2000 AMC 12 真题（P版）

分解质因数得 \(2001 = 3 \cdot 23 \cdot 29\)。

在乘积固定且三个正整数互不相同的条件下，要使和最大，取 \(I = 1\)，\(M = 3\)，且 \(O = 23 \cdot 29 = 667\)。

最大和为 \(1 + 3 + 667 = 671\)。

因此，正确答案是 E。

English

Factoring gives \(2001 = 3 \cdot 23 \cdot 29.\)

To maximize the sum of three distinct positive integers with this product, take \(I = 1,\) \(M = 3,\) and \(O = 23 \cdot 29 = 667.\)

The largest sum is \(1 + 3 + 667 = 671.\)

Thus, the correct answer is E.

将 \(2000 = 2000^1\) 得到 \[ \begin{gathered} 2000^1 \cdot 2000^{2000} \\ = 2000^{1 + 2000} \\ = 2000^{2001} \end{gathered}\text{。}\]

其他选项都比这个值大。

因此，正确答案是 A。

English

Writing \(2000 = 2000^1,\) we get \[ \begin{gathered} 2000^1 \cdot 2000^{2000} \\ = 2000^{1 + 2000} \\ = 2000^{2001}. \end{gathered} \]

All of the other options are larger than this.

Thus, the correct answer is A.

每天吃掉 \(20\%\)，所以每天结束时剩下 \(80\%\)。若原来有 \(x\) 颗，则 \[ (0.8)^2 x = 32\text{。}\]

解得 \(0.64x = 32\)，所以 \(x = 50\)。

因此，正确答案是 B。

English

Since \(20\%\) is eaten each day, \(80\%\) remains at the end of each day. If \(x\) is the original number, then \[ (0.8)^2 x = 32. \]

Solving gives \(0.64x = 32,\) so \(x = 50.\)

Thus, the correct answer is B.

个位数字序列开始为 \[ \begin{gathered} 1, 1, 2, 3, 5, 8, 3, 1, 4, 5, 9, \\ 4, 3, 7, 0, 7, 7, 4, 1, 5, 6, \ldots \end{gathered} \]

查看这个列表，数字 \(6\) 是十个数字中最后出现的。

因此，正确答案是 C。

English

The sequence of units digits begins \[ \begin{gathered} 1, 1, 2, 3, 5, 8, 3, 1, 4, 5, 9, \\ 4, 3, 7, 0, 7, 7, 4, 1, 5, 6, \ldots \end{gathered} \]

Scanning this list, the digit \(6\) is the last of the ten digits to appear.

Thus, the correct answer is C.

因为 \(x \lt 2\)，所以 \(|x - 2| = 2 - x = p\)，因此 \(x = 2 - p\)。

于是 \[ x - p = (2 - p) - p = 2 - 2p\text{。}\]

因此，正确答案是 C。

English

Since \(x \lt 2,\) we have \(|x - 2| = 2 - x = p,\) so \(x = 2 - p.\)

Then \[ x - p = (2 - p) - p = 2 - 2p. \]

Thus, the correct answer is C.

\(4\) 和 \(18\) 之间的质数是 \(5, 7, 11, 13\) 和 \(17\)。

对这样的两个质数，\(xy - (x + y)\) \(= (x - 1)(y - 1) - 1\) 是两个偶数的乘积减去 \(1\)，因此它同余于 \(3 \pmod 4\)。这样只剩下 \(119\) 和 \(231\)。后者要求 \((x-1)(y-1)=232\)，但 \(\{4,6,10,12,16\}\) 中没有两个不同的数的乘积等于 \(232\)。

实际上，\[ \begin{aligned} 11 \cdot 13 - (11 + 13) &= 143 - 24 \\ &= 119 \end{aligned}\text{。}\]

因此，正确答案是 C。

English

The primes between \(4\) and \(18\) are \(5, 7, 11, 13,\) and \(17.\)

For two such primes, \(xy - (x + y)\) \(= (x - 1)(y - 1) - 1\) is a product of two even numbers minus \(1,\) hence it is \(3 \pmod 4.\) This leaves \(119\) and \(231.\) The latter would require \((x-1)(y-1)=232,\) but no two distinct numbers in \(\{4,6,10,12,16\}\) have product \(232.\)

Indeed, \[ \begin{aligned} 11 \cdot 13 - (11 + 13) &= 143 - 24 \\ &= 119. \end{aligned} \]

Thus, the correct answer is C.

