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AMC10 2002 B

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AMC10 · 2002 (B)

Q1
The ratio $\frac{2^{2001} \cdot 3^{2003}}{6^{2002}}$ is
比例 $\frac{2^{2001} \cdot 3^{2003}}{6^{2002}}$ 是
Correct Answer: E
(E) We have $\dfrac{2^{2001}\cdot 3^{2003}}{6^{2002}}=\dfrac{2^{2001}\cdot 3^{2003}}{(2\cdot 3)^{2002}}=\dfrac{2^{2001}\cdot 3^{2003}}{2^{2002}\cdot 3^{2002}}=\dfrac{3}{2}$
（E）我们有 $\dfrac{2^{2001}\cdot 3^{2003}}{6^{2002}}=\dfrac{2^{2001}\cdot 3^{2003}}{(2\cdot 3)^{2002}}=\dfrac{2^{2001}\cdot 3^{2003}}{2^{2002}\cdot 3^{2002}}=\dfrac{3}{2}$
Q2
For the nonzero numbers $a$, $b$, and $c$, define $(a, b, c) = \frac{abc}{a + b + c}$. Find $(2, 4, 6)$.
对于非零数 $a$、$b$ 和 $c$，定义 $(a, b, c) = \frac{abc}{a + b + c}$。求 $(2, 4, 6)$。
Correct Answer: C
(C) We have $(2,4,6)=\dfrac{2\cdot4\cdot6}{2+4+6}=\dfrac{48}{12}=4.$
（C）我们有 $(2,4,6)=\dfrac{2\cdot4\cdot6}{2+4+6}=\dfrac{48}{12}=4.$
Q3
The arithmetic mean of the nine numbers in the set $\{9,99,999,9999,\dots ,999999999\}$ is a 9-digit number $M$, all of whose digits are distinct. The number $M$ does not contain the digit
集合 $\{9,99,999,9999,\dots ,999999999\}$ 中九个数的算术平均数是一个 9 位数 $M$，其所有数字均不同。数 $M$ 不包含数字
Correct Answer: A
(A) The number $M$ is equal to $\frac{1}{9}(9+99+999+\ldots+999,999,999)=1+11+111+\ldots+111,111,111=123,456,789.$ The number $M$ does not contain the digit $0$.
（A）数 $M$ 等于 $\frac{1}{9}(9+99+999+\ldots+999,999,999)=1+11+111+\ldots+111,111,111=123,456,789.$ 数 $M$ 不包含数字 $0$。
Q4
What is the value of $(3x -2)(4x + 1) - \frac{3x -2}{4x + 1}$ when $x = 4$?
当 $x = 4$ 时，$(3x -2)(4x + 1) - \frac{3x -2}{4x + 1}$ 的值是多少？
Correct Answer: D
(D) Since $(3x-2)(4x+1)-(3x-2)4x+1=(3x-2)(4x+1-4x)+1$ $=(3x-2)\cdot1+1=3x-1,$ when $x=4$ we have the value $3\cdot4-1=11.$
（D）因为 $(3x-2)(4x+1)-(3x-2)4x+1=(3x-2)(4x+1-4x)+1$ $=(3x-2)\cdot1+1=3x-1,$ 当 $x=4$ 时，其值为 $3\cdot4-1=11.$
Q5
Circles of radius 2 and 3 are externally tangent and are circumscribed by a third circle, as shown in the figure. Find the area of the shaded region.
半径为 2 和 3 的圆外部相切，并被第三个圆外接，如图所示。求阴影区域的面积。
stem
Correct Answer: E
(E) The diameter of the large circle is 6 + 4 = 10, so its radius is 5. Hence, the area of the shaded region is $\pi(5^2)-\pi(3^2)-\pi(2^2)=\pi(25-9-4)=12\pi.$
（E）大圆的直径是 $6+4=10$，所以半径是 $5$。因此，阴影部分的面积为 $\pi(5^2)-\pi(3^2)-\pi(2^2)=\pi(25-9-4)=12\pi.$
Q6
For how many positive integers $n$ is $n^2 -3n + 2$ a prime number?
