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AMC10 2003 A

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AMC10 · 2003 (A)

Q1
What is the difference between the sum of the first 2003 even counting numbers and the sum of the first 2003 odd counting numbers?
前2003个偶数计数数的和与前2003个奇数计数数的和的差是多少？
Correct Answer: D
Each even counting number, beginning with 2, is one more than the preceding odd counting number. Therefore the difference is $(1)(2003) = 2003$.
每个偶数计数数，从2开始，比前一个奇数计数数多1。因此差值为$(1)(2003) = 2003$。
Q2
Members of the Rockham Soccer League buy socks and T-shirts. Socks cost \$4 per pair and each T-shirt costs \$5 more than a pair of socks. Each member needs one pair of socks and a shirt for home games and another pair of socks and a shirt for away games. If the total cost is \$2366, how many members are in the League?
Rockham足球联盟的成员购买袜子和T恤。一双袜子4美元，每件T恤比一双袜子贵5美元。每位成员需要一双袜子和一件用于主场比赛的球衣，以及另一双袜子和一件用于客场比赛的球衣。如果总花费为2366美元，联盟有多少成员？
Correct Answer: B
(B) The cost for each member is the price of two pairs of socks, \$8, and two shirts, \$18, for a total of \$26. So there are $2366/26 = 91$ members.
（B）每位成员的费用是两双袜子，\$8，以及两件衬衫，\$18，总计\$26。因此人数为 $2366/26 = 91$ 人。
Q3
A solid box is 15 cm by 10 cm by 8 cm. A new solid is formed by removing a cube 3 cm on a side from each corner of this box. What percent of the original volume is removed?
一个实心盒子尺寸为15 cm × 10 cm × 8 cm。从盒子每个角切除一个边长3 cm的立方体，形成一个新实心体。切除的体积占原体积的百分之多少？
Correct Answer: D
(D) The total volume of the eight removed cubes is $8 \times 3^3 = 216$ cubic centimeters, and the volume of the original box is $15 \times 10 \times 8 = 1200$ cubic centimeters. Therefore the volume has been reduced by $\left(\frac{216}{1200}\right)(100\%) = 18\%$.
（D）被移除的 8 个小立方体的总体积为 $8 \times 3^3 = 216$ 立方厘米，原来的长方体盒子的体积为 $15 \times 10 \times 8 = 1200$ 立方厘米。因此体积减少了 $\left(\frac{216}{1200}\right)(100\%) = 18\%$。
Q4
It takes Mary 30 minutes to walk uphill 1 km from her home to school, but it takes her only 10 minutes to walk from school to home along the same route. What is her average speed, in km/hr, for the round trip?
Mary上坡从家走到学校1 km需要30分钟，但从学校沿同一路线走回家只需10分钟。她的往返平均速度是多少km/hr？
Correct Answer: A
Mary walks a total of 2 km in 40 minutes. Because 40 minutes is $\frac{2}{3}$ hr, her average speed, in km/hr, is $\frac{2}{\frac{2}{3}} = 3$.
Mary总共走2 km，用时40分钟。因为40分钟是$\frac{2}{3}$小时，她的平均速度为$\frac{2}{\frac{2}{3}} = 3$ km/hr。
Q5
Let $d$ and $e$ denote the solutions of $2x^2 + 3x -5 = 0$. What is the value of $(d -1)(e -1)$?
设$d$和$e$是方程$2x^2 + 3x -5 = 0$的根。求$(d -1)(e -1)$的值。
Correct Answer: B
(B) Since $0 = 2x^2 + 3x - 5 = (2x + 5)(x - 1)$ we have $d = -\frac{5}{2}$ and $e = 1$. So $(d - 1)(e - 1) = 0$.
（B）由于 $0 = 2x^2 + 3x - 5 = (2x + 5)(x - 1)$，我们有 $d = -\frac{5}{2}$ 且 $e = 1$。 所以 $(d - 1)(e - 1) = 0$。
Q6
Define $x\heartsuit y$ to be $|x - y|$ for all real numbers $x$ and $y$. Which of the following statements is not true?
