2005B AMC 12 真题（P版）

1 答案 A

小队买入 \(1000 \div 5 = 200\) 组五条装糖果棒，成本为 \(200 \cdot 2 = 400\) 美元。

他们卖出 \(1000 \div 2 = 500\) 对糖果棒，收入为 \(500 \cdot 1 = 500\) 美元。

利润为 \(500 - 400 = 100\) 美元。

所以正确答案是 A。

English

The troop buys \(1000 \div 5 = 200\) groups of five bars, costing \(200 \cdot 2 = 400\) dollars.

They sell \(1000 \div 2 = 500\) pairs of bars, earning \(500 \cdot 1 = 500\) dollars.

The profit is \(500 - 400 = 100\) dollars.

Thus, the correct answer is A.

2 答案 D

题意可化为 \[ \dfrac{x}{100}\cdot x = 4\text{，} \] 所以 \(x^2 = 400\)。

因为 \(x\) 为正，所以 \(x = 20\)。

所以正确答案是 D。

English

The statement translates to \[ \dfrac{x}{100}\cdot x = 4, \] so \(x^2 = 400.\)

Since \(x\) is positive, \(x = 20.\)

Thus, the correct answer is D.

3 答案 C

全部 CD 的费用是三分之一 CD 费用的三倍，即她的钱的 \(3\cdot\dfrac15 = \dfrac35\)。

她还剩下自己钱的 \(1 - \dfrac35 = \dfrac25\)。

所以正确答案是 C。

English

Buying all the CDs costs three times what one third of them cost, namely \(3\cdot\dfrac15 = \dfrac35\) of her money.

She has \(1 - \dfrac35 = \dfrac25\) of her money left.

Thus, the correct answer is C.

4 答案 B

Lisa 至少需要 \(0.8 \cdot 50 = 40\) 次 A。

她已经有 \(22\) 次，所以剩下 \(20\) 次中还需要 \(40 - 22 = 18\) 次 A。

因此最多可以有 \(20 - 18 = 2\) 次低于 A。

所以正确答案是 B。

English

Lisa needs an A on at least \(0.8 \cdot 50 = 40\) quizzes.

She has \(22\) already, so she needs \(40 - 22 = 18\) more of the remaining \(20\) quizzes.

She can earn a lower grade on at most \(20 - 18 = 2\) of them.

Thus, the correct answer is B.

5 答案 A

每块瓷砖的四个四分之一圆合成一个半径为 \(\dfrac12\) 的圆，面积为 \(\pi\left(\dfrac12\right)^2 = \dfrac{\pi}{4}\)。

每块瓷砖阴影面积为 \(1 - \dfrac{\pi}{4}\)。

共有 \(8 \cdot 10 = 80\) 块瓷砖，所以总阴影面积为 \[ 80\left(1 - \dfrac{\pi}{4}\right) = 80 - 20\pi\text{。}\]

所以正确答案是 A。

English

The four quarter circles in a tile together form one full circle of radius \(\dfrac12,\) with area \(\pi\left(\dfrac12\right)^2 = \dfrac{\pi}{4}.\)

So each tile has shaded area \(1 - \dfrac{\pi}{4}\) square feet.

There are \(8 \cdot 10 = 80\) tiles, so the total shaded area is \[ 80\left(1 - \dfrac{\pi}{4}\right) = 80 - 20\pi. \]

Thus, the correct answer is A.

6 答案 A

设 \(H\) 为从 \(C\) 到直线 \(AB\) 的垂足。由于 \(\triangle ABC\) 中 \(AC = BC\)，所以 \(H\) 是 \(AB\) 的中点，从而 \(AH = HB = 1\)。

于是 \(CH^2 = 7^2 - 1^2 = 48\)。对 \(\triangle CHD\) 应用勾股定理，其中 \(HD = HB + BD = 1 + BD\)，得 \[ 8^2 = 48 + (1 + BD)^2\text{，} \] 从而 \((1 + BD)^2 = 16\)。

于是由上式得 \(1 + BD = 4\)，所以 \(BD = 3\)。

所以正确答案是 A。

English

Let \(H\) be the foot of the altitude from \(C\) to line \(AB.\) Since \(\triangle ABC\) is isosceles with \(AC = BC,\) \(H\) is the midpoint of \(AB,\) so \(AH = HB = 1.\)

Then \(CH^2 = 7^2 - 1^2 = 48.\) Applying the Pythagorean Theorem to \(\triangle CHD\) with \(HD = HB + BD = 1 + BD\) gives \[ 8^2 = 48 + (1 + BD)^2, \] so \((1 + BD)^2 = 16.\)

Therefore \(1 + BD = 4,\) which means \(BD = 3.\)

Thus, the correct answer is A.

