概率为 \[ \frac{1+2\cdot 6+1}{28}=\frac{14}{28}=\frac{1}{2}. \]
Q16
The quadratic equation $x^2 + mx + n = 0$ has roots that are twice those of $x^2 + px + m = 0$, and none of $m$, $n$, and $p$ is zero. What is the value of $n/p$?
二次方程 $x^2 + mx + n = 0$ 的根是 $x^2 + px + m = 0$ 的根的两倍，且 $m$、$n$ 和 $p$ 均不为零。$n/p$ 的值是多少？
Correct Answer: D
(D) Let \(r_1\) and \(r_2\) be the roots of \(x^2+px+m=0\). Since the roots of \(x^2+mx+n=0\) are \(2r_1\) and \(2r_2\), we have the following relationships: \[ m=r_1r_2,\quad n=4r_1r_2,\quad p=-(r_1+r_2),\quad \text{and}\quad m=-2(r_1+r_2). \] So \[ n=4m,\quad p=\frac{1}{2}m,\quad \text{and}\quad \frac{n}{p}=\frac{4m}{\frac{1}{2}m}=8. \]
（D）设 \(r_1\) 和 \(r_2\) 为方程 \(x^2+px+m=0\) 的根。由于方程 \(x^2+mx+n=0\) 的根为 \(2r_1\) 和 \(2r_2\)，我们有如下关系： \[ m=r_1r_2,\quad n=4r_1r_2,\quad p=-(r_1+r_2),\quad \text{且}\quad m=-2(r_1+r_2)。 \] 因此 \[ n=4m,\quad p=\frac{1}{2}m,\quad \text{并且}\quad \frac{n}{p}=\frac{4m}{\frac{1}{2}m}=8。 \]
Q17
Suppose that $4^a = 5$, $5^b = 6$, $6^c = 7$, and $7^d = 8$. What is $a \cdot b \cdot c \cdot d$?
假设 $4^a = 5$，$5^b = 6$，$6^c = 7$，且 $7^d = 8$。$a \cdot b \cdot c \cdot d$ 是多少？
Correct Answer: B
(B) Because $4^{a\cdot b\cdot c\cdot d}=\left(\left(\left(4^a\right)^b\right)^c\right)^d=\left(\left(5^b\right)^c\right)^d=\left(6^c\right)^d=7^d=8=4^{3/2},$ we have $a\cdot b\cdot c\cdot d=3/2.$
（B）因为 $4^{a\cdot b\cdot c\cdot d}=\left(\left(\left(4^a\right)^b\right)^c\right)^d=\left(\left(5^b\right)^c\right)^d=\left(6^c\right)^d=7^d=8=4^{3/2},$ 所以有 $a\cdot b\cdot c\cdot d=3/2.$
Q18
All of David’s telephone numbers have the form 555–abc–defg, where a, b, c, d, e, f, and g are distinct digits and in increasing order, and none is either 0 or 1. How many different telephone numbers can David have?
David 的所有电话号码形式为 555–abc–defg，其中 a, b, c, d, e, f, g 是不同的数字且按升序排列，且都不为 0 或 1。David 可以有多少个不同的电话号码？
Correct Answer: D
The last seven digits of the phone number use seven of the eight digits ${2, 3, 4, 5, 6, 7, 8, 9}$, so all but one of these digits is used. The unused digit can be chosen in eight ways. The remaining seven digits are then placed in increasing order to obtain a possible phone number. Thus there are 8 possible phone numbers.
电话号码的后七位使用 {2, 3, 4, 5, 6, 7, 8, 9} 中的七个数字，因此漏掉其中一个数字。未使用的数字有 8 种选择。然后剩余的七个数字按升序排列即可得到一个可能的电话号码。因此共有 8 个可能的电话号码。
Q19
On a certain math exam, 10% of the students got 70 points, 25% got 80 points, 20% got 85 points, 15% got 90 points, and the rest got 95 points. What is the difference between the mean and the median score on this exam?
在某次数学考试中，10% 的学生得 70 分，25% 得 80 分，20% 得 85 分，15% 得 90 分，其余得 95 分。这次考试的平均分与中位数的差是多少？
Correct Answer: B
(B) The percentage of students getting 95 points is $100-10-25-20-15=30$, so the mean score on the exam is $0.10(70)+0.25(80)+0.20(85)+0.15(90)+0.30(95)=86.$ Since fewer than half of the scores were less than 85, and fewer than half of the scores were greater than 85, the median score is 85. The difference between the mean and the median score on this exam is $86-85=1$.
（B）得到95分的学生所占百分比为 $100-10-25-20-15=30$， 因此本次考试的平均分为 $0.10(70)+0.25(80)+0.20(85)+0.15(90)+0.30(95)=86.$ 由于低于85分的成绩不到一半，高于85分的成绩也不到一半，所以中位数为85。本次考试平均数与中位数之差为 $86-85=1$。
Q20
What is the average (mean) of all 5-digit numbers that can be formed by using each of the digits 1, 3, 5, 7, and 8 exactly once?
