2003B AMC 10 真题（P版）

1 答案 C

因式分解得 \[\dfrac{2(1-2+3-4+5-6+7)}{3(1-2+3-4+5-6+7)}\text{。}\]分子、分母中相同的交错和约去后，剩下 \(\dfrac{2}{3}\)。

所以正确答案是 C。

English

Factoring gives \[\dfrac{2(1-2+3-4+5-6+7)}{3(1-2+3-4+5-6+7)}.\] The identical alternating sums cancel, leaving \(\dfrac{2}{3}.\)

Thus, the correct answer is C.

2 答案 D

每天药丸费用为 \(546 \div 14 = 39\) 美元。设绿色药丸价格为 \(x\)，则粉色药丸价格为 \(x-1\)，所以 \(x+(x-1)=39\)，解得 \(x=20\)。

所以正确答案是 D。

English

Each day’s pills cost \(546 \div 14 = 39\) dollars. If \(x\) is the cost of a green pill, then the pink pill costs \(x-1,\) so \(x+(x-1)=39.\) Solving gives \(x=20.\)

Thus, the correct answer is D.

3 答案 B

前 \(8\) 个正奇数的和是 \(1+3+\cdots+15=64\)。

设最小偶数为 \(n\)，则 \[ \begin{gathered} n+(n+2)+(n+4) \\ {}+(n+6)+(n+8) \\ = 5n+20 = 60 \end{gathered}\text{，}\]因此 \(n=8\)。

所以正确答案是 B。

English

The first \(8\) odd counting numbers sum to \(1+3+\cdots+15=64.\)

Letting \(n\) be the smallest even integer, \[ \begin{gathered} n+(n+2)+(n+4) \\ {}+(n+6)+(n+8) \\ = 5n+20 = 60, \end{gathered} \] so \(n=8.\)

Thus, the correct answer is B.

4 答案 A

五个区域的面积分别为 \(4\)、\(6\)、\(15\)、\(20\) 和 \(21\) 平方英尺。

要使费用最小，应把最贵的花种在最小的区域，所以总费用为 \[ \begin{gathered} (3)(4)+(2.5)(6)+(2)(15) \\ {}+(1.5)(20)+(1)(21) \\ = 108 \end{gathered}\text{。}\]

所以正确答案是 A。

English

The five regions have areas \(4,\) \(6,\) \(15,\) \(20,\) and \(21\) square feet.

Cost is minimized by placing the most expensive flower in the smallest region, so the total is \[ \begin{gathered} (3)(4)+(2.5)(6)+(2)(15) \\ {}+(1.5)(20)+(1)(21) \\ = 108. \end{gathered} \]

Thus, the correct answer is A.

5 答案 C

草坪面积为 \(90 \cdot 150 = 13{,}500\) 平方英尺。

莫每走一英尺，有效割草宽度为 \(2\) 英尺，所以每小时割草面积为 \(2 \cdot 5000 = 10{,}000\) 平方英尺。因此所需时间为 \[\dfrac{13{,}500}{10{,}000}=1.35\] 小时。

所以正确答案是 C。

English

The lawn has area \(90 \cdot 150 = 13{,}500\) square feet.

Each foot Moe walks mows an effective strip \(2\) feet wide, so he mows \(2 \cdot 5000 = 10{,}000\) square feet per hour. The time needed is \[\dfrac{13{,}500}{10{,}000}=1.35\] hours.

Thus, the correct answer is C.

6 答案 D

因为长高比为 \(4:3\)，长、高、对角线组成 \(4:3:5\) 的直角三角形。对角线长为 \(27\)，所以水平长度为 \[\dfrac{4}{5}(27)=21.6\text{，}\]最接近 \(21.5\)。

所以正确答案是 D。

English

Since the length and height are in ratio \(4:3,\) the length, height, and diagonal form a \(4:3:5\) right triangle. The diagonal is \(27,\) so the horizontal length is \[\dfrac{4}{5}(27)=21.6,\] which is closest to \(21.5.\)

Thus, the correct answer is D.

7 答案 B

当 \(n=1,2,3\) 时，取整值为 \(1\)；当 \(n=4,\ldots,8\) 时，取整值为 \(2\)；当 \(n=9,\ldots,15\) 时，取整值为 \(3\)；当 \(n=16\) 时，取整值为 \(4\)。因此总和为 \[3\cdot 1 + 5\cdot 2 + 7\cdot 3 + 1\cdot 4 = 38\text{。}\]

所以正确答案是 B。

English

The value is \(1\) for \(n=1,2,3;\) it is \(2\) for \(n=4,\ldots,8;\) it is \(3\) for \(n=9,\ldots,15;\) and it is \(4\) for \(n=16.\) The sum is \[3\cdot 1 + 5\cdot 2 + 7\cdot 3 + 1\cdot 4 = 38.\]

Thus, the correct answer is B.