有且仅有几个正整数 $n$ 使得 $n^2 -3n + 2$ 是质数？
Correct Answer: B
(B) If $n \ge 4$, then $n^2 - 3n + 2 = (n - 1)(n - 2)$ is the product of two integers greater than 1, and thus is not prime. For $n = 1$, $2$, and $3$ we have, respectively, $(1 - 1)(1 - 2) = 0$, $(2 - 1)(2 - 2) = 0$, and $(3 - 1)(3 - 2) = 2$. Therefore, $n^2 - 3n + 2$ is prime only when $n = 3$.
(B) 若 $n \ge 4$，则 $n^2 - 3n + 2 = (n - 1)(n - 2)$ 是两个大于 1 的整数的乘积，因此不是素数。对 $n = 1、2、3$ 分别有：$(1 - 1)(1 - 2) = 0$，$(2 - 1)(2 - 2) = 0$，以及 $(3 - 1)(3 - 2) = 2$。因此，$n^2 - 3n + 2$ 只有在 $n = 3$ 时才是素数。
Q7
Let $n$ be a positive integer such that $\frac{1}{2} + \frac{1}{3} + \frac{1}{7} + \frac{1}{n}$ is an integer. Which of the following statements is not true:
设 $n$ 是正整数，使得 $\frac{1}{2} + \frac{1}{3} + \frac{1}{7} + \frac{1}{n}$ 是整数。以下哪个陈述不正确：
Correct Answer: E
(E) The number $\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{n}$ is greater than 0 and less than $\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{1}<2$. Hence, $\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{n}=\frac{41}{42}+\frac{1}{n}$ is an integer precisely when it is equal to 1. This implies that $n=42$, so the answer is (E).
（E）数 $\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{n}$ 大于 0 且小于 $\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{1}<2$。因此， $\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{n}=\frac{41}{42}+\frac{1}{n}$ 恰好在其等于 1 时为整数。这推出 $n=42$，所以答案是（E）。
Q8
Suppose July of year $N$ has five Mondays. Which of the following must occur five times in August of year $N$? (Note: Both months have 31 days.)
假设 $N$ 年的 7 月有五个星期一。那么 $N$ 年的 8 月中以下哪个一定出现五次？（注：两个月都有 31 天。）
Correct Answer: D
(D) Since July has 31 days, Monday must be one of the last three days of July. Therefore, Thursday must be one of the first three days of August, which also has 31 days. So Thursday must occur five times in August.
（D）由于七月有31天，星期一必定落在七月的最后三天之一。因此，星期四必定落在八月的前三天之一，而八月同样有31天。所以，八月中星期四必定出现五次。
Q9
Using the letters A, M, O, S, and U, we can form 120 five-letter “words”. If these “words” are arranged in alphabetical order, then the “word” USAMO occupies position
使用字母 A, M, O, S, 和 U，可以形成 120 个五字母“单词”。如果这些“单词”按字母顺序排列，那么“单词” USAMO 占据的位置是
Correct Answer: D
(D) The last “word,” which occupies position 120, is $\textit{USOMA}$. Immediately preceding this we have $\textit{USOAM}$, $\textit{USMOA}$, $\textit{USMAO}$, $\textit{USAOM}$, and $\textit{USAMO}$. The alphabetic position of the word $\textit{USAMO}$ is consequently 115.
(D) 最后一个“单词”位于第 120 位，是 $\textit{USOMA}$。紧接在它之前的依次是 $\textit{USOAM}$、$\textit{USMOA}$、$\textit{USMAO}$、$\textit{USAOM}$ 和 $\textit{USAMO}$。因此，单词 $\textit{USAMO}$ 的字母序位置是 115。
Q10
Suppose that $a$ and $b$ are nonzero real numbers, and that the equation $x^2 + ax + b = 0$ has solutions $a$ and $b$. Then the pair $(a, b)$ is
假设 $a$ 和 $b$ 是非零实数，且方程 $x^2 + ax + b = 0$ 的解是 $a$ 和 $b$。则对 $(a, b)$ 是
Correct Answer: C
(C) The given conditions imply that \[ x^2+ax+b=(x-a)(x-b)=x^2-(a+b)x+ab, \] so \[ a+b=-a \quad \text{and} \quad ab=b. \] Since $b\ne 0$, the second equation implies that $a=1$. The first equation gives $b=-2$, so $(a,b)=(1,-2)$.