7 答案 D

令 \(y = 0\)，得 \(|3x| = 12\)，所以 \(x = \pm 4\)。令 \(x = 0\)，得 \(|4y| = 12\)，所以 \(y = \pm 3\)。

该图形是一个菱形，顶点为 \((\pm 4, 0)\) 和 \((0, \pm 3)\)，所以两条对角线的长度分别为 \(8\) 和 \(6\)。

它的面积为 \(\dfrac12 \cdot 8 \cdot 6 = 24\)。

所以正确答案是 D。

English

Setting \(y = 0\) gives \(|3x| = 12,\) so \(x = \pm 4.\) Setting \(x = 0\) gives \(|4y| = 12,\) so \(y = \pm 3.\)

The graph is a rhombus with vertices \((\pm 4, 0)\) and \((0, \pm 3),\) so its diagonals have lengths \(8\) and \(6.\)

Its area is \(\dfrac12 \cdot 8 \cdot 6 = 24.\)

Thus, the correct answer is D.

8 答案 C

抛物线 \(y = x^2 + a^2\) 的顶点为 \((0, a^2)\)。

直线 \(y = x + a\) 经过该点当且仅当 \(a^2 = 0 + a\)，即 \(a^2 - a = 0\)。

所以 \(a = 0\) 或 \(a = 1\)，共有 \(2\) 个值。

所以正确答案是 C。

English

The vertex of the parabola \(y = x^2 + a^2\) is \((0, a^2).\)

The line \(y = x + a\) passes through it exactly when \(a^2 = 0 + a,\) that is \(a^2 - a = 0.\)

This gives \(a = 0\) or \(a = 1,\) so there are \(2\) values.

Thus, the correct answer is C.

9 答案 B

得 \(95\) 分的学生百分比是 \(100 - 10 - 25 - 20 - 15 = 30\)。

平均分为 \[ \begin{aligned} &0.10(70) + 0.25(80) \\ &\quad {}+ 0.20(85) + 0.15(90) \\ &\quad {}+ 0.30(95) = 86 \end{aligned}\text{。}\]

累计看，\(10\%\) 的学生低于 \(80\) 分，\(35\%\) 的学生不高于 \(80\) 分，\(55\%\) 的学生不高于 \(85\) 分。中间位置落在 \(85\) 分，所以中位数为 \(85\)。

差为 \(86 - 85 = 1\)。

所以正确答案是 B。

English

The percentage scoring \(95\) is \(100 - 10 - 25 - 20 - 15 = 30.\)

The mean is \[ \begin{aligned} &0.10(70) + 0.25(80) \\ &\quad {}+ 0.20(85) + 0.15(90) \\ &\quad {}+ 0.30(95) = 86. \end{aligned} \]

Cumulatively, \(10\%\) are below \(80,\) \(35\%\) are at or below \(80,\) and \(55\%\) are at or below \(85.\) The middle scores fall at \(85,\) so the median is \(85.\)

The difference is \(86 - 85 = 1.\)

Thus, the correct answer is B.

10 答案 E

数列开始为 \(2005, 133, 55, 250, 133, \ldots\)，因为 \(2^3 + 0^3 + 0^3 + 5^3 = 133\)，\(1^3 + 3^3 + 3^3 = 55\)，\(5^3 + 5^3 = 250\)，且 \(2^3 + 5^3 + 0^3 = 133\)。

在首项 \(2005\) 之后，数列以 \(133, 55, 250\) 为周期 \(3\) 循环。

当 \(n \ge 2\) 时，第 \(n\) 项是序列 \(133, 55, 250\) 中索引为 \(((n-2)\bmod 3)\) 的一项。因为 \(2005 - 2 = 2003 \equiv 2 \pmod 3\)，所以第 \(2005\) 项是 \(250\)。

所以正确答案是 E。

English

The sequence begins \(2005, 133, 55, 250, 133, \ldots\) since \(2^3 + 0^3 + 0^3 + 5^3 = 133,\) \(1^3 + 3^3 + 3^3 = 55,\) \(5^3 + 5^3 = 250,\) and \(2^3 + 5^3 + 0^3 = 133.\)

After the initial \(2005,\) the terms cycle through \(133, 55, 250\) with period \(3.\)

Term \(n\) for \(n \ge 2\) is the \(((n-2)\bmod 3)\)th entry of \(133, 55, 250.\) Since \(2005 - 2 = 2003 \equiv 2 \pmod 3,\) the \(2005\)th term is \(250.\)

Thus, the correct answer is E.

