2003A AMC 10 真题（P版）

1 答案 D

第 \(k\) 个正偶数 \(2k\) 正好比第 \(k\) 个正奇数 \(2k-1\) 大 \(1\)。

对 \(2003\) 对求和，差为 \(2003 \cdot 1 = 2003\)。

所以正确答案是 D。

English

The \(k\)th even number \(2k\) is exactly \(1\) more than the \(k\)th odd number \(2k-1.\)

Summing this difference over all \(2003\) pairs gives \(2003 \cdot 1 = 2003.\)

Thus, the correct answer is D.

2 答案 B

每件 T 恤价格为 \(\$4 + \$5 = \$9\)。

每位成员需要两双袜子和两件 T 恤，费用为 \(2 \cdot 4 + 2 \cdot 9 = \$26\)。

成员数为 \(2366 \div 26 = 91\)。

所以正确答案是 B。

English

Each T-shirt costs \(\$4 + \$5 = \$9.\)

Each member needs two pairs of socks and two shirts, costing \(2 \cdot 4 + 2 \cdot 9 = \$26.\)

The number of members is \(2366 \div 26 = 91.\)

Thus, the correct answer is B.

3 答案 D

被切去的八个立方体总体积为 \(8 \cdot 3^3 = 216\) 立方厘米。

原盒子体积为 \(15 \cdot 10 \cdot 8 = 1200\) 立方厘米。

切去的百分比为 \(\dfrac{216}{1200} \cdot 100\% = 18\%\)。

所以正确答案是 D。

English

The eight removed cubes have total volume \(8 \cdot 3^3 = 216\) cubic centimeters.

The original box has volume \(15 \cdot 10 \cdot 8 = 1200\) cubic centimeters.

The percent removed is \(\dfrac{216}{1200} \cdot 100\% = 18\%.\)

Thus, the correct answer is D.

4 答案 A

玛丽总共走了 \(2\) 千米，用时 \(30 + 10 = 40\) 分钟。

因为 \(40\) 分钟是 \(\dfrac{2}{3}\) 小时，平均速度为 \(2 \div \dfrac{2}{3} = 3\) 千米每小时。

所以正确答案是 A。

English

Mary walks a total of \(2\) km in \(30 + 10 = 40\) minutes.

Since \(40\) minutes is \(\dfrac{2}{3}\) hour, her average speed is \(2 \div \dfrac{2}{3} = 3\) km/hr.

Thus, the correct answer is A.

5 答案 B

因式分解得 \(2x^2 + 3x - 5 = (2x + 5)(x - 1)\)，所以两个根为 \(-\dfrac{5}{2}\) 和 \(1\)。

因为一个根等于 \(1\)，所以 \(d-1\) 和 \(e-1\) 这两个因子中有一个等于 \(0\)，乘积为 \(0\)。

所以正确答案是 B。

English

Factoring gives \(2x^2 + 3x - 5 = (2x + 5)(x - 1),\) so the roots are \(-\dfrac{5}{2}\) and \(1.\)

Since one root equals \(1,\) one of the two factors \(d-1\) and \(e-1\) equals \(0,\) making the product \(0.\)

Thus, the correct answer is B.

6 答案 C

选项 C 声称 \(x \heartsuit 0 = x\)，但 \(x \heartsuit 0 = |x - 0| = |x|\)，当 \(x\) 为负数时不成立。例如 \(-1 \heartsuit 0 = 1 \ne -1\)。

其他选项都直接来自绝对值的性质。

所以正确答案是 C。

English

Statement (C) claims \(x \heartsuit 0 = x,\) but \(x \heartsuit 0 = |x - 0| = |x|,\) which fails for negative \(x.\) For example, \(-1 \heartsuit 0 = 1 \ne -1.\)

The remaining statements all follow directly from the properties of absolute value.

Thus, the correct answer is C.