（C）给定条件推出 \[ x^2+ax+b=(x-a)(x-b)=x^2-(a+b)x+ab, \] 因此 \[ a+b=-a \quad \text{且} \quad ab=b. \] 由于 $b\ne 0$，第二个方程推出 $a=1$。第一个方程得到 $b=-2$，所以 $(a,b)=(1,-2)$。
Q11
The product of three consecutive positive integers is 8 times their sum. What is the sum of their squares?
三个连续正整数的乘积是它们和的8倍。它们平方和是多少？
Correct Answer: B
(B) Let $n-1$, $n$, and $n+1$ denote the three integers. Then $$(n-1)n(n+1)=8(3n).$$ Since $n\neq 0$, we have $n^2-1=24$. It follows that $n^2=25$ and $n=5$. Thus, $$(n-1)^2+n^2+(n+1)^2=16+25+36=77.$$
（B）令 $n-1$、$n$ 和 $n+1$ 表示这三个整数。则 $$(n-1)n(n+1)=8(3n)。$$ 由于 $n\neq 0$，有 $n^2-1=24$。因此 $n^2=25$ 且 $n=5$。于是 $$(n-1)^2+n^2+(n+1)^2=16+25+36=77。$$
Q12
For which of the following values of $k$ does the equation $\frac{x-1}{x-2} = \frac{x-k}{x-6}$ have no solution for $x$?
对于下列哪个$k$值，方程$\frac{x-1}{x-2}=\frac{x-k}{x-6}$没有$x$的解？
Correct Answer: E
(E) From the given equation we have $(x-1)(x-6)=(x-2)(x-k)$. This implies that $x^2-7x+6=x^2-(2+k)x+2k$, so $(k-5)x=2k-6$ and $x=\dfrac{2k-6}{k-5}$. Hence a value of $x$ satisfying the equation occurs unless $k=5$. Note that when $k=6$ there is also no solution for $x$, but this is not one of the answer choices.
（E）由已知方程可得 $(x-1)(x-6)=(x-2)(x-k)$。这意味着 $x^2-7x+6=x^2-(2+k)x+2k$， 所以 $(k-5)x=2k-6$ 且 $x=\dfrac{2k-6}{k-5}$。 因此，除非 $k=5$，否则存在满足该方程的 $x$ 值。 注意当 $k=6$ 时，$x$ 也无解，但这不在给出的答案选项中。
Q13
Find the value(s) of $x$ such that $8xy -12y + 2x -3 = 0$ is true for all values of $y$.
求使得$8xy-12y+2x-3=0$对所有$y$值都成立的$x$值。
Correct Answer: D
(D) The given equation can be factored as $0=8xy-12y+2x-3=4y(2x-3)+(2x-3)=(4y+1)(2x-3).$ For this equation to be true for all values of $y$ we must have $2x-3=0$, that is, $x=\dfrac{3}{2}.$
（D）给定方程可因式分解为 $0=8xy-12y+2x-3=4y(2x-3)+(2x-3)=(4y+1)(2x-3).$ 要使该方程对所有 $y$ 的取值都成立，必须有 $2x-3=0$，即 $x=\dfrac{3}{2}.$
Q14
The number $256^4 \cdot 64^{25}$ is the square of a positive integer $N$. In decimal representation, the sum of the digits of $N$ is
数$256^4\cdot64^{25}$是一个正整数$N$的平方。在十进制表示中，$N$的各位数字之和是
Correct Answer: B
(B) We have $N=\sqrt{(5^2)^{64}\cdot(2^6)^{25}}=5^{64}\cdot2^{3\cdot25}=(5\cdot2)^{64}\cdot2^{11}=10^{64}\cdot2048=2048\,000\cdots0.$ $64$ digits The zeros do not contribute to the sum, so the sum of the digits of $N$ is $2+4+8=14$.