定义对于所有实数 $x$ 和 $y$，$x\heartsuit y$ 为 $|x - y|$。以下哪个陈述不正确？
Correct Answer: C
(C) For example, \(-1 \heartsuit 0 = |-1-0| = 1 \ne -1\). All the other statements are true: (A) \(x \heartsuit y = |x-y| = |-(y-x)| = |y-x| = y \heartsuit x\) for all \(x\) and \(y\). (B) \(2(x \heartsuit y) = 2|x-y| = |2x-2y| = (2x) \heartsuit (2y)\) for all \(x\) and \(y\). (D) \(x \heartsuit x = |x-x| = 0\) for all \(x\). (E) \(x \heartsuit y = |x-y| > 0\) if \(x \ne y\).
（C）例如，\(-1 \heartsuit 0 = |-1-0| = 1 \ne -1\)。其他所有陈述都为真： （A）对所有 \(x\) 和 \(y\)，有 \(x \heartsuit y = |x-y| = |-(y-x)| = |y-x| = y \heartsuit x\)。 （B）对所有 \(x\) 和 \(y\)，有 \(2(x \heartsuit y) = 2|x-y| = |2x-2y| = (2x) \heartsuit (2y)\)。 （D）对所有 \(x\)，有 \(x \heartsuit x = |x-x| = 0\)。 （E）若 \(x \ne y\)，则 \(x \heartsuit y = |x-y| > 0\)。
Q7
How many non-congruent triangles with perimeter 7 have integer side lengths?
周长为 7 的非全等整数边三角形有多少个？
Correct Answer: B
The longest side cannot be greater than 3, since otherwise the remaining two sides would not be long enough to form a triangle. The only possible triangles have side lengths 1–3–3 or 2–2–3.
最长边不能大于 3，否则其余两边不足以形成三角形。唯一可能的三角形边长为 1–3–3 或 2–2–3。
Q8
What is the probability that a randomly drawn positive factor of 60 is less than 7?
随机抽取 60 的一个正因数小于 7 的概率是多少？
Correct Answer: E
(E) The factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60. Six of the twelve factors are less than 7, so the probability is 1/2.
（E）60 的因数是 1，2，3，4，5，6，10，12，15，20，30 和 60。 12 个因数中有 6 个小于 7，所以概率是 $1/2$。
Q9
Simplify $\sqrt[3]{\sqrt[3]{x^3\sqrt[3]{x\sqrt{x}}}}$ .
化简 $\sqrt[3]{\sqrt[3]{x^3\sqrt[3]{x\sqrt{x}}}}$ 。
Correct Answer: A
(A) We have $$\sqrt[3]{x\sqrt[3]{x\sqrt[3]{x\sqrt{x}}}}=\left(x\left(x\left(x\cdot x^{\frac12}\right)^{\frac13}\right)^{\frac13}\right)^{\frac13}$$ $$=\left(x\left(x\left(x^{\frac32}\right)^{\frac13}\right)^{\frac13}\right)^{\frac13}$$ $$=\left(x\left(x\cdot x^{\frac12}\right)^{\frac13}\right)^{\frac13}$$ $$=\left(x\left(x^{\frac32}\right)^{\frac13}\right)^{\frac13}=\left(x\cdot x^{\frac12}\right)^{\frac13}=\left(x^{\frac32}\right)^{\frac13}=x^{\frac12}=\sqrt{x}.$$
(A) 我们有 $$\sqrt[3]{x\sqrt[3]{x\sqrt[3]{x\sqrt{x}}}}=\left(x\left(x\left(x\cdot x^{\frac12}\right)^{\frac13}\right)^{\frac13}\right)^{\frac13}$$ $$=\left(x\left(x\left(x^{\frac32}\right)^{\frac13}\right)^{\frac13}\right)^{\frac13}$$ $$=\left(x\left(x\cdot x^{\frac12}\right)^{\frac13}\right)^{\frac13}$$ $$=\left(x\left(x^{\frac32}\right)^{\frac13}\right)^{\frac13}=\left(x\cdot x^{\frac12}\right)^{\frac13}=\left(x^{\frac32}\right)^{\frac13}=x^{\frac12}=\sqrt{x}.$$
Q10
The polygon enclosed by the solid lines in the figure consists of 4 congruent squares joined edge-to-edge. One more congruent square is attached to an edge at one of the nine positions indicated. How many of the nine resulting polygons can be folded to form a cube with one face missing?