11 答案 D

共有 \(\binom{8}{2} = 28\) 对等可能的纸币。

总额达到 \(\$20\) 或更多的情况有：两张二十（\(1\) 种），一张二十配六张较小纸币之一（\(2 \cdot 6 = 12\) 种），以及两张十（\(1\) 种）。

有利情况共 \(1 + 12 + 1 = 14\) 种，所以概率为 \(\dfrac{14}{28} = \dfrac12\)。

所以正确答案是 D。

English

There are \(\binom{8}{2} = 28\) equally likely pairs of bills.

The sum is \(\$20\) or more in these cases: both twenties (\(1\) way), one twenty with one of the six smaller bills (\(2 \cdot 6 = 12\) ways), or both tens (\(1\) way).

That is \(1 + 12 + 1 = 14\) favorable pairs, so the probability is \(\dfrac{14}{28} = \dfrac12.\)

Thus, the correct answer is D.

12 答案 D

设 \(r_1\) 和 \(r_2\) 是 \(x^2 + px + m = 0\) 的根，则 \(m = r_1 r_2\)，且 \(p = -(r_1 + r_2)\)。

方程 \(x^2 + mx + n = 0\) 的根为 \(2r_1\) 和 \(2r_2\)，所以 \(n = 4r_1 r_2\)，且 \(m = -2(r_1 + r_2)\)。

于是 \(n = 4m\)，且 \(m = 2p\)，即 \(p = \dfrac{m}{2}\)，所以 \[ \dfrac{n}{p} = \dfrac{4m}{\tfrac{m}{2}} = 8\text{。}\]

所以正确答案是 D。

English

Let \(r_1\) and \(r_2\) be the roots of \(x^2 + px + m = 0,\) so \(m = r_1 r_2\) and \(p = -(r_1 + r_2).\)

The roots of \(x^2 + mx + n = 0\) are \(2r_1\) and \(2r_2,\) so \(n = 4r_1 r_2\) and \(m = -2(r_1 + r_2).\)

Then \(n = 4m\) and \(m = 2p,\) which gives \(p = \dfrac{m}{2},\) so \[ \dfrac{n}{p} = \dfrac{4m}{\tfrac{m}{2}} = 8. \]

Thus, the correct answer is D.

13 答案 D

由 \(4^{x_1} = 5\) 得 \(x_1 = \log_4 5\)，一般地 \(x_k = \log_{k+3}(k+4)\)。

这个乘积会裂项相消：\[ \begin{aligned} &x_1 x_2 \cdots x_{124} \\ &= \log_4 5 \cdot \log_5 6 \cdots \log_{127} 128 \\ &= \log_4 128 \end{aligned}\text{。}\]

因为 \(128 = 2^7\) 且 \(4 = 2^2\)，结果为 \(\dfrac{7\log 2}{2\log 2} = \dfrac72\)。

所以正确答案是 D。

English

From \(4^{x_1} = 5\) we get \(x_1 = \log_4 5,\) and in general \(x_k = \log_{k+3}(k+4).\)

The product telescopes: \[ \begin{aligned} &x_1 x_2 \cdots x_{124} \\ &= \log_4 5 \cdot \log_5 6 \cdots \log_{127} 128 \\ &= \log_4 128. \end{aligned} \]

Since \(128 = 2^7\) and \(4 = 2^2,\) this equals \(\dfrac{7\log 2}{2\log 2} = \dfrac72.\)

Thus, the correct answer is D.

14 答案 E

圆与 \(y = 6\) 相切，且圆心 \((0, k)\) 在其上方，所以半径为 \(r = k - 6\)。

点 \((0, k)\) 到直线 \(x - y = 0\) 的距离为 \(\dfrac{|0 - k|}{\sqrt2} = \dfrac{k}{\sqrt2}\)，这也等于 \(r\)。

令 \(\dfrac{k}{\sqrt2} = k - 6\)，得 \(k = \dfrac{6\sqrt2}{\sqrt2 - 1}\) \(= 6\sqrt2\,(\sqrt2+1)\) \(= 12 + 6\sqrt2\)。

因此 \(r = k - 6 = 6 + 6\sqrt2\)。

所以正确答案是 E。

English

Since the circle is tangent to \(y = 6\) and its center \((0, k)\) is above that line, the radius is \(r = k - 6.\)

The distance from \((0, k)\) to the line \(x - y = 0\) is \(\dfrac{|0 - k|}{\sqrt2} = \dfrac{k}{\sqrt2},\) and this must also equal \(r.\)

Setting \(\dfrac{k}{\sqrt2} = k - 6\) gives \(k = \dfrac{6\sqrt2}{\sqrt2 - 1}\) \(= 6\sqrt2\,(\sqrt2+1)\) \(= 12 + 6\sqrt2.\)

Then \(r = k - 6 = 6 + 6\sqrt2.\)

Thus, the correct answer is E.