7 答案 B

最长边不能超过 \(3\)，否则另外两边之和不能大于它。

可行的边长只有 \(1\)、\(3\)、\(3\) 和 \(2\)、\(2\)、\(3\)，因此共有 \(2\) 个三角形。

所以正确答案是 B。

English

The longest side cannot exceed \(3,\) since otherwise the other two sides could not reach it.

The only possibilities are side lengths \(1,\) \(3,\) \(3\) and \(2,\) \(2,\) \(3,\) giving \(2\) triangles.

Thus, the correct answer is B.

8 答案 E

\(60\) 的因数为 \(1\)、\(2\)、\(3\)、\(4\)、\(5\)、\(6\)、\(10\)、\(12\)、\(15\)、\(20\)、\(30\) 和 \(60\)。

其中十二个因数有六个小于 \(7\)，所以概率为 \(\dfrac{6}{12} = \dfrac{1}{2}\)。

所以正确答案是 E。

English

The factors of \(60\) are \(1,\) \(2,\) \(3,\) \(4,\) \(5,\) \(6,\) \(10,\) \(12,\) \(15,\) \(20,\) \(30,\) and \(60.\)

Six of these twelve factors are less than \(7,\) so the probability is \(\dfrac{6}{12} = \dfrac{1}{2}.\)

Thus, the correct answer is E.

9 答案 A

从内向外计算，\(x\sqrt{x} = x^{\frac{3}{2}}\)，它的立方根为 \(x^{\frac{1}{2}}\)。

接着 \(x \cdot x^{\frac{1}{2}} = x^{\frac{3}{2}}\)，它的立方根仍为 \(x^{\frac{1}{2}}\)。

于是下一层又是 \(x \cdot x^{\frac{1}{2}} = x^{\frac{3}{2}}\)，开立方后仍为 \(x^{\frac{1}{2}} = \sqrt{x}\)。

所以正确答案是 A。

English

Working outward, \(x\sqrt{x} = x^{\frac{3}{2}},\) and its cube root is \(x^{\frac{1}{2}}.\)

Then \(x \cdot x^{\frac{1}{2}} = x^{\frac{3}{2}},\) whose cube root is again \(x^{\frac{1}{2}}.\)

Repeating once more, \(x \cdot x^{\frac{1}{2}} = x^{\frac{3}{2}},\) whose cube root is \(x^{\frac{1}{2}} = \sqrt{x}.\)

Thus, the correct answer is A.

10 答案 E

先折起四个阴影正方形。它们占据立方体的四个不同面，留下两个空面。

逐一检查标号位置：位置 \(1\)、\(2\) 和 \(3\) 会使新正方形折到已被阴影正方形占据的面上。位置 \(4\)、\(5\)、\(6\)、\(7\)、\(8\) 和 \(9\) 会使它折到两个空面之一。因此 \(9\) 个多边形中恰有 \(6\) 个可行。

所以正确答案是 E。

English

Fold the four shaded squares first. They occupy four distinct faces of the cube, leaving two faces open.

Checking the numbered attachments, positions \(1,\) \(2,\) and \(3\) fold the new square onto a face already occupied by one of the shaded squares. Positions \(4,\) \(5,\) \(6,\) \(7,\) \(8,\) and \(9\) fold it onto one of the two open faces. Therefore exactly \(6\) of the \(9\) polygons work.

Thus, the correct answer is E.

11 答案 E

两个数分别为 \(100 \cdot \overline{AMC} + 10\) 和 \(100 \cdot \overline{AMC} + 12\)，所以 \(200 \cdot \overline{AMC} + 22 = 123422\)。

所以 \(200 \cdot \overline{AMC} = 123400\)，\(\overline{AMC} = 617\)。

\(A + M + C = 6 + 1 + 7 = 14\)。

所以正确答案是 E。

English

The two numbers equal \(100 \cdot \overline{AMC} + 10\) and \(100 \cdot \overline{AMC} + 12,\) so their sum is \(200 \cdot \overline{AMC} + 22 = 123422.\)

Then \(200 \cdot \overline{AMC} = 123400,\) so \(\overline{AMC} = 617.\)

Therefore \(A + M + C = 6 + 1 + 7 = 14.\)

Thus, the correct answer is E.