图中实线围成的多边形由 4 个全等的正方形边对边连接而成。再在 9 个指示位置之一的边上附加一个全等的正方形。其中 9 个结果多边形有多少个可以折叠成缺一面立方体？
stem
Correct Answer: E
(E) If the polygon is folded before the fifth square is attached, then edges $a$ and $a'$ must be joined, as must $b$ and $b'$. The fifth face of the cube can be attached at any of the six remaining edges.
（E）如果在连接第五个正方形之前先把多边形折起来，那么边 $a$ 和 $a'$ 必须相接，边 $b$ 和 $b'$ 也必须相接。立方体的第五个面可以连接在剩下的六条边中的任意一条上。
solution
Q11
The sum of the two 5-digit numbers AMC10 and AMC12 is 123422. What is A + M + C?
两个5位数 AMC10 和 AMC12 的和是 123422。A + M + C 的值是多少？
Correct Answer: E
(E) Since the last two digits of $AMC10$ and $AMC12$ sum to 22, we have $AMC + AMC = 2(AMC) = 1234.$ Hence $AMC = 617$, so $A = 6$, $M = 1$, $C = 7$, and $A + M + C = 6 + 1 + 7 = 14$.
（E）由于 $AMC10$ 和 $AMC12$ 的后两位数字之和为 22，我们有 $AMC + AMC = 2(AMC) = 1234.$ 因此 $AMC = 617$，所以 $A = 6$，$M = 1$，$C = 7$，并且 $A + M + C = 6 + 1 + 7 = 14$。
Q12
A point $(x, y)$ is randomly picked from inside the rectangle with vertices $(0, 0)$, $(4, 0)$, $(4, 1)$, and $(0, 1)$. What is the probability that $x < y$?
从矩形内部随机选取一点 $(x, y)$，该矩形的顶点为 $(0, 0)$、$(4, 0)$、$(4, 1)$ 和 $(0, 1)$。$x < y$ 的概率是多少？
Correct Answer: A
(A) The point $(x,y)$ satisfies $x<y$ if and only if it belongs to the shaded triangle bounded by the lines $x=y$, $y=1$, and $x=0$, the area of which is $1/2$. The ratio of the area of the triangle to the area of the rectangle is $\frac{1/2}{4}=\frac{1}{8}$.
（A）点$(x,y)$满足$x<y$当且仅当它位于由直线$x=y$、$y=1$和$x=0$围成的阴影三角形内，该三角形的面积为$1/2$。三角形面积与矩形面积之比为$\frac{1/2}{4}=\frac{1}{8}$。
solution
Q13
The sum of three numbers is 20. The first is 4 times the sum of the other two. The second is seven times the third. What is the product of all three?
三个数的和是 20。第一数是另外两个数之和的 4 倍。第二数是第三数的 7 倍。三者乘积是多少？
Correct Answer: A
(A) Let $a$, $b$, and $c$ be the three numbers. Replace $a$ by four times the sum of the other two to get $4(b+c)+b+c=20$, so $b+c=4$. Then replace $b$ with $7c$ to get $7c+c=4$, so $c=\frac{1}{2}$. The other two numbers are $b=7/2$ and $a=16$, and the product of the three is $16\cdot 7/2\cdot 1/2=28$.
（A）设 $a$、$b$、$c$ 为这三个数。把 $a$ 替换为另外两个数之和的 4 倍，得到 $4(b+c)+b+c=20$，所以 $b+c=4$。 再把 $b$ 替换为 $7c$，得到 $7c+c=4$，所以 $c=\frac{1}{2}$。 另外两个数是 $b=7/2$ 和 $a=16$，三个数的乘积为 $16\cdot 7/2\cdot 1/2=28$。
Q14
Let $n$ be the largest integer that is the product of exactly 3 distinct prime numbers, $d$, $e$ and $10d + e$, where $d$ and $e$ are single digits. What is the sum of the digits of $n$?