15 答案 D

八个数字来自 \(1\) 到 \(9\)，全部非零数字的和为 \(45\)，所以这八个数字的总和在 \(45 - 9 = 36\) 与 \(45 - 1 = 44\) 之间。

设个位数字和为 \(U\)，十位数字和为 \(T\)，则 \(10T + U = 221\)，所以 \(U\) 的个位为 \(1\)。又 \(1+2+3+4 = 10\) \(\le U \le\) \(6+7+8+9 = 30\)，故 \(U = 11\) 或 \(U = 21\)。

若 \(U = 11\) 则 \(10T = 210\)，所以 \(T = 21\)，八个数字总和为 \(32\)，小于 \(36\)，不可能。因此 \(U = 21\)，\(T = 20\)，总和为 \(41\)。

缺失数字为 \(45 - 41 = 4\)。例如 \(13 + 25 + 86 + 97 = 221\)。

所以正确答案是 D。

English

The eight digits are distinct and chosen from \(1\) through \(9,\) whose total is \(45.\) So the eight used digits sum to between \(45 - 9 = 36\) and \(45 - 1 = 44.\)

Let the four units digits sum to \(U\) and the four tens digits sum to \(T.\) Then \(10T + U = 221,\) so \(U\) ends in \(1.\) Since \(1+2+3+4 = 10\) \(\le U \le\) \(6+7+8+9 = 30,\) we have \(U = 11\) or \(U = 21.\)

If \(U = 11,\) then \(10T = 210,\) so \(T = 21\) and the eight digits sum to \(32,\) which is below \(36.\) So \(U = 21,\) giving \(T = 20\) and total \(41.\)

The missing digit is \(45 - 41 = 4.\) For example, \(13 + 25 + 86 + 97 = 221.\)

Thus, the correct answer is D.

16 答案 D

在某个八分体中，与三个坐标平面都相切、半径为 \(1\) 的球，其球心可取为 \((1, 1, 1)\) 这样的点，于是 \[ \sqrt{1^2 + 1^2 + 1^2} = \sqrt3 \] 就是球心到原点的距离。

这个球上离原点最远的点距离为 \(\sqrt3 + 1\)，所以包含八个球的最小球半径为 \(1 + \sqrt3\)。

所以正确答案是 D。

English

A sphere of radius \(1\) tangent to the three coordinate planes in one octant has its center at a point like \((1, 1, 1),\) at distance \[ \sqrt{1^2 + 1^2 + 1^2} = \sqrt3 \] from the origin.

The farthest point of that sphere from the origin is at distance \(\sqrt3 + 1,\) so the containing sphere has radius \(1 + \sqrt3.\)

Thus, the correct answer is D.

17 答案 B

原方程等价于 \(\log_{10}\left(2^a 3^b 5^c 7^d\right) = 2005\)，所以 \[ 2^a 3^b 5^c 7^d = 10^{2005} = 2^{2005} \cdot 5^{2005}\text{。}\]

把 \(a, b, c, d\) 的分母乘以同一个整数消去，再用质因数分解的唯一性比较指数，可得 \(a = 2005\)、\(b = 0\)、\(c = 2005\)、\(d = 0\)。

所以恰有 \(1\) 个这样的四元组。

所以正确答案是 B。

English

The equation is equivalent to \(\log_{10}\left(2^a 3^b 5^c 7^d\right) = 2005,\) so \[ 2^a 3^b 5^c 7^d = 10^{2005} = 2^{2005} \cdot 5^{2005}. \]

Clearing the denominators of \(a, b, c, d\) with a common integer multiplier and using the uniqueness of prime factorization, the exponents must match: \(a = 2005,\) \(b = 0,\) \(c = 2005,\) and \(d = 0.\)

So there is exactly \(1\) such four-tuple.

Thus, the correct answer is B.