12 答案 A

条件 \(x \lt y\) 对应由 \(y = x\)、\(y = 1\) 和 \(x = 0\) 围成的三角形，顶点为 \((0, 0)\)、\((0, 1)\) 和 \((1, 1)\)。

这个三角形面积为 \(\dfrac{1}{2}\)，长方形面积为 \(4\)。

概率为 \(\dfrac{\frac{1}{2}}{4} = \dfrac{1}{8}\)。

所以正确答案是 A。

English

The condition \(x \lt y\) holds in the triangle bounded by \(y = x,\) \(y = 1,\) and \(x = 0,\) which has vertices \((0, 0),\) \((0, 1),\) and \((1, 1).\)

This triangle has area \(\dfrac{1}{2},\) while the rectangle has area \(4.\)

The probability is \(\dfrac{\frac{1}{2}}{4} = \dfrac{1}{8}.\)

Thus, the correct answer is A.

13 答案 A

设三个数为 \(a\)、\(b\)、\(c\)。由于 \(a = 4(b + c)\)，可得 \(4(b + c) + (b + c) = 20\)，所以 \(b + c = 4\)，且 \(a = 16\)。

因为 \(b = 7c\)，所以 \(7c + c = 4\)，得 \(c = \dfrac{1}{2}\)，\(b = \dfrac{7}{2}\)。

乘积为 \(16 \cdot \dfrac{7}{2} \cdot \dfrac{1}{2} = 28\)。

所以正确答案是 A。

English

Let the numbers be \(a,\) \(b,\) \(c.\) Since \(a = 4(b + c),\) we get \(4(b + c) + (b + c) = 20,\) so \(b + c = 4\) and \(a = 16.\)

With \(b = 7c,\) we have \(7c + c = 4,\) so \(c = \dfrac{1}{2}\) and \(b = \dfrac{7}{2}.\)

The product is \(16 \cdot \dfrac{7}{2} \cdot \dfrac{1}{2} = 28.\)

Thus, the correct answer is A.

14 答案 A

\(d\) 和 \(e\) 是集合 \(\{2,3,5,7\}\) 中两个不同的元素，而且 \(10d+e\) 也必须是质数。

从最大的十位数字开始。若 \(d=7\)，取 \(e=5\) 或 \(e=2\) 分别得到合数 \(75\) 和 \(72\)，而取 \(e=3\) 得到质数 \(73\)。于是 \(n=7\cdot3\cdot73=1533\)。

若 \(d=5\)，取 \(e=7\) 或 \(e=2\) 分别得到合数 \(57\) 和 \(52\)，而取 \(e=3\) 只能得到 \(5\cdot3\cdot53=795\)。所有 \(d\le3\) 的情况都不超过 \(3\cdot7\cdot37=777\)。所以 \(1533\) 是最大的有效值。

它的各位数字之和为 \(1 + 5 + 3 + 3 = 12\)。

所以正确答案是 A。

English

Both \(d\) and \(e\) are distinct members of \(\{2,3,5,7\},\) and \(10d+e\) must also be prime.

Start with the largest possible tens digit. For \(d=7,\) the choices \(e=5\) and \(e=2\) give the composite numbers \(75\) and \(72,\) while \(e=3\) gives the prime \(73.\) This produces \(n=7\cdot3\cdot73=1533.\)

For \(d=5,\) the choices \(e=7\) and \(e=2\) give the composite numbers \(57\) and \(52,\) while \(e=3\) gives only \(5\cdot3\cdot53=795.\) Every case with \(d\le3\) is at most \(3\cdot7\cdot37=777.\) Hence \(1533\) is the largest valid value.

The sum of its digits is \(1 + 5 + 3 + 3 = 12.\)

Thus, the correct answer is A.