设 $n$ 是恰为 3 个不同素数 $d$、$e$ 和 $10d + e$ 的乘积的最大整数，其中 $d$ 和 $e$ 是单个数字。$n$ 的各位数字之和是多少？
Correct Answer: A
The largest single-digit primes are 5 and 7, but neither 75 nor 57 is prime. Using 3, 7, and 73 gives 1533, whose digits have a sum of 12.
最大的个位素数是 5 和 7，但 75 和 57 都不是素数。使用 3、7 和 73 得到 1533，其各位数字之和为 12。
Q15
What is the probability that an integer in the set $\lbrace 1, 2, 3, \dots , 100\rbrace$ is divisible by 2 and not divisible by 3?
集合 $\lbrace 1, 2, 3, \dots , 100\rbrace$ 中的整数能被 2 整除且不能被 3 整除的概率是多少？
Correct Answer: C
(C) Of the $\frac{100}{2}=50$ integers that are divisible by $2$, there are $\left\lfloor\frac{100}{6}\right\rfloor=16$ that are divisible by both $2$ and $3$. So there are $50-16=34$ that are divisible by $2$ and not by $3$, and $\frac{34}{100}=\frac{17}{50}$.
（C）在能被 $2$ 整除的 $\frac{100}{2}=50$ 个整数中，有 $\left\lfloor\frac{100}{6}\right\rfloor=16$ 个同时能被 $2$ 和 $3$ 整除。因此，能被 $2$ 整除但不能被 $3$ 整除的有 $50-16=34$ 个，并且 $\frac{34}{100}=\frac{17}{50}$。
Q16
What is the units digit of $13^{2003}$?
$13^{2003}$ 的单位数字是多少？
Correct Answer: C
(C) Powers of 13 have the same units digit as the corresponding powers of 3; and $3^1=3,\quad 3^2=9,\quad 3^3=27,\quad 3^4=81,$ and $3^5=243.$ Since the units digit of $3^1$ is the same as the units digit of $3^5$, units digits of powers of 3 cycle through $3,9,7,$ and $1$. Hence the units digit of $3^{2000}$ is $1$, so the units digit of $3^{2003}$ is $7$. The same is true of the units digit of $13^{2003}$.
（C）$13$ 的幂与对应的 $3$ 的幂具有相同的个位数字；并且 $3^1=3,\quad 3^2=9,\quad 3^3=27,\quad 3^4=81,$ 且 $3^5=243.$ 由于 $3^1$ 的个位数与 $3^5$ 的个位数相同，$3$ 的各次幂的个位数按 $3,9,7,1$ 循环。因此 $3^{2000}$ 的个位数是 $1$，所以 $3^{2003}$ 的个位数是 $7$。$13^{2003}$ 的个位数也同样如此。
Q17
The number of inches in the perimeter of an equilateral triangle equals the number of square inches in the area of its circumscribed circle. What is the radius, in inches, of the circle?
一个正三角形的周长（英寸数）等于其外接圆面积（平方英寸数）。求该圆的半径（英寸）。
Correct Answer: B
(B) Let the triangle have vertices $A$, $B$, and $C$, let $O$ be the center of the circle, and let $D$ be the midpoint of $\overline{BC}$. Triangle $COD$ is a 30-60-90 degree triangle. If $r$ is the radius of the circle, then the sides of $\triangle COD$ are $r$, $r/2$, and $r\sqrt{3}/2$. The perimeter of $\triangle ABC$ is $6\left(r\sqrt{3}/2\right)=3r\sqrt{3}$, and the area of the circle is $\pi r^2$. Thus $3r\sqrt{3}=\pi r^2$, and $r=(3\sqrt{3})/\pi$.