8 答案 B

设数列为 \(a\)、\(ar\)、\(ar^2\)、\(ar^3\)、\(\dots\)，其中 \(ar=2\) 且 \(ar^3=6\)。相除得 \(r^2=3\)，所以 \(r=\pm\sqrt3\)。

于是 \(a=\dfrac{2}{r}=\pm\dfrac{2}{\sqrt3}=\pm\dfrac{2\sqrt3}{3}\)。选项中对应的负值是 \(-\dfrac{2\sqrt3}{3}\)。

所以正确答案是 B。

English

Let the terms be \(a,\) \(ar,\) \(ar^2,\) \(ar^3,\) \(\dots\) with \(ar=2\) and \(ar^3=6.\) Dividing gives \(r^2=3,\) so \(r=\pm\sqrt3.\)

Then \(a=\dfrac{2}{r}=\pm\dfrac{2}{\sqrt3}=\pm\dfrac{2\sqrt3}{3}.\) The choice \(-\dfrac{2\sqrt3}{3}\) matches the negative case.

Thus, the correct answer is B.

9 答案 B

全部写成以 \(5\) 为底的幂，左边是 \(5^{-4}\)，右边是 \[5^{\frac{48-26-34}{x}}=5^{-\frac{12}{x}}\text{。}\]令指数相等，得到 \(-4=-\dfrac{12}{x}\)，所以 \(x=3\)。

所以正确答案是 B。

English

Writing everything base \(5,\) the left side is \(5^{-4}\) and the right side is \[5^{\frac{48-26-34}{x}}=5^{-\frac{12}{x}}.\] Setting exponents equal, \(-4=-\dfrac{12}{x},\) so \(x=3.\)

Thus, the correct answer is B.

10 答案 C

旧方案有 \(26 \cdot 10^4\) 种车牌，新方案有 \(26^3 \cdot 10^3\) 种。增加倍数为 \[\dfrac{26^3 \cdot 10^3}{26 \cdot 10^4}=\dfrac{26^2}{10}\text{。}\]

所以正确答案是 C。

English

The old scheme allows \(26 \cdot 10^4\) plates and the new scheme allows \(26^3 \cdot 10^3\) plates. The increase factor is \[\dfrac{26^3 \cdot 10^3}{26 \cdot 10^4}=\dfrac{26^2}{10}.\]

Thus, the correct answer is C.

11 答案 A

两条直线分别为 \(y-15=3(x-10)\) 和 \(y-15=5(x-10)\)。

令 \(y=0\)，得到 \(x\) 轴截距分别为 \(x=5\) 和 \(x=7\)。两点 \((5,0)\) 与 \((7,0)\) 的距离为 \(2\)。

所以正确答案是 A。

English

The lines are \(y-15=3(x-10)\) and \(y-15=5(x-10).\)

Setting \(y=0\) gives \(x\)-intercepts \(x=5\) and \(x=7.\) The distance between \((5,0)\) and \((7,0)\) is \(2.\)

Thus, the correct answer is A.

12 答案 C

设三人的原始金额分别为 \(a\)、\(b\) 和 \(c\)。则 \(a+b+c=1000\)，且 \((a-100)+2(b+c)=1500\)。

将 \(b+c=1000-a\) 代入第二个方程，得到 \[a-100+2(1000-a)=1500\text{，}\]所以 \(a=400\)。

所以正确答案是 C。

English

Let \(a,\) \(b,\) and \(c\) be the original portions. Then \(a+b+c=1000\) and \((a-100)+2(b+c)=1500.\)

Substituting \(b+c=1000-a\) into the second equation, \[a-100+2(1000-a)=1500,\] so \(a=400.\)

Thus, the correct answer is C.