若 \(\log_b 729 = n\)，则 \(b^n = 729 = 3^6\)，所以 \(n\) 必须是 \(6\) 的正因数。

\(n = 1, 2, 3, 6\) 分别给出 \(b = 3^6, 3^3, 3^2, 3^1\)，也就是 \(729, 27, 9\) 和 \(3\)。

共有 \(4\) 个这样的 \(b\)。

因此，正确答案是 E。

English

If \(\log_b 729 = n,\) then \(b^n = 729 = 3^6,\) so \(n\) must be a positive divisor of \(6.\)

The possibilities \(n = 1, 2, 3, 6\) give \(b = 3^6, 3^3, 3^2, 3^1,\) that is, \(729, 27, 9,\) and \(3.\)

There are \(4\) such values of \(b.\)

Thus, the correct answer is E.

图形 \(n\) 是一个菱形，各行长度按奇数增大后再减小，所以共有 \(n^2 + (n + 1)^2\) 个单位正方形。这与 \(n = 0, 1, 2, 3\) 时的 \(1, 5, 13, 25\) 相符。

因此图形 \(100\) 中单位正方形的个数为 \[ \begin{aligned} 100^2 + 101^2 &= 10000 + 10201 \\ &= 20201 \end{aligned} \] 个。

因此，正确答案是 C。

English

Figure \(n\) is a diamond whose row lengths increase through the odd numbers and back down, giving a total of \(n^2 + (n + 1)^2\) unit squares. This matches \(1, 5, 13, 25\) for \(n = 0, 1, 2, 3.\)

Therefore figure \(100\) has \[ \begin{aligned} 100^2 + 101^2 &= 10000 + 10201 \\ &= 20201 \end{aligned} \] unit squares.

Thus, the correct answer is C.

总和为 \(71 + 76 + 80 + 82 + 91 = 400\)，可被 \(5\) 整除。前三个成绩的和必须能被 \(3\) 整除。

模 \(3\)，这些成绩的余数是 \(2, 1, 2, 1, 1\)。唯一和为 \(3\) 的倍数的三元组是 \(76 + 82 + 91 = 249\)，所以它们是前三个成绩（其中 \(91\) 第三个，因为前两个 \(76\) 和 \(82\)，必须奇偶性相同）。

因为 \(249 \equiv 1 \pmod 4\)，第四个成绩必须 \(\equiv 3 \pmod 4\)，即 \(71\)。于是最后输入的是 \(80\)。

因此，正确答案是 C。

English

The total is \(71 + 76 + 80 + 82 + 91 = 400,\) which is divisible by \(5.\) The sum of the first three scores must be divisible by \(3.\)

Modulo \(3,\) the scores are \(2, 1, 2, 1, 1.\) The only triple summing to a multiple of \(3\) is \(76 + 82 + 91 = 249,\) so these are the first three (with \(91\) third, since the first two, \(76\) and \(82,\) must have equal parity).

Since \(249 \equiv 1 \pmod 4,\) the fourth score must be \(\equiv 3 \pmod 4,\) which is \(71.\) That leaves \(80\) as the last score entered.

Thus, the correct answer is C.

\((1, 2, 3)\) 关于 \(xy\)-平面反射得到 \(Q = (1, 2, -3)\)。

绕 \(x\)-轴旋转 \(180^\circ\) 会使 \(y\) 和 \(z\) 变号，得到 \(R = (1, -2, 3)\)。

沿正 \(y\) 方向平移 \(5\) 个单位得到 \(S = (1, 3, 3)\)。

因此，正确答案是 E。

English

Reflecting \((1, 2, 3)\) in the \(xy\)-plane gives \(Q = (1, 2, -3).\)

Rotating \(180^\circ\) about the \(x\)-axis negates \(y\) and \(z,\) giving \(R = (1, -2, 3).\)

Translating \(5\) units in the positive \(y\) direction gives \(S = (1, 3, 3).\)

Thus, the correct answer is E.