（B）设三角形的顶点为 $A$、$B$、$C$，$O$ 为圆心，$D$ 为线段 $\overline{BC}$ 的中点。三角形 $COD$ 是一个 $30^\circ\!-\!60^\circ\!-\!90^\circ$ 三角形。若 $r$ 为圆的半径，则 $\triangle COD$ 的三边分别为 $r$、$r/2$、$r\sqrt{3}/2$。$\triangle ABC$ 的周长为 $6\left(r\sqrt{3}/2\right)=3r\sqrt{3}$，圆的面积为 $\pi r^2$。因此 $3r\sqrt{3}=\pi r^2$，从而 $r=(3\sqrt{3})/\pi$。
solution
Q18
What is the sum of the reciprocals of the roots of the equation $\frac{2003}{2004x + 1} + \frac{1}{x} = 0$?
方程 $\frac{2003}{2004x + 1} + \frac{1}{x} = 0$ 的根的倒数之和是多少？
Correct Answer: B
(B) Let $a=2003/2004$. The given equation is equivalent to $ax^2+x+1=0$. If the roots of this equation are denoted $r$ and $s$, then $rs=\frac{1}{a}$ and $r+s=-\frac{1}{a}$, so $\frac{1}{r}+\frac{1}{s}=\frac{r+s}{rs}=-1$.
（B）令 $a=2003/2004$。所给方程等价于 $ax^2+x+1=0$。 若此方程的根记为 $r$ 和 $s$，则 $rs=\frac{1}{a}$ 且 $r+s=-\frac{1}{a}$， 所以 $\frac{1}{r}+\frac{1}{s}=\frac{r+s}{rs}=-1$。
Q19
A semicircle of diameter 1 sits at the top of a semicircle of diameter 2, as shown. The shaded area inside the smaller semicircle and outside the larger semicircle is called a lune. Determine the area of this lune.
直径为 1 的半圆位于直径为 2 的半圆顶部，如图所示。小半圆内、大半圆外的阴影区域称为月形区。求此月形区的面积。
stem
Correct Answer: C
(C) First note that the area of the region determined by the triangle topped by the semicircle of diameter 1 is $\dfrac{1}{2}\cdot\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\pi\left(\dfrac{1}{2}\right)^{2}=\dfrac{\sqrt{3}}{4}+\dfrac{1}{8}\pi.$ The area of the lune results from subtracting from this the area of the sector of the larger semicircle, $\dfrac{1}{6}\pi(1)^{2}=\dfrac{1}{6}\pi.$ So the area of the lune is $\dfrac{\sqrt{3}}{4}+\dfrac{1}{8}\pi-\dfrac{1}{6}\pi=\dfrac{\sqrt{3}}{4}-\dfrac{1}{24}\pi.$
（C）首先注意，由边长为 1 的半圆（直径为 1）覆盖在三角形上所确定的区域面积为 $\dfrac{1}{2}\cdot\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\pi\left(\dfrac{1}{2}\right)^{2}=\dfrac{\sqrt{3}}{4}+\dfrac{1}{8}\pi.$ 月牙形（lune）的面积是从该面积中减去较大半圆的扇形面积得到的，即 $\dfrac{1}{6}\pi(1)^{2}=\dfrac{1}{6}\pi.$ 因此月牙形的面积为 $\dfrac{\sqrt{3}}{4}+\dfrac{1}{8}\pi-\dfrac{1}{6}\pi=\dfrac{\sqrt{3}}{4}-\dfrac{1}{24}\pi.$
solution
Q20
A base-10 three-digit number $n$ is selected at random. Which of the following is closest to the probability that the base-9 representation and the base-11 representation of $n$ are both three-digit numerals?
随机选取一个十进制的三位数 $n$。$n$ 的九进制表示和十一进制表示均为三位数表示法的概率最接近于以下哪一项？
Correct Answer: E
(E) The largest base-9 three-digit number is $9^3-1=728$ and the smallest base-11 three-digit number is $11^2=121$. There are $608$ integers that satisfy $121\le n\le 728$, and $900$ three-digit numbers altogether, so the probability is $608/900\approx 0.7$.