（B）我们有 $N=\sqrt{(5^2)^{64}\cdot(2^6)^{25}}=5^{64}\cdot2^{3\cdot25}=(5\cdot2)^{64}\cdot2^{11}=10^{64}\cdot2048=2048\,000\cdots0.$ $64$ 位数字 这些 0 不会对数字和有贡献，因此 $N$ 的各位数字之和为 $2+4+8=14$。
Q15
The positive integers $A$, $B$, $A - B$, and $A + B$ are all prime numbers. The sum of these four primes is
正整数$A$、$B$、$A-B$和$A+B$都是素数。这四个素数的和是
Correct Answer: E
(E) The numbers $A-B$ and $A+B$ are both odd or both even. However, they are also both prime, so they must both be odd. Therefore, one of $A$ and $B$ is odd and the other even. Because $A$ is a prime between $A-B$ and $A+B$, $A$ must be the odd prime. Therefore, $B=2$, the only even prime. So $A-2$, $A$, and $A+2$ are consecutive odd primes and thus must be 3, 5, and 7. The sum of the four primes 2, 3, 5, and 7 is the prime number 17.
（E）数 $A-B$ 和 $A+B$ 要么同为奇数，要么同为偶数。然而它们也都是素数，因此它们必须都为奇数。所以 $A$ 与 $B$ 其中一个是奇数，另一个是偶数。因为 $A$ 是介于 $A-B$ 与 $A+B$ 之间的一个素数，所以 $A$ 必须是奇素数。因此 $B=2$，唯一的偶素数。于是 $A-2$、$A$、$A+2$ 是连续的奇素数，因此只能是 3、5、7。四个素数 2、3、5、7 的和是素数 17。
Q16
For how many integers $n$ is $\frac{n}{20-n}$ the square of an integer?
有几个整数 $n$ 使得 $\frac{n}{20-n}$ 是一个整数的平方？
Correct Answer: D
(D) If $\frac{n}{20-n}=k^2$, for some $k\ge 0$, then $n=\frac{20k^2}{k^2+1}$. Since $k^2$ and $k^2+1$ have no common factors and $n$ is an integer, $k^2+1$ must be a factor of 20. This occurs only when $k=0,1,2,$ or $3$. The corresponding values of $n$ are 0, 10, 16, and 18.
(D) 若对某个 $k\ge 0$ 有 $\frac{n}{20-n}=k^2$，则 $n=\frac{20k^2}{k^2+1}$。由于 $k^2$ 与 $k^2+1$ 互质且 $n$ 为整数，因此 $k^2+1$ 必须是 20 的因数。这只在 $k=0,1,2$ 或 $3$ 时发生。对应的 $n$ 值为 0、10、16 和 18。
Q17
A regular octagon ABCDEFGH has sides of length two. Find the area of $\triangle ADG$.
一个边长为二的正八边形 ABCDEFGH，求 $\triangle ADG$ 的面积。
Correct Answer: C
(C) Construct the right triangle $\triangle AOB$ as shown in the figure. Since $AB=2$, we have $AO=\sqrt{2}$ and $AD=2+2\sqrt{2}$. Similarly, we have $OG=2+\sqrt{2}$, so $$ \text{Area}(\triangle ADG)=\frac{1}{2}(2+2\sqrt{2})(2+\sqrt{2})=(1+\sqrt{2})(2+\sqrt{2})=4+3\sqrt{2}. $$
（C）按图所示作直角三角形 $\triangle AOB$。由于 $AB=2$，可得 $AO=\sqrt{2}$，且 $AD=2+2\sqrt{2}$。同理，$OG=2+\sqrt{2}$，因此 $$ \text{面积}(\triangle ADG)=\frac{1}{2}(2+2\sqrt{2})(2+\sqrt{2})=(1+\sqrt{2})(2+\sqrt{2})=4+3\sqrt{2}. $$
solution
Q18
Four distinct circles are drawn in a plane. What is the maximum number of points where at least two of the circles intersect?