13 答案 E

令 \(y=\clubsuit(x)\)。因为 \(x\le 99\)，所以 \(y\le 18\)；要满足 \(\clubsuit(y)=3\)，只能有 \(y=3\) 或 \(y=12\)。

数字和为 \(3\) 的两位数是 \(12\)、\(21\) 和 \(30\)（共 \(3\) 个）。数字和为 \(12\) 的两位数是 \(39\)、\(48\)、\(57\)、\(66\)、\(75\)、\(84\) 和 \(93\)（共 \(7\) 个）。总共有 \(10\) 个。

所以正确答案是 E。

English

Let \(y=\clubsuit(x).\) Since \(x\le 99,\) we have \(y\le 18,\) so \(\clubsuit(y)=3\) forces \(y=3\) or \(y=12.\)

The two-digit numbers with digit sum \(3\) are \(12,\) \(21,\) and \(30\) (\(3\) of them). Those with digit sum \(12\) are \(39,\) \(48,\) \(57,\) \(66,\) \(75,\) \(84,\) and \(93\) (\(7\) of them). In all there are \(10.\)

Thus, the correct answer is E.

14 答案 D

因为 \(a\) 必须含因子 \(5\)，而 \(3^8 \cdot 5^2\) 被 \(5^2\) 整除但不被 \(5^3\) 整除，所以 \(b\le 2\)。

取 \(b=2\)，得 \(a=\sqrt{3^8 \cdot 5^2}=3^4 \cdot 5=405\)，因此 \(a+b=407\)。若 \(b=1\)，则 \(a+b=164{,}026\)，更大。

所以正确答案是 D。

English

Because \(a\) must be divisible by \(5,\) and \(3^8 \cdot 5^2\) is divisible by \(5^2\) but not \(5^3,\) we need \(b\le 2.\)

Taking \(b=2\) gives \(a=\sqrt{3^8 \cdot 5^2}=3^4 \cdot 5=405,\) so \(a+b=407.\) This beats \(b=1,\) which gives \(a+b=164{,}026.\)

Thus, the correct answer is D.

15 答案 E

每场比赛淘汰一名选手。共有 \(100\) 名选手开始，除冠军外都被淘汰，所以有 \(99\) 场比赛。

因为 \(99=9 \cdot 11\)，所以比赛场数能被 \(11\) 整除，而其他选项都不满足。

所以正确答案是 E。

English

Each match eliminates exactly one player. Since \(100\) players start and all but the champion are eliminated, there are \(99\) matches.

Because \(99=9 \cdot 11,\) it is divisible by \(11\) but satisfies none of the other options.

Thus, the correct answer is E.

16 答案 E

若有 \(m\) 种主菜，则前菜种数是主菜的两倍，晚餐总数为 \(3 \cdot m \cdot 2m = 6m^2\)。这个数至少需要达到 \(365\)。

\(m^2 \ge \dfrac{365}{6}\approx 60.8\)。因为 \(7^2=49\) 太小，而 \(8^2=64\) 可行，所以 \(m=8\)。

所以正确答案是 E。

English

With \(m\) main courses, the number of dinners is \(3 \cdot m \cdot 2m = 6m^2.\) This must be at least \(365.\)

So \(m^2 \ge \dfrac{365}{6}\approx 60.8.\) Since \(7^2=49\) is too small but \(8^2=64\) works, \(m=8.\)

Thus, the correct answer is E.

17 答案 B

融化后的体积等于圆锥体积，所以 \[\dfrac34 \cdot \dfrac43 \pi r^3 = \dfrac13 \pi r^2 h\text{。}\]

化简得 \(\pi r^3 = \dfrac13 \pi r^2 h\)，所以 \(h=3r\)，高与半径之比为 \(3:1\)。

所以正确答案是 B。

English

The melted volume equals the cone’s volume, so \[\dfrac34 \cdot \dfrac43 \pi r^3 = \dfrac13 \pi r^2 h.\]

Simplifying gives \(\pi r^3 = \dfrac13 \pi r^2 h,\) so \(h=3r.\) The ratio of height to radius is \(3:1.\)

Thus, the correct answer is B.

18 答案 D

当 \(n\) 为偶数时，这五个因数是连续五个奇数。其中至少有一个能被 \(3\) 整除，且恰有一个能被 \(5\) 整除，所以乘积总能被 \(15\) 整除。

要证明不存在更大的固定公因数，比较以下三种情形：\[ \begin{aligned} n=2 &: 3\cdot5\cdot7\cdot9\cdot11,\\ n=10 &: 11\cdot13\cdot15\cdot17\cdot19,\\ n=12 &: 13\cdot15\cdot17\cdot19\cdot21 \end{aligned}\text{。}\]这三个乘积的最大公因数恰为 \(15\)，所以能整除所有情形的整数不可能更大。

所以正确答案是 D。

English

When \(n\) is even, the factors are five consecutive odd numbers. Among any five consecutive odd numbers, at least one is divisible by \(3\) and exactly one by \(5,\) so the product is always divisible by \(15.\)

To prove that no larger fixed divisor is forced, compare three cases: \[ \begin{aligned} n=2 &: 3\cdot5\cdot7\cdot9\cdot11,\\ n=10 &: 11\cdot13\cdot15\cdot17\cdot19,\\ n=12 &: 13\cdot15\cdot17\cdot19\cdot21. \end{aligned} \] The greatest common divisor of these three products is exactly \(15,\) so a divisor common to every case cannot be any larger.