15 答案 C

\(100\) 个整数中有 \(50\) 个能被 \(2\) 整除。

其中同时能被 \(3\) 整除的是 \(6\) 的倍数，共 \(16\) 个。

因此符合条件的数有 \(50 - 16 = 34\) 个，概率为 \(\dfrac{34}{100} = \dfrac{17}{50}\)。

所以正确答案是 C。

English

Of the \(100\) integers, \(50\) are divisible by \(2.\)

Among those, the ones also divisible by \(3\) are the multiples of \(6,\) of which there are \(16.\)

So \(50 - 16 = 34\) qualify, giving probability \(\dfrac{34}{100} = \dfrac{17}{50}.\)

Thus, the correct answer is C.

16 答案 C

\(13^{2003}\) 的个位数字等同于 \(3^{2003}\) 的个位数字。

\(3\) 的幂个位数字按 \(3\)、\(9\)、\(7\)、\(1\) 循环，周期为 \(4\)。

因为 \(2003 = 4 \cdot 500 + 3\)，所以个位数字是循环中的第三个，即 \(7\)。

所以正确答案是 C。

English

The units digit of \(13^{2003}\) matches that of \(3^{2003}.\)

Powers of \(3\) have units digits cycling \(3,\) \(9,\) \(7,\) \(1\) with period \(4.\)

Since \(2003 = 4 \cdot 500 + 3,\) the units digit is the third in the cycle, which is \(7.\)

Thus, the correct answer is C.

17 答案 B

设边长为 \(s\)，外接圆半径为 \(R\)。由等边三角形中的 \(30\)-\(60\)-\(90\) 三角形，\(R = \dfrac{s}{\sqrt{3}}\)，所以 \(s = R\sqrt{3}\)。

周长为 \(3s = 3R\sqrt{3}\)，圆面积为 \(\pi R^2\)。

令二者相等，\(3R\sqrt{3} = \pi R^2\)，所以 \(R = \dfrac{3\sqrt{3}}{\pi}\)。

所以正确答案是 B。

English

Let the side length be \(s\) and the circumradius be \(R.\) From a \(30\)-\(60\)-\(90\) triangle formed by the center and a side, \(R = \dfrac{s}{\sqrt{3}},\) so \(s = R\sqrt{3}.\)

The perimeter is \(3s = 3R\sqrt{3}\) and the circle’s area is \(\pi R^2.\)

Setting them equal, \(3R\sqrt{3} = \pi R^2,\) so \(R = \dfrac{3\sqrt{3}}{\pi}.\)

Thus, the correct answer is B.

18 答案 B

设 \(a = \dfrac{2003}{2004}\)。两边乘以 \(x\)，得 \(ax^2 + x + 1 = 0\)。

若根为 \(r\) 和 \(s\)，则由韦达定理 \(r + s = -\dfrac{1}{a}\)，且 \(rs = \dfrac{1}{a}\)。

倒数和为 \(\dfrac{1}{r} + \dfrac{1}{s} = \dfrac{r + s}{rs} = \dfrac{-\frac{1}{a}}{\frac{1}{a}} = -1\)。

所以正确答案是 B。

English

Let \(a = \dfrac{2003}{2004}.\) Multiplying the equation by \(x\) gives \(ax^2 + x + 1 = 0.\)

If the roots are \(r\) and \(s,\) then by Vieta’s formulas \(r + s = -\dfrac{1}{a}\) and \(rs = \dfrac{1}{a}.\)

The sum of reciprocals is \(\dfrac{1}{r} + \dfrac{1}{s} = \dfrac{r + s}{rs} = \dfrac{-\frac{1}{a}}{\frac{1}{a}} = -1.\)

Thus, the correct answer is B.