18 答案 C

直线 \(AB\) 的斜率为 \(1\)。要使 \(\angle A\) 为锐角，\(C\) 必须在过 \(A\) 且垂直于 \(AB\) 的直线远侧；在第一象限中，这条线连接 \(P(4, 0)\) 与 \(Q(0, 4)\)。要使 \(\angle B\) 为锐角，\(C\) 必须在过 \(B\) 且垂直于 \(AB\) 的直线近侧，这条线连接 \(S(14, 0)\) 与 \(T(0, 14)\)。

为了使 \(\angle C\) 为锐角，\(C\) 必须在以 \(AB\) 为直径的圆 \(U\) 外，其半径为 \(\dfrac{AB}{2} = \dfrac{5\sqrt2}{2}\)。

圆完全位于这条带内并且在第一象限中。所以所求区域是大直角三角形 \(OST\) 减去小直角三角形 \(OPQ\) 和整个圆 \(U\)：\[ \begin{aligned} &\dfrac12 \cdot 14^2 - \dfrac12 \cdot 4^2 \\ &\quad {}- \pi\left(\dfrac{5\sqrt2}{2}\right)^2 \\ &= 98 - 8 - \dfrac{25\pi}{2} \\ &= 90 - \dfrac{25\pi}{2} \approx 51 \end{aligned}\text{。}\]

所以正确答案是 C。

English

Line \(AB\) has slope \(1.\) For \(\angle A\) to be acute, \(C\) must lie beyond the line through \(A\) perpendicular to \(AB;\) in the first quadrant that line runs between \(P(4, 0)\) and \(Q(0, 4).\) For \(\angle B\) to be acute, \(C\) must lie before the line through \(B\) perpendicular to \(AB,\) between \(S(14, 0)\) and \(T(0, 14).\)

For \(\angle C\) to be acute, \(C\) must lie outside the circle \(U\) with diameter \(AB,\) whose radius is \(\dfrac{AB}{2} = \dfrac{5\sqrt2}{2}.\)

The circle lies entirely inside this strip and in the first quadrant. Thus the region is the large right triangle \(OST\) minus the small right triangle \(OPQ\) and the full circle \(U:\) \[ \begin{aligned} &\dfrac12 \cdot 14^2 - \dfrac12 \cdot 4^2 \\ &\quad {}- \pi\left(\dfrac{5\sqrt2}{2}\right)^2 \\ &= 98 - 8 - \dfrac{25\pi}{2} \\ &= 90 - \dfrac{25\pi}{2} \approx 51. \end{aligned} \]

Thus, the correct answer is C.

19 答案 E

设 \(x = 10a + b\)、\(y = 10b + a\)，其中 \(a \gt b\)。则 \[ \begin{aligned} &x^2 - y^2 = (10a+b)^2 \\ &\quad {}- (10b+a)^2 \\ &= 99(a^2 - b^2) \\ &= 99(a+b)(a-b) \end{aligned}\text{。}\]

因为 \(99 = 9 \cdot 11\)，要使它成为完全平方数，必须有 \((a+b)(a-b)\) 是 \(11\) 的倍数。由于 \(a + b \le 17\)，且 \(a - b \le 8\)，可用的唯一 \(11\) 的倍数是 \(a + b = 11\)。

此时 \(x^2 - y^2 = 9 \cdot 11^2 (a - b)\)，它为完全平方数，当且仅当 \(a - b\) 是完全平方数。因为 \(a+b=11\) 是奇数，所以 \(a-b\) 是奇数；又因为 \(1 \le a-b \le 8\)，唯一可能的平方值为 \(1\)。因此 \((a, b) = (6, 5)\)。

所以 \(x = 65\)，\(y = 56\)，且 \(m = \sqrt{65^2 - 56^2}\) \(= \sqrt{1089} = 33\)。因此 \(x + y + m\) \(= 65 + 56 + 33 = 154\)。

所以正确答案是 E。

English

Let \(x = 10a + b\) and \(y = 10b + a\) with \(a \gt b.\) Then \[ \begin{aligned} &x^2 - y^2 = (10a+b)^2 \\ &\quad {}- (10b+a)^2 \\ &= 99(a^2 - b^2) \\ &= 99(a+b)(a-b). \end{aligned} \]

Since \(99 = 9 \cdot 11,\) for this to be a perfect square we need \((a+b)(a-b)\) to be a multiple of \(11.\) As \(a + b \le 17\) and \(a - b \le 8,\) the only multiple of \(11\) available is \(a + b = 11.\)

Then \(x^2 - y^2 = 9 \cdot 11^2 (a - b),\) which is a perfect square exactly when \(a - b\) is a perfect square. Because \(a+b=11\) is odd, \(a-b\) is odd; and because \(1 \le a-b \le 8,\) its only possible square value is \(1.\) Hence \((a, b) = (6, 5).\)

So \(x = 65,\) \(y = 56,\) and \(m = \sqrt{65^2 - 56^2}\) \(= \sqrt{1089} = 33.\) Thus \(x + y + m\) \(= 65 + 56 + 33 = 154.\)

Thus, the correct answer is E.