使用数字 1、3、5、7、8 各一次可以形成的全部五位数的平均数（均值）是多少？
Correct Answer: C
(C) Each digit appears the same number of times in the 1’s place, the 10’s place, $\ldots$, and the 10,000’s place. The average of the digits in each place is $$\frac{1}{5}(1+3+5+7+8)=\frac{24}{5}=4.8.$$ Hence the average of all the numbers is $$4.8(1+10+100+1000+10000)=4.8(11111)=53332.8.$$
（C）每个数字在个位、十位、$\ldots$、万位上出现的次数相同。每一位上数字的平均值为 $$\frac{1}{5}(1+3+5+7+8)=\frac{24}{5}=4.8.$$ 因此所有这些数的平均值为 $$4.8(1+10+100+1000+10000)=4.8(11111)=53332.8.$$
Q21
Forty slips are placed into a hat, each bearing a number 1, 2, 3, 4, 5, 6, 7, 8, 9, or 10, with each number entered on four slips. Four slips are drawn from the hat at random and without replacement. Let $p$ be the probability that all four slips bear the same number. Let $q$ be the probability that two of the slips bear a number $a$ and the other two bear a number $b \ne a$. What is the value of $q/p$?
帽子里放入40张纸条，每张纸条上写着数字1、2、3、4、5、6、7、8、9或10，每个数字有四张纸条。从帽子里随机无放回抽取四张纸条。设$p$为四张纸条上数字都相同的概率。设$q$为其中两张纸条上数字为$a$，另外两张为$b\ne a$的概率。$q/p$的值是多少？
Correct Answer: A
(A) The total number of ways that the numbers can be chosen is $\binom{40}{4}$. Exactly 10 of these possibilities result in the four slips having the same number. Now we need to determine the number of ways that two slips can have a number $a$ and the other two slips have a number $b$, with $b \ne a$. There are $\binom{10}{2}$ ways to choose the distinct numbers $a$ and $b$. For each value of $a$ there are $\binom{4}{2}$ to choose the two slips with $a$ and for each value of $b$ there are $\binom{4}{2}$ to choose the two slips with $b$. Hence the number of ways that two slips have some number $a$ and the other two slips have some distinct number $b$ is $$ \binom{10}{2}\cdot\binom{4}{2}\cdot\binom{4}{2}=45\cdot6\cdot6=1620. $$ So the probabilities $q$ and $p$ are $\frac{10}{\binom{40}{4}}$ and $\frac{1620}{\binom{40}{4}}$, respectively, which implies that $$ \frac{p}{q}=\frac{1620}{10}=162. $$
（A）选择这些数字的总方法数是 $\binom{40}{4}$。其中恰有 10 种情况会使四张纸条上的数字相同。 现在需要确定这样的情况数：两张纸条上的数字为 $a$，另外两张纸条上的数字为 $b$，且 $b \ne a$。选择不同的数字 $a$ 和 $b$ 有 $\binom{10}{2}$ 种方法。对每个 $a$，选出两张写 $a$ 的纸条有 $\binom{4}{2}$ 种方法；对每个 $b$，选出两张写 $b$ 的纸条也有 $\binom{4}{2}$ 种方法。因此，两张纸条为某个数字 $a$ 且另外两张为不同数字 $b$ 的方法数为 $$ \binom{10}{2}\cdot\binom{4}{2}\cdot\binom{4}{2}=45\cdot6\cdot6=1620. $$ 所以概率 $q$ 和 $p$ 分别为 $\frac{10}{\binom{40}{4}}$ 和 $\frac{1620}{\binom{40}{4}}$，从而 $$ \frac{p}{q}=\frac{1620}{10}=162. $$
Q22
For how many positive integers $n$ less than or equal to 24 is $n!$ evenly divisible by $1 + 2 + \cdots + n$?
对于多少个正整数$n\le 24$，$n!$能被$1 + 2 + \cdots + n$整除？
Correct Answer: C
(C) Since $1+2+\cdots+n=\dfrac{n(n+1)}{2}$, the condition is equivalent to having an integer value for $\dfrac{n!}{n(n+1)/2}$. This reduces, when $n\ge 1$, to having an integer value for $\dfrac{2(n-1)!}{n+1}$. This fraction is an integer unless $n+1$ is an odd prime. There are 8 odd primes less than or equal to 25, so there are $24-8=16$ numbers less than or equal to 24 that satisfy the condition.
（C）因为 $1+2+\cdots+n=\dfrac{n(n+1)}{2}$， 所以该条件等价于下面这个式子的值为整数： $\dfrac{n!}{n(n+1)/2}$。 当 $n\ge 1$ 时，这可化简为要求下式为整数： $\dfrac{2(n-1)!}{n+1}$。 除非 $n+1$ 是奇素数，否则该分式为整数。不超过 25 的奇素数有 8 个，因此不超过 24 且满足条件的数有 $24-8=16$ 个。
Q23
In trapezoid ABCD we have AB parallel to DC, E as the midpoint of BC, and F as the midpoint of DA. The area of ABEF is twice the area of FECD. What is AB/DC?