（E）九进制的最大三位数是 $9^3-1=728$，十一进制的最小三位数是 $11^2=121$。满足 $121\le n\le 728$ 的整数有 $608$ 个，而所有三位数一共有 $900$ 个，所以概率是 $608/900\approx 0.7$。
Q21
Pat is to select six cookies from a tray containing only chocolate chip, oatmeal, and peanut butter cookies. There are at least six of each of these three kinds of cookies on the tray. How many different assortments of six cookies can be selected?
Pat 要从一个盘子里选择六块饼干，盘子里只有巧克力碎片、燕麦和花生酱饼干，每种至少有六块。有多少种不同的六块饼干组合可以选择？
Correct Answer: D
(D) The numbers of the three types of cookies must have a sum of six. Possible sets of whole numbers whose sum is six are $0,0,6;\ 0,1,5;\ 0,2,4;\ 0,3,3;\ 1,1,4;\ 1,2,3;\ \text{and}\ 2,2,2.$ Every ordering of each of these sets determines a different assortment of cookies. There are 3 orders for each of the sets $0,0,6;\ 0,3,3;\ \text{and}\ 1,1,4.$ There are 6 orders for each of the sets $0,1,5;\ 0,2,4;\ \text{and}\ 1,2,3.$ There is only one order for $2,2,2$. Therefore the total number of assortments of six cookies is $3\cdot 3+3\cdot 6+1=28$.
（D）三种类型饼干的数量之和必须为 6。所有和为 6 的非负整数集合可以是 $0,0,6;\ 0,1,5;\ 0,2,4;\ 0,3,3;\ 1,1,4;\ 1,2,3;\ \text{以及}\ 2,2,2.$ 上述每个集合的不同排列对应一种不同的饼干组合。以下每个集合各有 3 种排列： $0,0,6;\ 0,3,3;\ \text{以及}\ 1,1,4.$ 以下每个集合各有 6 种排列： $0,1,5;\ 0,2,4;\ \text{以及}\ 1,2,3.$ $2,2,2$ 只有一种排列。因此，6 块饼干的组合总数为 $3\cdot 3+3\cdot 6+1=28$。
Q22
In rectangle ABCD, we have AB = 8, BC = 9, H is on BC with BH = 6, E is on AD with DE = 4, line EC intersects line AH at G, and F is on line AD with GF $\perp$ AF. Find the length GF.
在矩形 ABCD 中，AB = 8，BC = 9，H 在 BC 上且 BH = 6，E 在 AD 上且 DE = 4，直线 EC 与直线 AH 相交于 G，F 在直线 AD 上且 GF $\perp$ AF。求 GF 的长度。
stem
Correct Answer: B
(B) We have \(EA = 5\) and \(CH = 3\). Triangles \(GCH\) and \(GEA\) are similar, so \[ \frac{GC}{GE}=\frac{3}{5} \quad\text{and}\quad \frac{CE}{GE}=\frac{GE-GC}{GE}=1-\frac{3}{5}=\frac{2}{5}. \] Triangles \(GFE\) and \(CDE\) are similar, so \[ \frac{GF}{8}=\frac{CE}{GE}=\frac{5}{2} \] and \(FG=20\).
（B）我们有 \(EA = 5\) 且 \(CH = 3\)。三角形 \(GCH\) 与 \(GEA\) 相似，所以 \[ \frac{GC}{GE}=\frac{3}{5} \quad\text{并且}\quad \frac{CE}{GE}=\frac{GE-GC}{GE}=1-\frac{3}{5}=\frac{2}{5}. \] 三角形 \(GFE\) 与 \(CDE\) 相似，所以 \[ \frac{GF}{8}=\frac{CE}{GE}=\frac{5}{2} \] 且 \(FG=20\)。
Q23
A large equilateral triangle is constructed by using toothpicks to create rows of small equilateral triangles. For example, in the figure we have 3 rows of small congruent equilateral triangles, with 5 small triangles in the base row. How many toothpicks would be needed to construct a large equilateral triangle if the base row of the triangle consists of 2003 small equilateral triangles?