通分得 \[ \frac{a}{b} + \frac{b}{a} - ab = \frac{a^2 + b^2 - (ab)^2}{ab}\text{。}\]

在分子中用 \(a - b\) 替换 \(ab\)，得 \[ \begin{gathered} a^2 + b^2 - (a - b)^2 \\ = a^2 + b^2 \\ {}- (a^2 - 2ab + b^2) \\ = 2ab \end{gathered}\text{。}\]

因此原式等于 \(\dfrac{2ab}{ab} = 2\)。

因此，正确答案是 E。

English

Combining over a common denominator, \[ \frac{a}{b} + \frac{b}{a} - ab = \frac{a^2 + b^2 - (ab)^2}{ab}. \]

Replacing \(ab\) with \(a - b\) in the numerator, \[ \begin{gathered} a^2 + b^2 - (a - b)^2 \\ = a^2 + b^2 \\ {}- (a^2 - 2ab + b^2) \\ = 2ab. \end{gathered} \]

Therefore the expression equals \(\dfrac{2ab}{ab} = 2.\)

Thus, the correct answer is E.

注意 \[ \begin{aligned} &AMC + AM + MC + CA \\ &\quad = (A + 1)(M + 1)(C + 1) \\ &\quad {}- (A + M + C) - 1 \end{aligned}\text{。}\]

因为 \(A + M + C = 12\)，原式等于 \((A + 1)(M + 1)(C + 1) - 13\)。这三个因数的和为 \(15\)，乘积在它们都等于 \(5\) 时最大，等于 \(5^3 = 125\)。

最大值为 \(125 - 13 = 112\)。

因此，正确答案是 E。

English

Observe that \[ \begin{aligned} &AMC + AM + MC + CA \\ &\quad = (A + 1)(M + 1)(C + 1) \\ &\quad {}- (A + M + C) - 1. \end{aligned} \]

Since \(A + M + C = 12,\) this equals \((A + 1)(M + 1)(C + 1) - 13.\) The three factors sum to \(15,\) so their product is maximized when each equals \(5,\) giving \(5^3 = 125.\)

The maximum value is \(125 - 13 = 112.\)

Thus, the correct answer is E.

以 \(8\)-盎司的一杯为单位计量，所以 Angela 的杯中有 \(c\) 杯咖啡和 \(m\) 杯牛奶，且 \(c + m = 1\)。

因为 Angela 喝掉了咖啡总量的六分之一，所以咖啡总量为 \(6c\)；因为她喝掉了牛奶总量的四分之一，所以牛奶总量为 \(4m\)。人数等于总杯数：\[ \begin{aligned} 6c + 4m &= 6c + 4(1 - c) \\ &= 4 + 2c \end{aligned}\text{。}\]

这必须是整数，所以 \(2c\) 是整数。又因 \(0 \lt c \lt 1\)，只能有 \(c = \tfrac12\)，从而人数为 \(4 + 1 = 5\)。

因此，正确答案是 C。

English

Measure amounts in \(8\)-ounce cups, so Angela’s cup holds \(c\) coffee and \(m\) milk with \(c + m = 1.\)

Since Angela drank a sixth of the coffee, the total coffee is \(6c\); since she drank a quarter of the milk, the total milk is \(4m.\) The number of people equals the total number of cups, \[ \begin{aligned} 6c + 4m &= 6c + 4(1 - c) \\ &= 4 + 2c. \end{aligned} \]

This is an integer only when \(2c\) is an integer, and since \(0 \lt c \lt 1\) this forces \(c = \tfrac12,\) giving \(4 + 1 = 5\) people.

Thus, the correct answer is C.