19 答案 C

小半圆的直径是大圆中的一条长为 \(1\) 的弦。将这条弦的两端与大圆圆心相连，得到边长为 \(1\) 的等边三角形，面积为 \(\dfrac{\sqrt{3}}{4}\)。

弦与小弧之间的区域连同这个三角形，面积为 \(\dfrac{\sqrt{3}}{4} + \dfrac{1}{2}\pi\left(\dfrac{1}{2}\right)^2 = \dfrac{\sqrt{3}}{4} + \dfrac{\pi}{8}\)。

减去大圆中 \(60^\circ\) 扇形面积 \(\dfrac{1}{6}\pi(1)^2 = \dfrac{\pi}{6}\)，得到弓月形面积为 \(\dfrac{\sqrt{3}}{4} + \dfrac{\pi}{8} - \dfrac{\pi}{6} = \dfrac{\sqrt{3}}{4} - \dfrac{\pi}{24}\)。

所以正确答案是 C。

English

The small semicircle’s diameter is a chord of length \(1\) in the large circle. Joining its endpoints to the large circle’s center gives an equilateral triangle of side \(1\) and area \(\dfrac{\sqrt{3}}{4}.\)

The region between the chord and the small arc, taken together with that triangle, has area \(\dfrac{\sqrt{3}}{4} + \dfrac{1}{2}\pi\left(\dfrac{1}{2}\right)^2 = \dfrac{\sqrt{3}}{4} + \dfrac{\pi}{8}.\)

Subtracting the \(60^\circ\) sector of the large circle, of area \(\dfrac{1}{6}\pi(1)^2 = \dfrac{\pi}{6},\) leaves the lune: \(\dfrac{\sqrt{3}}{4} + \dfrac{\pi}{8} - \dfrac{\pi}{6} = \dfrac{\sqrt{3}}{4} - \dfrac{\pi}{24}.\)

Thus, the correct answer is C.

20 答案 E

最大的 \(9\) 进制三位数是 \(9^3 - 1 = 728\)，最小的 \(11\) 进制三位数是 \(11^2 = 121\)。

所以两种表示都为三位数当且仅当 \(121 \le n \le 728\)，共有 \(608\) 个整数。

全部三位数共有 \(900\) 个，概率为 \(\dfrac{608}{900} \approx 0.7\)。

所以正确答案是 E。

English

The largest three-digit base-\(9\) number is \(9^3 - 1 = 728,\) and the smallest three-digit base-\(11\) number is \(11^2 = 121.\)

So both conditions hold exactly when \(121 \le n \le 728,\) giving \(608\) integers.

Out of \(900\) three-digit numbers, the probability is \(\dfrac{608}{900} \approx 0.7.\)

Thus, the correct answer is E.

21 答案 D

一种组合由三种饼干各取多少块决定，所以要数 \(a + b + c = 6\) 的非负整数解。

由插板法，在 \(8\) 个位置中放 \(2\) 个隔板，个数为 \(\dbinom{8}{2} = 28\)。

所以正确答案是 D。

English

An assortment is determined by how many of each type are chosen, so we count nonnegative integer solutions to \(a + b + c = 6.\)

By stars and bars, placing \(2\) dividers among \(8\) slots gives \(\dbinom{8}{2} = 28\) assortments.

Thus, the correct answer is D.

22 答案 B

取坐标 \(D = (0, 0)\)、\(A = (9, 0)\)、\(B = (9, 8)\)、\(C = (0, 8)\)、\(H = (3, 8)\)、\(E = (4, 0)\)。

直线 \(AH\) 的方程为 \(y = -\dfrac{4}{3}x + 12\)，直线 \(EC\) 的方程为 \(y = -2x + 8\)。

联立得 \(x = -6\)、\(y = 20\)，所以 \(G = (-6, 20)\)。因为 \(\overline{GF}\) 垂直于直线 \(AD\)，也就是垂直于 \(x\) 轴，所以其长度就是高度 \(20\)。

所以正确答案是 B。

English

Place \(D = (0, 0),\) \(A = (9, 0),\) \(B = (9, 8),\) \(C = (0, 8),\) \(H = (3, 8),\) and \(E = (4, 0).\)

Line \(AH\) has equation \(y = -\dfrac{4}{3}x + 12,\) and line \(EC\) has equation \(y = -2x + 8.\)

Setting them equal gives \(x = -6\) and \(y = 20,\) so \(G = (-6, 20).\) Since \(\overline{GF}\) is perpendicular to line \(AD\) (the \(x\)-axis), its length is the height \(20.\)

Thus, the correct answer is B.