平面内画四个不同的圆，最多有多少个至少有两个圆相交的点？
Correct Answer: D
(D) Each pair of circles has at most two intersection points. There are $\binom{4}{2}=6$ pairs of circles, so there are at most $6\times 2=12$ points of intersection. The following configuration shows that 12 points of intersection are indeed possible:
（D）每一对圆最多有两个交点。有 $\binom{4}{2}=6$ 对圆，因此最多有 $6\times 2=12$ 个交点。下面的构型表明，确实可以达到 12 个交点：
solution
Q19
Suppose that $\{a_n\}$ is an arithmetic sequence with $a_1 + a_2 + \cdots + a_{100} = 100$ and $a_{101} + a_{102} + \cdots + a_{200} = 200$. What is the value of $a_2 - a_1$?
设 $\{a_n\}$ 是等差数列，且 $a_1 + a_2 + \cdots + a_{100} = 100$，$a_{101} + a_{102} + \cdots + a_{200} = 200$。求 $a_2 - a_1$ 的值。
Correct Answer: C
(C) Let $d=a_2-a_1$. Then $a_{k+100}=a_k+100d$, and $a_{101}+a_{102}+\cdots+a_{200}=(a_1+100d)+(a_2+100d)+\cdots+(a_{100}+100d)$ $=a_1+a_2+\cdots+a_{100}+10{,}000d$. Thus $200=100+10{,}000d$ and $d=\frac{100}{10{,}000}=0.01$.
（C）令 $d=a_2-a_1$。则 $a_{k+100}=a_k+100d$，并且 $a_{101}+a_{102}+\cdots+a_{200}=(a_1+100d)+(a_2+100d)+\cdots+(a_{100}+100d)$ $=a_1+a_2+\cdots+a_{100}+10{,}000d$。 因此 $200=100+10{,}000d$，并且 $d=\frac{100}{10{,}000}=0.01$。
Q20
Let $a$, $b$, and $c$ be real numbers such that $a -7b + 8c = 4$ and $8a + 4b -c = 7$. Then $a^2 - b^2 + c^2$ is
设 $a$、$b$ 和 $c$ 是实数，使得 $a -7b + 8c = 4$ 和 $8a + 4b -c = 7$。则 $a^2 - b^2 + c^2$ 是
Correct Answer: B
(B) We have $a+8c=4+7b$ and $8a-c=7-4b$. Squaring both equations and adding the results yields $$(a+8c)^2+(8a-c)^2=(4+7b)^2+(7-4b)^2.$$ Expanding gives $65(a^2+c^2)=65(1+b^2)$. So $a^2+c^2=1+b^2$, and $a^2-b^2+c^2=1$.
（B）我们有 $a+8c=4+7b$ 和 $8a-c=7-4b$。将两式分别平方并把结果相加得到 $$(a+8c)^2+(8a-c)^2=(4+7b)^2+(7-4b)^2.$$ 展开可得 $65(a^2+c^2)=65(1+b^2)$。因此 $a^2+c^2=1+b^2$，并且 $a^2-b^2+c^2=1$。
Q21
Andy’s lawn has twice as much area as Beth’s lawn and three times as much area as Carlos’ lawn. Carlos’ lawn mower cuts half as fast as Beth’s mower and one third as fast as Andy’s mower. If they all start to mow their lawns at the same time, who will finish first?
安迪的草坪面积是贝丝草坪面积的两倍，是卡洛斯草坪面积的三倍。卡洛斯的割草机割草速度是贝丝割草机的一半，是安迪割草机三分之一。他们同时开始修剪草坪，谁会最先完成？
Correct Answer: B
(B) Let $A$ be the number of square feet in Andy’s lawn. Then $A/2$ and $A/3$ are the areas of Beth’s lawn and Carlos’ lawn, respectively, in square feet. Let $R$ be the rate, in square feet per minute, that Carlos’ lawn mower cuts. Then Beth’s mower and Andy’s mower cut at rates of $2R$ and $3R$ square feet per minute, respectively. Thus, Andy takes $\frac{A}{3R}$ minutes to mow his lawn, Beth takes $\frac{A/2}{2R}=\frac{A}{4R}$ minutes to mow hers, and Carlos takes $\frac{A/3}{R}=\frac{A}{3R}$ minutes to mow his. Since $\frac{A}{4R}<\frac{A}{3R}$, Beth will finish first.