六个固定数的和为 \(25\)，所以平均数是 \(\dfrac{25 + x}{7}\)，众数是 \(2\)。若 \(x \le 2\)，则 \(2\) 同时是中位数和众数，会迫使等差数列为常数，所以 \(x \gt 2\)。

情形 \(2 \lt x \lt 4\)： 中位数为 \(x\)。要求 \(2, x, \dfrac{25 + x}{7}\) 构成等差数列，得到此范围内唯一的值 \(x = 3\)。

情形 \(x \ge 4\)： 中位数为 \(4\)，等差数列 \(2, 4, 6\) 迫使平均数为 \(6\)，所以 \(\dfrac{25 + x}{7} = 6\)，得 \(x = 17\)。

所有可能值的和为 \(3 + 17 = 20\)。

因此，正确答案是 E。

English

The six fixed numbers sum to \(25,\) so the mean is \(\dfrac{25 + x}{7},\) and the mode is \(2.\) If \(x \le 2,\) then \(2\) is both median and mode, forcing a constant progression, so \(x \gt 2.\)

Case \(2 \lt x \lt 4\): the median is \(x.\) Requiring \(2, x, \dfrac{25 + x}{7}\) to form an arithmetic progression yields \(x = 3\) as the only value in this range.

Case \(x \ge 4\): the median is \(4,\) and the progression \(2, 4, 6\) forces the mean to be \(6,\) so \(\dfrac{25 + x}{7} = 6,\) giving \(x = 17.\)

The sum of all possible values is \(3 + 17 = 20.\)

Thus, the correct answer is E.

令 \(\dfrac{x}{3} = 3z\) 得到 \(x = 9z\)，所以 \[ \begin{aligned} f(3z) &= (9z)^2 + 9z + 1 \\ &= 81z^2 + 9z + 1 \\ &= 7 \end{aligned}\text{。}\]

整理得 \(81z^2 + 9z - 6 = 0\)。

由根和公式，所有 \(z\) 的和为 \(-\dfrac{9}{81} = -\dfrac{1}{9}\)。

因此，正确答案是 B。

English

Setting \(\dfrac{x}{3} = 3z\) gives \(x = 9z,\) so \[ \begin{aligned} f(3z) &= (9z)^2 + 9z + 1 \\ &= 81z^2 + 9z + 1 \\ &= 7. \end{aligned} \]

This rearranges to \(81z^2 + 9z - 6 = 0.\)

By the sum-of-roots formula, the sum of the values of \(z\) is \(-\dfrac{9}{81} = -\dfrac{1}{9}.\)

Thus, the correct answer is B.

方格 \((m, n)\) 原来的编号为 \(17(m - 1) + n\)，重新编号后为 \(13(n - 1) + m\)。令它们相等，得 \[ 4m - 3n = 1\text{。}\]

满足 \(1 \le m \le 13\) 且 \(1 \le n \le 17\) 的解为 \((1, 1)\)，\((4, 5)\)，\((7, 9)\)，\((10, 13)\)，和 \((13, 17)\)。

这些方格中的数为 \(1, 56, 111, 166\) 和 \(221\)，总和为 \(555\)。

因此，正确答案是 D。

English

The square \((m, n)\) is numbered \(17(m - 1) + n\) originally and \(13(n - 1) + m\) after renumbering. Setting these equal gives \[ 4m - 3n = 1. \]

The solutions with \(1 \le m \le 13\) and \(1 \le n \le 17\) are \((1, 1),\) \((4, 5),\) \((7, 9),\) \((10, 13),\) and \((13, 17).\)

These squares hold the numbers \(1, 56, 111, 166,\) and \(221,\) whose sum is \(555.\)

Thus, the correct answer is D.

因为 \(OA = 1\) 且 \(AB\) 在 \(A\) 点与圆相切，角 \(OAB\) 为直角，所以 \[ BA = \tan\theta, \qquad OB = \sec\theta\text{。}\]

因为 \(BC\) 平分 \(\angle ABO\)，由角平分线定理得 \(\dfrac{OC}{CA} = \dfrac{OB}{BA}\)。又 \(OC + CA = OA = 1\)，\[ \begin{aligned} OC &= \frac{OB}{OB + BA} \\ &= \frac{\sec\theta}{\sec\theta + \tan\theta} \end{aligned}\text{。}\]

分子分母同乘 \(\cos\theta\) 得到 \(OC = \dfrac{1}{1 + \sin\theta}\)。

因此，正确答案是 D。

English

Because \(OA = 1\) and \(AB\) is tangent at \(A,\) angle \(OAB\) is right, so \[ BA = \tan\theta, \qquad OB = \sec\theta. \]