20 答案 C

所有元素的和为 \(8\)。若 \(a + b + c + d = x\)，则 \(e + f + g + h = 8 - x\)，所以 \[ x^2 + (8 - x)^2 = 2(x - 4)^2 + 32\text{。}\]

该式在 \(x = 4\) 时取得最小值 \(32\)。但是 \(13\) 必须在某一组中，而其余元素中没有三个数能与 \(13\) 相加得到 \(4\)（这要求三个数之和为 \(-9\)）。若包含 \(-7\)，另两个数需要和为 \(-2\)，但没有可用数对满足；若不包含 \(-7\)，把 \(-5-3-2=-10\) 中任一项替换都会使和超过 \(-9\)。所以 \(x = 4\) 无法达到，且 \((x - 4)^2 \ge 1\)。

最小值为 \(2(1) + 32 = 34\)，例如分组 \(\{-7, -5, 2, 13\}\)（和为 \(3\)）与 \(\{-3, -2, 4, 6\}\)（和为 \(5\)）可以达到。

所以正确答案是 C。

English

The elements sum to \(8.\) If \(a + b + c + d = x,\) then \(e + f + g + h = 8 - x,\) so \[ x^2 + (8 - x)^2 = 2(x - 4)^2 + 32. \]

This is minimized when \(x = 4,\) giving \(32.\) But \(13\) must lie in one group, and no three of the remaining elements add with \(13\) to make \(4\) (that would need three of them to sum to \(-9\)). With \(-7,\) the other two would need to sum to \(-2,\) which no available pair does; without \(-7,\) replacing any term in \(-5-3-2=-10\) raises the sum past \(-9.\) So \(x = 4\) is unattainable and \((x - 4)^2 \ge 1.\)

The minimum is \(2(1) + 32 = 34,\) achieved for instance by \(\{-7, -5, 2, 13\}\) (sum \(3\)) and \(\{-3, -2, 4, 6\}\) (sum \(5\)).

Thus, the correct answer is C.

21 答案 C

写成 \(n = 7^k Q\)，其中 \(Q\) 不是 \(7\) 的倍数，并设 \(Q\) 有 \(d\) 个因数。那么 \(n\) 有 \((k + 1)d = 60\) 个因数，而 \(7n = 7^{k+1}Q\) 有 \((k + 2)d = 80\) 个因数。

两式相除，得到 \(\dfrac{k + 2}{k + 1} = \dfrac{80}{60} = \dfrac43\)，所以 \(3(k + 2) = 4(k + 1)\)，从而 \(k = 2\)。

所以正确答案是 C。

English

Write \(n = 7^k Q\) where \(Q\) is not divisible by \(7,\) and let \(d\) be the number of divisors of \(Q.\) Then \(n\) has \((k + 1)d = 60\) divisors and \(7n = 7^{k+1}Q\) has \((k + 2)d = 80\) divisors.

Dividing, \(\dfrac{k + 2}{k + 1} = \dfrac{80}{60} = \dfrac43,\) so \(3(k + 2) = 4(k + 1),\) giving \(k = 2.\)

Thus, the correct answer is C.

22 答案 E

因为 \(|z_0| = 1\)，每个 \(|z_n| = 1\)，所以 \(\overline{z_n} = \dfrac{1}{z_n}\)，于是 \[ z_{n+1} = \dfrac{i z_n}{\overline{z_n}} = i z_n^2\text{。}\]

迭代得 \(z_1 = i z_0^2\)，\(z_2 = i(i z_0^2)^2 = -i z_0^4\)。此外，若 \(z_n=-i z_0^{2^n}\)，则 \(z_{n+1}=i(-i)^2z_0^{2^{n+1}}=-i z_0^{2^{n+1}}\)。于是对每个 \(n\ge2\) 都有 \(z_n=-i z_0^{2^n}\)。