(B) 设 $A$ 为 Andy 的草坪面积（单位：平方英尺）。则 Beth 的草坪面积为 $A/2$，Carlos 的草坪面积为 $A/3$（单位：平方英尺）。设 $R$ 为 Carlos 的割草机割草速率（单位：平方英尺/分钟）。则 Beth 和 Andy 的割草机速率分别为 $2R$ 和 $3R$ 平方英尺/分钟。因此， Andy 割完草坪需要 $\frac{A}{3R}$ 分钟， Beth 割完草坪需要 $\frac{A/2}{2R}=\frac{A}{4R}$ 分钟， 并且 Carlos 割完草坪需要 $\frac{A/3}{R}=\frac{A}{3R}$ 分钟。 由于 $\frac{A}{4R}<\frac{A}{3R}$，所以 Beth 会最先完成。
Q22
Let $\triangle XOY$ be a right-angled triangle with $\angle XOY = 90^\circ$. Let $M$ and $N$ be the midpoints of legs $OX$ and $OY$, respectively. Given that $XN = 19$ and $YM = 22$, find $XY$.
设 $\triangle XOY$ 为直角三角形，$\angle XOY = 90^\circ$。$M$ 和 $N$ 分别为腿 $OX$ 和 $OY$ 的中点。已知 $XN = 19$ 和 $YM = 22$，求 $XY$。
Correct Answer: B
(B) Let $OM=a$ and $ON=b$. Then \[ 19^2=(2a)^2+b^2 \quad \text{and} \quad 22^2=a^2+(2b)^2. \] Hence \[ 5(a^2+b^2)=19^2+22^2=845. \] It follows that \[ MN=\sqrt{a^2+b^2}=\sqrt{169}=13. \] Since $\triangle XOY$ is similar to $\triangle MON$ and $XO=2\cdot MO$, we have $XY=2\cdot MN=26$. \[ h=\sqrt{5^2-\left(\frac{25}{13}\right)^2}=5\sqrt{1-\frac{25}{169}}=5\sqrt{\frac{144}{169}}=\frac{60}{13}. \]
（B）设 $OM=a$，$ON=b$。则 \[ 19^2=(2a)^2+b^2 \quad \text{且} \quad 22^2=a^2+(2b)^2。 \] 因此 \[ 5(a^2+b^2)=19^2+22^2=845。 \] 从而 \[ MN=\sqrt{a^2+b^2}=\sqrt{169}=13。 \] 由于 $\triangle XOY$ 与 $\triangle MON$ 相似，且 $XO=2\cdot MO$，所以 $XY=2\cdot MN=26$。 \[ h=\sqrt{5^2-\left(\frac{25}{13}\right)^2}=5\sqrt{1-\frac{25}{169}}=5\sqrt{\frac{144}{169}}=\frac{60}{13}。 \]
solution solution
Q23
Let $\{a_k\}$ be a sequence of integers such that $a_1 = 1$ and $a_{m+n} = a_m + a_n + mn$, for all positive integers $m$ and $n$. Then $a_{12}$ is
设 $\{a_k\}$ 为整数序列，满足 $a_1 = 1$ 且 $a_{m+n} = a_m + a_n + mn$，对所有正整数 $m$ 和 $n$ 成立。则 $a_{12}$ 是
Correct Answer: D
(D) By setting $n=1$ in the given recursive equation, we obtain $a_{m+1}=a_m+a_1+m$, for all positive integers $m$. So $a_{m+1}-a_m=m+1$ for each $m=1,2,3,\ldots$ Hence, $a_{12}-a_{11}=12,\ a_{11}-a_{10}=11,\ \ldots,\ a_2-a_1=2.$ Summing these equalities yields $a_{12}-a_1=12+11+\cdots+2$. So $a_{12}=12+11+\cdots+2+1=\dfrac{12(12+1)}{2}=78.$ OR We have $a_2=a_{1+1}=a_1+a_1+1\cdot1=1+1+1=3,$ $a_3=a_{2+1}=a_2+a_1+2\cdot1=3+1+2=6,$ $a_6=a_{3+3}=a_3+a_3+3\cdot3=6+6+9=21,$ and $a_{12}=a_{6+6}=a_6+a_6+6\cdot6=21+21+36=78.$