一个大正三角形由牙签构成小正三角形的行来构建。例如，图中 3 行小正三角形，底行有 5 个小三角形。如果底行由 2003 个小正三角形构成，需要多少牙签来构建这个大正三角形？
stem
Correct Answer: C
(C) The base row of the large equilateral triangle has 1001 triangles pointing downward and 1002 pointing upward. This base row requires $3(1002)$ toothpicks since the downward pointing triangles require no additional toothpicks. Each succeeding row will require one less set of $3$ toothpicks, so the total number of toothpicks required is $3(1002 + 1001 + 1000 + \cdots + 2 + 1)=3\cdot\frac{1002\cdot1003}{2}=1,507,509.$
（C）这个大等边三角形的底边一行有 1001 个倒三角形和 1002 个正三角形。由于倒三角形不需要额外的牙签，这一行需要 $3(1002)$ 根牙签。之后每一行所需的牙签组数都会少一组 $3$ 根牙签，因此所需牙签总数为 $3(1002 + 1001 + 1000 + \cdots + 2 + 1)=3\cdot\frac{1002\cdot1003}{2}=1,507,509.$
solution
Q24
Sally has five red cards numbered 1 through 5 and four blue cards numbered 3 through 6. She stacks the cards so that the colors alternate and so that the number on each red card divides evenly into the number on each neighboring blue card. What is the sum of the numbers on the middle three cards?
Sally 有五张红牌，编号 1 到 5，和四张蓝牌，编号 3 到 6。她将牌堆叠，使得颜色交替，并且每张红牌的数字整除相邻蓝牌的数字。中间三张牌的数字之和是多少？
Correct Answer: E
(E) Let R1, ..., R5 and B3, ..., B6 denote the numbers on the red and blue cards, respectively. Note that R4 and R5 divide evenly into only B4 and B5, respectively. Thus the stack must be R4, B4, ..., B5, R5, or the reverse. Since R2 divides evenly into only B4 and B6, we must have R4, B4, R2, B6, ..., B5, R5, or the reverse. Since R3 divides evenly into only B3 and B6, the stack must be R4, B4, R2, B6, R3, B3, R1, B5, R5, or the reverse. In either case, the sum of the middle three cards is 12.
（E）令 R1，…，R5 和 B3，…，B6 分别表示红色和蓝色卡片上的数字。注意到 R4 和 R5 分别只能整除 B4 和 B5。因此，这一叠必须是 R4，B4，…，B5，R5，或其逆序。由于 R2 只能整除 B4 和 B6，我们必须有 R4，B4，R2，B6，…，B5，R5，或其逆序。由于 R3 只能整除 B3 和 B6，这一叠必须是 R4，B4，R2，B6，R3，B3，R1，B5，R5，或其逆序。无论哪种情况，中间三张卡片的和都是 12。
Q25
Let $n$ be a 5-digit number, and let $q$ and $r$ be the quotient and remainder, respectively, when $n$ is divided by 100. For how many values of $n$ is $q + r$ divisible by 11?
设 $n$ 是一个五位数，当 $n$ 除以 100 得到商 $q$ 和余数 $r$。有几个 $n$ 使得 $q + r$ 能被 11 整除？
Correct Answer: B
(B) Note that $n = 100q + r = q + r + 99q$. Hence $q + r$ is divisible by $11$ if and only if $n$ is divisible by $11$. Since $10,000 \le n \le 99,999$, there are $$ \left\lfloor \frac{99999}{11} \right\rfloor - \left\lfloor \frac{9999}{11} \right\rfloor = 9090 - 909 = 8181 $$ such numbers.
（B）注意 $n = 100q + r = q + r + 99q$。因此，当且仅当 $n$ 能被 $11$ 整除时，$q + r$ 才能被 $11$ 整除。由于 $10,000 \le n \le 99,999$，这样的数有 $$ \left\lfloor \frac{99999}{11} \right\rfloor - \left\lfloor \frac{9999}{11} \right\rfloor = 9090 - 909 = 8181 $$ 个。