于是条件 \(z_{2005} = 1\) 就化为 \(z_0^{2^{2005}} = i\)。而任何非零复数方程 \(z_0^{N} = i\) 恰好有 \(N\) 个不同的解，且它们都在单位圆上。

这里 \(N = 2^{2005}\)，所以 \(z_0\) 有 \(2^{2005}\) 个可能值。

所以正确答案是 E。

English

Because \(|z_0| = 1,\) every \(|z_n| = 1,\) so \(\overline{z_n} = \dfrac{1}{z_n}\) and \[ z_{n+1} = \dfrac{i z_n}{\overline{z_n}} = i z_n^2. \]

Iterating, \(z_1 = i z_0^2\) and \(z_2 = i(i z_0^2)^2 = -i z_0^4.\) Moreover, if \(z_n=-i z_0^{2^n},\) then \(z_{n+1}=i(-i)^2z_0^{2^{n+1}}=-i z_0^{2^{n+1}}.\) Thus \(z_n=-i z_0^{2^n}\) for every \(n\ge2.\)

The condition \(z_{2005} = 1\) is therefore \(z_0^{2^{2005}} = i.\) Every nonzero complex equation \(z_0^{N} = i\) has exactly \(N\) distinct solutions, all on the unit circle.

Here \(N = 2^{2005},\) so there are \(2^{2005}\) possible values for \(z_0.\)

Thus, the correct answer is E.

23 答案 B

条件给出 \(x + y = 10^z\) 和 \(x^2 + y^2 = 10 \cdot 10^z\)。于是 \[ \begin{aligned} &2xy = (x+y)^2 - (x^2+y^2) \\ &= 10^{2z} - 10 \cdot 10^z \end{aligned}\text{，}\]所以 \(xy = \dfrac12\left(10^{2z} - 10 \cdot 10^z\right)\)。

使用 \(x^3 + y^3\) \(= (x+y)^3 - 3xy(x+y)\)，得到 \[ \begin{aligned} &x^3 + y^3 = 10^{3z} \\ &\quad {}- \dfrac32\left(10^{2z} - 10 \cdot 10^z\right)10^z \\ &= -\dfrac12 \cdot 10^{3z} + 15 \cdot 10^{2z} \end{aligned}\text{。}\]

所以 \(a = -\dfrac12\)、\(b = 15\)，从而 \(a + b = \dfrac{29}{2}\)。这些系数是唯一确定的：令 \(z=0\)，可得实数解 \(x,y=\dfrac{1\pm\sqrt{19}}{2}\)，其方程迫使 \(a+b\) 仍取同一值。

所以正确答案是 B。

English

The conditions give \(x + y = 10^z\) and \(x^2 + y^2 = 10 \cdot 10^z.\) Then \[ \begin{aligned} &2xy = (x+y)^2 - (x^2+y^2) \\ &= 10^{2z} - 10 \cdot 10^z, \end{aligned} \] so \(xy = \dfrac12\left(10^{2z} - 10 \cdot 10^z\right).\)

Using \(x^3 + y^3\) \(= (x+y)^3 - 3xy(x+y),\) \[ \begin{aligned} &x^3 + y^3 = 10^{3z} \\ &\quad {}- \dfrac32\left(10^{2z} - 10 \cdot 10^z\right)10^z \\ &= -\dfrac12 \cdot 10^{3z} + 15 \cdot 10^{2z}. \end{aligned} \]

So \(a = -\dfrac12\) and \(b = 15,\) giving \(a + b = \dfrac{29}{2}.\) These coefficients are determined: setting \(z=0\) gives real solutions \(x,y=\dfrac{1\pm\sqrt{19}}{2},\) and their equation forces this same value of \(a+b.\)

Thus, the correct answer is B.

24 答案 A

对顶点 \((a, a^2), (b, b^2), (c, c^2)\)，一条边的斜率为 \(\dfrac{b^2 - a^2}{b - a} = a + b\)。三条边斜率相加得 \[ \begin{aligned} &(a+b) + (b+c) + (c+a) \\ &= 2(a + b + c) \\ &= 2 \cdot \dfrac{m}{n} \end{aligned}\text{。}\]

若一条边的方向角为 \(\theta\)，其斜率为 \(2 = \tan\theta\)。等边三角形另外两条边的方向角为 \(\theta \pm 60^\circ\)，所以它们的斜率为 \[ \begin{aligned} &\tan(\theta \pm 60^\circ) = \dfrac{2 \pm \sqrt3}{1 \mp 2\sqrt3} \\ &= -\dfrac{8 \pm 5\sqrt3}{11} \end{aligned}\text{。}\]