Since \(BC\) bisects \(\angle ABO,\) the angle bisector theorem gives \(\dfrac{OC}{CA} = \dfrac{OB}{BA}.\) Using \(OC + CA = OA = 1,\) \[ \begin{aligned} OC &= \frac{OB}{OB + BA} \\ &= \frac{\sec\theta}{\sec\theta + \tan\theta}. \end{aligned} \]

Multiplying numerator and denominator by \(\cos\theta\) gives \(OC = \dfrac{1}{1 + \sin\theta}.\)

Thus, the correct answer is D.

从 \(N\) 年第 \(300\) 天到 \(N + 1\) 年第 \(200\) 天，若 \(N\) 不是闰年，相隔 \(365 - 300 + 200 = 265\) 天。但 \(265 = 7 \cdot 37 + 6\)，会落在星期一，而不是星期二。

所以 \(N\) 是闰年，相隔 \(266 = 7 \cdot 38\) 天，正好仍为星期二。因此 \(N - 1\) 年不是闰年。

\(N - 1\) 年第 \(100\) 天比 \(N\) 年第 \(300\) 天早 \(365 - 100 + 300 = 565\) 天。因为 \(565 = 7 \cdot 80 + 5\)，所以那一天比星期二早 \(5\) 天，即星期四。

因此，正确答案是 A。

English

From day \(300\) of year \(N\) to day \(200\) of year \(N + 1,\) the number of days is \(365 - 300 + 200 = 265\) if \(N\) is not a leap year. But \(265 = 7 \cdot 37 + 6,\) which would land on a Monday, not a Tuesday.

So year \(N\) is a leap year, and the gap is \(266 = 7 \cdot 38\) days, giving a Tuesday as stated. It follows that year \(N - 1\) is not a leap year.

The \(100\)th day of year \(N - 1\) precedes the Tuesday on day \(300\) of year \(N\) by \(365 - 100 + 300 = 565\) days. Since \(565 = 7 \cdot 80 + 5,\) that day is \(5\) weekdays before Tuesday, namely Thursday.

Thus, the correct answer is A.

由海伦公式，\(\triangle ABC\) 的面积为 \(\sqrt{21 \cdot 8 \cdot 7 \cdot 6} = 84\)，所以从 \(A\) 到 \(BC\) 的高为 \(\dfrac{2 \cdot 84}{14} = 12\)。

中点 \(D\) 到 \(B\) 的距离为 \(7\)。从 \(A\) 作出的角平分线与 \(BC\) 相交于 \(E\)，且 \(BE : EC = AB : AC = 13 : 15\)，所以 \(BE = 14 \cdot \dfrac{13}{28} = 6.5\)。

\(D\) 和 \(E\) 都在 \(BC\) 上，所以 \(\triangle ADE\) 的底为 \(DE = 7 - 6.5 = 0.5\)，高为 \(12\)，面积为 \[ \tfrac12 \cdot 0.5 \cdot 12 = 3\text{。}\]

因此，正确答案是 C。

English

By Heron’s formula, the area of \(\triangle ABC\) is \(\sqrt{21 \cdot 8 \cdot 7 \cdot 6} = 84,\) so the altitude from \(A\) to \(BC\) is \(\dfrac{2 \cdot 84}{14} = 12.\)

The midpoint \(D\) is \(7\) from \(B.\) The bisector from \(A\) meets \(BC\) at \(E\) with \(BE : EC = AB : AC = 13 : 15,\) so \(BE = 14 \cdot \dfrac{13}{28} = 6.5.\)

Both \(D\) and \(E\) lie on \(BC,\) so \(\triangle ADE\) has base \(DE = 7 - 6.5 = 0.5\) and altitude \(12,\) giving area \[ \tfrac12 \cdot 0.5 \cdot 12 = 3. \]

Thus, the correct answer is C.