Thus, the correct answer is D.

19 答案 E

大半圆面积为 \(\dfrac12 \pi (2)^2 = 2\pi\)。

删去的部分等于五个半径为 \(1\) 的 \(60^\circ\) 扇形和两个边长为 \(1\) 的等边三角形；每个扇形面积为 \(\dfrac{\pi}{6}\)，每个等边三角形面积为 \(\dfrac{\sqrt3}{4}\)。

阴影面积为 \[ \begin{gathered} 2\pi - 5 \cdot \dfrac{\pi}{6} - 2 \cdot \dfrac{\sqrt3}{4} \\ = \dfrac{7}{6}\pi - \dfrac{\sqrt3}{2} \end{gathered}\text{。}\]

所以正确答案是 E。

English

The large semicircle has area \(\dfrac12 \pi (2)^2 = 2\pi.\)

Removing the small semicircles deletes a region equal to five congruent \(60^\circ\) sectors of radius \(1\) plus two equilateral triangles of side \(1.\) Each sector has area \(\dfrac{\pi}{6}\) and each triangle has area \(\dfrac{\sqrt3}{4}.\)

The shaded area is \[ \begin{gathered} 2\pi - 5 \cdot \dfrac{\pi}{6} - 2 \cdot \dfrac{\sqrt3}{4} \\ = \dfrac{7}{6}\pi - \dfrac{\sqrt3}{2}. \end{gathered} \]

Thus, the correct answer is E.

20 答案 D

先计算 \(FG = CD - DF - GC\) \(= 5 - 1 - 2\) \(= 2\)。设 \(E\) 到直线 \(CD\) 的距离为 \(h\)。由于 \(\triangle FEG \sim \triangle AEB\)，相似比为 \(\dfrac{FG}{AB}=\dfrac25\)，所以 \[\dfrac{h}{h+3}=\dfrac25\text{，}\]解得 \(h=2\)。

\(\triangle AEB\) 从 \(E\) 到 \(AB\) 的高为 \(h+3=5\)，所以面积为 \(\dfrac12 \cdot 5 \cdot 5 = \dfrac{25}{2}\)。

所以正确答案是 D。

English

Here \(FG = CD - DF - GC\) \(= 5 - 1 - 2\) \(= 2.\) Let \(h\) be the distance from \(E\) down to line \(CD.\) Since \(\triangle FEG \sim \triangle AEB\) with ratio \(\dfrac{FG}{AB}=\dfrac25,\) we have \[\dfrac{h}{h+3}=\dfrac25,\] so \(h=2.\)

The height of \(\triangle AEB\) from \(E\) to \(AB\) is \(h+3=5,\) giving area \(\dfrac12 \cdot 5 \cdot 5 = \dfrac{25}{2}.\)

Thus, the correct answer is D.

21 答案 C

袋中始终有 \(4\) 颗珠子。

最终全为红色，正好表示两颗绿珠都被抽到。先绿后绿的概率为 \(\dfrac24 \cdot \dfrac14 = \dfrac18\)。绿、红、绿的概率为 \(\dfrac24 \cdot \dfrac34 \cdot \dfrac14 = \dfrac{3}{32}\)。红、绿、绿的概率为 \(\dfrac24 \cdot \dfrac24 \cdot \dfrac14 = \dfrac{1}{16}\)。

把三种情况相加，得到 \[\dfrac18 + \dfrac{3}{32} + \dfrac{1}{16} = \dfrac{9}{32}\text{。}\]

所以正确答案是 C。

English

The bag always holds \(4\) beads. All are red at the end precisely when both greens are drawn.

Drawing green then green has probability \(\dfrac24 \cdot \dfrac14 = \dfrac18.\) Green, red, green has probability \(\dfrac24 \cdot \dfrac34 \cdot \dfrac14 = \dfrac{3}{32}.\) Red, green, green has probability \(\dfrac24 \cdot \dfrac24 \cdot \dfrac14 = \dfrac{1}{16}.\)

The total is \[\dfrac18 + \dfrac{3}{32} + \dfrac{1}{16} = \dfrac{9}{32}.\]

Thus, the correct answer is C.