三个斜率和为 \(2 - \dfrac{8 + 5\sqrt3}{11}\) \(- \dfrac{8 - 5\sqrt3}{11} =\) \(\dfrac{22 - 16}{11} = \dfrac{6}{11}\)。

因此 \(a + b + c = \dfrac12 \cdot \dfrac{6}{11} = \dfrac{3}{11}\)，所以 \(m + n = 3 + 11 = 14\)。

所以正确答案是 A。

English

For vertices \((a, a^2), (b, b^2), (c, c^2),\) the slope of a side is \(\dfrac{b^2 - a^2}{b - a} = a + b.\) Adding the three side slopes, \[ \begin{aligned} &(a+b) + (b+c) + (c+a) \\ &= 2(a + b + c) \\ &= 2 \cdot \dfrac{m}{n}. \end{aligned} \]

One side has slope \(2 = \tan\theta.\) Because the triangle is equilateral, its sides make angles \(\theta\) and \(\theta \pm 60^\circ,\) so the other two slopes are \[ \begin{aligned} &\tan(\theta \pm 60^\circ) = \dfrac{2 \pm \sqrt3}{1 \mp 2\sqrt3} \\ &= -\dfrac{8 \pm 5\sqrt3}{11}. \end{aligned} \]

The sum of the three slopes is \(2 - \dfrac{8 + 5\sqrt3}{11}\) \(- \dfrac{8 - 5\sqrt3}{11} =\) \(\dfrac{22 - 16}{11} = \dfrac{6}{11}.\)

Thus \(a + b + c = \dfrac12 \cdot \dfrac{6}{11} = \dfrac{3}{11},\) so \(m + n = 3 + 11 = 14.\)

Thus, the correct answer is A.

25 答案 A

共有 \(4^6\) 种等可能移动。把顶点标为 \(A, B, C, A', B', C'\)，其中带撇号的是对应对顶点。有效结果是一个排列 \(f\)，且 \(f(A) \notin \{A, A'\}\)，其他顶点同理。

有序对 \((f(A), f(A'))\) 有 \(4 \cdot 3 = 12\) 个选择。其中 \(f(A)\)、\(f(A')\) 互为对顶点的有 \(4\) 种，相邻的有 \(8\) 种。

若 \(f(A), f(A')\) 互为对顶点，例如 \(B, B'\)，则 \(\{f(C), f(C')\} = \{A, A'\}\)，且 \(\{f(B), f(B')\} = \{C, C'\}\)，给出 \(4 \cdot 2 \cdot 2 = 16\) 种。

若 \(f(A), f(A')\) 相邻，例如 \(B, C\)，则 \(f(B), f(B')\) 中必须有一个是 \(C'\)，有 \(4\) 个有序选择 \((f(B), f(B'))\)，每个给 \((f(C), f(C'))\) 留下 \(2\) 种，共 \(8 \cdot 4 \cdot 2 = 64\) 种。

概率为 \[ \dfrac{16 + 64}{4^6} = \dfrac{80}{4096} = \dfrac{5}{256}\text{。}\]

所以正确答案是 A。

English

There are \(4^6\) equally likely combinations of moves. Label the vertices \(A, B, C, A', B', C',\) where primed vertices are opposite the corresponding unprimed ones. An ant cannot move to its own vertex or the opposite one, so a valid outcome is a permutation \(f\) with \(f(A) \notin \{A, A'\},\) and similarly for each pair.

There are \(4 \cdot 3 = 12\) ordered choices for \((f(A), f(A')).\) Of these, \(f(A)\) and \(f(A')\) are opposite in \(4\) cases and adjacent in \(8.\)

If \(f(A), f(A')\) are opposite, say \(B, B',\) then \(\{f(C), f(C')\} = \{A, A'\}\) and \(\{f(B), f(B')\} = \{C, C'\},\) giving \(4 \cdot 2 \cdot 2 = 16\) valid combinations.

If \(f(A), f(A')\) are adjacent, say \(B, C,\) then one of \(f(B), f(B')\) must be \(C'\) and there are \(4\) ordered choices for \((f(B), f(B')),\) each leaving \(2\) for \((f(C), f(C')):\) that is \(8 \cdot 4 \cdot 2 = 64\) valid combinations.

Hence the probability is \[ \dfrac{16 + 64}{4^6} = \dfrac{80}{4096} = \dfrac{5}{256}. \]

Thus, the correct answer is A.