23 答案 C

有 \(n\) 行时，底行小三角形个数为 \(2n - 1\)，所以 \(2n - 1 = 2003\)，得 \(n = 1002\)。

第 \(k\) 行需要 \(3k\) 根牙签，所以总数为 \(3(1 + 2 + \cdots + 1002)\)。

这等于 \(3 \cdot \dfrac{1002 \cdot 1003}{2} = 1{,}507{,}509\)。

所以正确答案是 C。

English

A triangle with \(n\) rows has \(2n - 1\) small triangles in its base row, so \(2n - 1 = 2003\) gives \(n = 1002.\)

Each row \(k\) requires \(3k\) toothpicks, so the total is \(3(1 + 2 + \cdots + 1002).\)

This equals \(3 \cdot \dfrac{1002 \cdot 1003}{2} = 1{,}507{,}509.\)

Thus, the correct answer is C.

24 答案 E

在蓝牌 \(3\)、\(4\)、\(5\)、\(6\) 中，红 \(5\) 只能整除 \(5\)，红 \(4\) 只能整除 \(4\)，所以这些配对必须位于两端。

红 \(2\) 只能整除 \(4\) 和 \(6\)，红 \(3\) 只能整除 \(3\) 和 \(6\)。继续连接会迫使牌列为 \(R4\)、\(B4\)、\(R2\)、\(B6\)、\(R3\)、\(B3\)、\(R1\)、\(B5\)、\(R5\)。

中间三张是 \(B6\)、\(R3\)、\(B3\)，和为 \(6 + 3 + 3 = 12\)。

所以正确答案是 E。

English

Among blue cards \(3,\) \(4,\) \(5,\) \(6,\) red \(5\) divides only \(5\) and red \(4\) divides only \(4,\) so those pairs must sit at the ends.

Red \(2\) divides only \(4\) and \(6,\) and red \(3\) divides only \(3\) and \(6.\) Chaining these forces the stack \(R4,\) \(B4,\) \(R2,\) \(B6,\) \(R3,\) \(B3,\) \(R1,\) \(B5,\) \(R5.\)

The middle three cards are \(B6,\) \(R3,\) \(B3,\) summing to \(6 + 3 + 3 = 12.\)

Thus, the correct answer is E.

25 答案 B

写成 \(n = 100q + r = (q + r) + 99q\)。

因为 \(99q\) 是 \(11\) 的倍数，所以 \(q + r\) 能被 \(11\) 整除当且仅当 \(n\) 也能被整除。

\(5\) 位数中，\(11\) 的倍数满足 \(10000 \le n \le 99999\)，共有

\[ \begin{aligned} &\left\lfloor\dfrac{99999}{11}\right\rfloor - \left\lfloor\dfrac{9999}{11}\right\rfloor \\ &= 9090 - 909 \\ &= 8181 \end{aligned}\text{。}\]

所以正确答案是 B。

English

Write \(n = 100q + r = (q + r) + 99q.\)

Since \(99q\) is a multiple of \(11,\) \(q + r\) is divisible by \(11\) if and only if \(n\) is.

The \(5\)-digit multiples of \(11\) satisfy \(10000 \le n \le 99999,\) and there are

\[ \begin{aligned} &\left\lfloor\dfrac{99999}{11}\right\rfloor - \left\lfloor\dfrac{9999}{11}\right\rfloor \\ &= 9090 - 909 \\ &= 8181. \end{aligned} \]

Thus, the correct answer is B.