22 答案 B

每 \(12\) 小时有 \(12 + 78 = 90\) 次钟声，所以每天有 \(180\) 次钟声。

从二月 \(26\) 日上午 \(11{:}15\) 之后，到当天结束有 \(91\) 次钟声。之后每天增加 \(180\) 次，到三月 \(8\) 日结束时累计达到 \(1891\) 次。

剩余钟声发生在三月 \(9\) 日；第 \(2003\) 次钟声是当天第 \(112\) 次钟声，发生在三月 \(9\) 日下午 \(3{:}30\)。

所以正确答案是 B。

English

Each \(12\)-hour period has \(12 + 78 = 90\) chimes, so each full day has \(180.\)

After \(11{:}15\) AM on February \(26,\) the rest of that day has \(91\) chimes. Adding \(180\) for each following full day, the count reaches \(1891\) by the end of March \(8.\)

The remaining chimes fall on March \(9:\) the \(2003\)rd chime is the \(112\)th chime of that day, occurring at \(3{:}30\) PM on March \(9.\)

Thus, the correct answer is B.

23 答案 D

设 \(O\) 是中心。八边形可由 \(O\) 分成 \(8\) 个全等三角形，所以 \(\triangle AOB\) 的面积为 \(\dfrac18\)。

由于 \(O\) 是 \(AE\) 的中点，三角形 \(AOB\) 与 \(BOE\) 面积相等，所以 \(\triangle ABE\) 面积为 \(\dfrac14\)。长方形 \(ABEF\) 的面积是它的两倍，即 \(\dfrac12\)。

所以正确答案是 D。

English

Let \(O\) be the center. The octagon splits into \(8\) congruent triangles from \(O,\) so \(\triangle AOB\) has area \(\dfrac18.\)

Since \(O\) is the midpoint of \(AE,\) triangles \(AOB\) and \(BOE\) have equal area, so \(\triangle ABE\) has area \(\dfrac14.\) The rectangle \(ABEF\) is twice this, namely \(\dfrac12.\)

Thus, the correct answer is D.

24 答案 E

公差为 \(-2y\)，所以第三项和第四项应为 \(x-3y\) 和 \(x-5y\)。因此 \(xy=x-3y\) 且 \(\dfrac{x}{y}=x-5y\)。

因此由 \(\dfrac{x}{y}=x-5y\) 得 \(x=xy-5y^2\)，再用 \(xy=x-3y\) 得 \(-3y-5y^2=0\)。因为 \(y\ne 0\)，所以 \(y=-\dfrac35\)，进而 \(x=-\dfrac98\)。

第五项为 \(x-7y=-\dfrac98 + \dfrac{21}{5}=\dfrac{123}{40}\)。

所以正确答案是 E。

English

The common difference is \(-2y,\) so the third and fourth terms must be \(x-3y\) and \(x-5y.\) Thus \(xy=x-3y\) and \(\dfrac{x}{y}=x-5y.\)

From \(\dfrac{x}{y}=x-5y\) we get \(x=xy-5y^2,\) and substituting \(xy=x-3y\) gives \(-3y-5y^2=0.\) Since \(y\ne 0,\) \(y=-\dfrac35\) and then \(x=-\dfrac98.\)

The fifth term is \(x-7y=-\dfrac98 + \dfrac{21}{5}=\dfrac{123}{40}.\)

Thus, the correct answer is E.

25 答案 B

将该数写成 \(\overline{ab23}\)。它能被 \(3\) 整除，当且仅当数字和 \(a+b+2+3=a+b+5\) 能被 \(3\) 整除，也就是 \(a+b\equiv 1 \pmod 3\)。

两位前缀 \(\overline{ab}\) 从 \(10\) 到 \(99\)，共有 \(90\) 个，其中恰好三分之一满足这个同余条件，因此共有 \(\dfrac{90}{3}=30\) 个。

所以正确答案是 B。

English

Write the number as \(\overline{ab23}.\) It is divisible by \(3\) when \(a+b+2+3=a+b+5\) is divisible by \(3,\) that is, when \(a+b\equiv 1 \pmod 3.\)

The two-digit prefix \(\overline{ab}\) ranges over the \(90\) values from \(10\) to \(99,\) and exactly one third of them satisfy this, giving \(\dfrac{90}{3}=30.\)

Thus, the correct answer is B.