（D）在给定的递推式中令 $n=1$，得到对所有正整数 $m$，有 $a_{m+1}=a_m+a_1+m$。因此对每个 $m=1,2,3,\ldots$，有 $a_{m+1}-a_m=m+1$。于是， $a_{12}-a_{11}=12,\ a_{11}-a_{10}=11,\ \ldots,\ a_2-a_1=2。$ 把这些等式相加得 $a_{12}-a_1=12+11+\cdots+2$。所以 $a_{12}=12+11+\cdots+2+1=\dfrac{12(12+1)}{2}=78。$ 或者 我们有 $a_2=a_{1+1}=a_1+a_1+1\cdot1=1+1+1=3，$ $a_3=a_{2+1}=a_2+a_1+2\cdot1=3+1+2=6，$ $a_6=a_{3+3}=a_3+a_3+3\cdot3=6+6+9=21，$ 并且 $a_{12}=a_{6+6}=a_6+a_6+6\cdot6=21+21+36=78。$
Q24
Riders on a Ferris wheel travel in a circle in a vertical plane. A particular wheel has radius 20 feet and revolves at the constant rate of one revolution per minute. How many seconds does it take a rider to travel from the bottom of the wheel to a point 10 vertical feet above the bottom?
摩天轮上的乘客在垂直平面内沿圆周运动。该摩天轮半径为 20 英尺，每分钟匀速转一圈。乘客从轮底到轮底上方 10 英尺垂直高度处需要多少秒？
Correct Answer: D
(D) In the figure, the center of the wheel is at $O$, and the rider travels from $A$ to $B$. Since $AC = 10$ and $OB = OA = 20$, the point $C$ is the midpoint of $\overline{OA}$. In the right $\triangle OCB$, we have $OC$ half of the length of the hypotenuse $OB$, so $m\angle COB = 60^\circ$. Since the wheel turns through an angle of $360^\circ$ in 60 seconds, the time required to turn through an angle of $60^\circ$ is $$ 60\left(\frac{60}{360}\right)=10 \text{ seconds.} $$
（D）在图中，车轮的圆心在 $O$，骑手从 $A$ 移动到 $B$。由于 $AC = 10$ 且 $OB = OA = 20$，点 $C$ 是线段 $\overline{OA}$ 的中点。在直角三角形 $\triangle OCB$ 中，$OC$ 是斜边 $OB$ 长度的一半，因此 $m\angle COB = 60^\circ$。由于车轮在 60 秒内转过 $360^\circ$，转过 $60^\circ$ 所需的时间为 $$ 60\left(\frac{60}{360}\right)=10 \text{秒。} $$
solution
Q25
When 15 is appended to a list of integers, the mean is increased by 2. When 1 is appended to the enlarged list, the mean of the enlarged list is decreased by 1. How many integers were in the original list?
将 15 添加到整数列表后，均值增加 2。将 1 添加到扩大列表后，扩大列表的均值减少 1。原列表中有多少个整数？
Correct Answer: A
(A) Let $n$ denote the number of integers in the original list, and $m$ the original mean. Then the sum of the original numbers is $mn$. After $15$ is appended to the list, we have the sum $$(m+2)(n+1)=mn+15,\quad \text{so}\quad m+2n=13.$$ After $1$ is appended to the enlarged list, we have the sum $$(m+1)(n+2)=mn+16,\quad \text{so}\quad 2m+n=14.$$ Solving $m+2n=13$ and $2m+n=14$ gives $m=5$ and $n=4$.
（A）设 $n$ 表示原始列表中整数的个数，$m$ 表示原始平均数。则原始数的总和为 $mn$。将 $15$ 添加到列表后，总和为 $$(m+2)(n+1)=mn+15,\quad \text{因此}\quad m+2n=13.$$ 在扩大的列表中再添加 $1$ 后，总和为 $$(m+1)(n+2)=mn+16,\quad \text{因此}\quad 2m+n=14.$$ 解方程组 $m+2n=13$ 与 $2m+n=14$，得 $m=5$、$n=4$。