梯形ABCD中，AB平行的DC，E为BC中点，F为DA中点。区域ABEF的面积是FECD面积的两倍。AB/DC是多少？
Correct Answer: C
(C) First note that $FE=\dfrac{AB+DC}{2}$. Because trapezoids $ABEF$ and $FECD$ have the same height, the ratio of their areas is equal to the ratio of the averages of their parallel sides. Since \[ AB+\dfrac{AB+DC}{2}=\dfrac{3AB+DC}{2} \] and \[ \dfrac{AB+DC}{2}+DC=\dfrac{AB+3DC}{2}, \] we have \[ 3AB+DC=2(AB+3DC)=2AB+6DC,\quad \text{and}\quad \dfrac{AB}{DC}=5. \]
（C）首先注意到 $FE=\dfrac{AB+DC}{2}$。因为梯形 $ABEF$ 和 $FECD$ 的高相同，它们的面积之比等于它们平行边平均数之比。由于 \[ AB+\dfrac{AB+DC}{2}=\dfrac{3AB+DC}{2} \] 并且 \[ \dfrac{AB+DC}{2}+DC=\dfrac{AB+3DC}{2}, \] 因此有 \[ 3AB+DC=2(AB+3DC)=2AB+6DC,\quad \text{并且}\quad \dfrac{AB}{DC}=5. \]
solution
Q24
Let x and y be two-digit integers such that y is obtained by reversing the digits of x. The integers x and y satisfy $x^2 - y^2 = m^2$ for some positive integer m. What is x + y + m?
设$x$和$y$是两位整数，$y$由$x$的数字反转得到。整数$x$和$y$满足$x^2 - y^2 = m^2$，其中$m$为正整数。$x + y + m$是多少？
Correct Answer: E
(E) By the given conditions, it follows that $x>y$. Let $x=10a+b$ and $y=10b+a$, where $a>b$. Then $m^2=x^2-y^2=(10a+b)^2-(10b+a)^2=99a^2-99b^2=99(a^2-b^2).$ Since $99(a^2-b^2)$ must be a perfect square, $a^2-b^2=(a+b)(a-b)=11k^2,$ for some positive integer $k$. Because $a$ and $b$ are distinct digits, we have $a-b\le 9-1=8$ and $a+b\le 9+8=17$. It follows that $a+b=11$, $a-b=k^2$, and $k$ is either 1 or 2. If $k=2$, then $(a,b)=(15/2,7/2)$, which is impossible. Thus $k=1$ and $(a,b)=(6,5)$. This gives $x=65$, $y=56$, $m=33$, and $x+y+m=154$.
（E）由已知条件可得 $x>y$。令 $x=10a+b$，$y=10b+a$，其中 $a>b$。则 $m^2=x^2-y^2=(10a+b)^2-(10b+a)^2=99a^2-99b^2=99(a^2-b^2)。$ 由于 $99(a^2-b^2)$ 必须是完全平方数， $a^2-b^2=(a+b)(a-b)=11k^2,$ 其中 $k$ 为某个正整数。因为 $a$ 与 $b$ 是不同的数字，所以 $a-b\le 9-1=8$ 且 $a+b\le 9+8=17$。因此可得 $a+b=11$，$a-b=k^2$，并且 $k$ 只能是 1 或 2。 若 $k=2$，则 $(a,b)=(15/2,7/2)$，不可能。故 $k=1$，$(a,b)=(6,5)$。于是 $x=65$，$y=56$，$m=33$，且 $x+y+m=154$。
Q25
A subset B of the set of integers from 1 to 100, inclusive, has the property that no two elements of B sum to 125. What is the maximum possible number of elements in B?
{1,2,…,100}的一个子集B，具有B中无两个元素和为125的性质。B的最大可能元素个数是多少？
Correct Answer: C
(C) Several pairs of numbers from 1 to 100 sum to 125. These pairs are (25, 100), (26, 99), . . . , (62, 63). Set $B$ can have at most one number from each of these $62-25+1=38$ pairs. In addition, $B$ can contain all of the numbers $1, 2, . . . , 24$ since these cannot be paired with any of the available numbers to sum to 125. So $B$ has at most $38+24=62$ numbers. The set containing the first 62 positive integers, for example, is one of these maximum sets.
（C）从 1 到 100 的数中，有若干对数的和为 125。这些数对是 (25, 100)、(26, 99)、……、(62, 63)。集合 $B$ 在这些 $62-25+1=38$ 对数中，每一对最多只能取一个数。此外，$B$ 还可以包含所有的 $1, 2, \ldots, 24$，因为它们无法与任何可用的数配对使和为 125。因此，$B$ 最多有 $38+24=62$ 个数。例如，包含前 62 个正整数的集合就是这样的一个最大集合。