三个方程相加得 \[ \begin{gathered} \left(x + \tfrac1y\right) + \left(y + \tfrac1z\right) \\ {}+ \left(z + \tfrac1x\right) \\ = 4 + 1 + \tfrac73 \\ = \tfrac{22}{3}\text{。} \end{gathered} \]

三个方程相乘得 \[ 4 \cdot 1 \cdot \tfrac73 = \tfrac{28}{3}\text{。} \]

展开乘积，得到 \[ \begin{aligned} &\left(x + \tfrac1y\right) \\ &\quad {}\cdot \left(y + \tfrac1z\right) \\ &\quad {}\cdot \left(z + \tfrac1x\right) \\ &= xyz \\ &\quad {}+ \left(x + y + z + \tfrac1x + \tfrac1y + \tfrac1z\right) \\ &\quad {}+ \frac{1}{xyz}\text{。} \end{aligned} \] 中间括号中的一组之和就是 \(\tfrac{22}{3}\)，所以 \(xyz + \dfrac{1}{xyz} = \tfrac{28}{3} - \tfrac{22}{3} = 2\)。

因此 \((xyz - 1)^2 = 0\)，所以 \(xyz = 1\)。

因此，正确答案是 B。

English

Adding the three equations gives \[ \begin{gathered} \left(x + \tfrac1y\right) + \left(y + \tfrac1z\right) \\ {}+ \left(z + \tfrac1x\right) \\ = 4 + 1 + \tfrac73 \\ = \tfrac{22}{3}. \end{gathered} \]

Multiplying them gives \[ 4 \cdot 1 \cdot \tfrac73 = \tfrac{28}{3}. \]

Expanding the product, \[ \begin{aligned} &\left(x + \tfrac1y\right) \\ &\quad {}\cdot \left(y + \tfrac1z\right) \\ &\quad {}\cdot \left(z + \tfrac1x\right) \\ &= xyz \\ &\quad {}+ \left(x + y + z + \tfrac1x + \tfrac1y + \tfrac1z\right) \\ &\quad {}+ \frac{1}{xyz}. \end{aligned} \] The middle group is the sum \(\tfrac{22}{3},\) so \(xyz + \dfrac{1}{xyz} = \tfrac{28}{3} - \tfrac{22}{3} = 2.\)

Hence \((xyz - 1)^2 = 0,\) so \(xyz = 1.\)

Thus, the correct answer is B.

设正方形边长为 \(1\)。与正方形一边相邻的小三角形有一条长为 \(r\) 的垂直边，所以它的面积为 \(\tfrac12 \cdot 1 \cdot r = m\)，从而 \(r = 2m\)。

两个小三角形相似，所以另一个三角形沿正方形的直角边长为 \(\dfrac1r\)，面积为 \[ \frac12 \cdot 1 \cdot \frac1r = \frac{1}{2r} = \frac{1}{4m}\text{。}\]

因此，正确答案是 D。

English

Let the square have side \(1.\) The small triangle sharing one side of the square has a perpendicular leg \(r,\) so its area is \(\tfrac12 \cdot 1 \cdot r = m,\) giving \(r = 2m.\)

The two small triangles are similar, so the other triangle’s leg along the square is \(\dfrac1r,\) and its area is \[ \frac12 \cdot 1 \cdot \frac1r = \frac{1}{2r} = \frac{1}{4m}. \]

Thus, the correct answer is D.

图像与 \(x\)-轴恰好相交两次，且两次都在正数处，所以 \(P\) 有两个实零点和两个非实（复共轭）零点。

从图像读出：系数之和为 \(P(1) \gt 3\)；\(P(-1) \gt 4\)；实零点之和大于 \(4.5\)；所有零点的乘积为 \(d\)，即 \(y\)-截距，小于 \(6\)。

设实零点的乘积为 \(R\)，则 \(R \gt 4.5\)。非实零点的乘积为 \(\dfrac dR\)，它小于 \(\dfrac{6}{4.5} \lt 2\)。

这比列出的其他每一个量都小。

因此，正确答案是 C。

English

The graph crosses the \(x\)-axis exactly twice, both times at positive values, so \(P\) has two real zeros and two non-real (complex conjugate) zeros.

Reading off the graph: the sum of the coefficients is \(P(1) \gt 3;\) \(P(-1) \gt 4;\) the sum of the real zeros is greater than \(4.5;\) and the product of all zeros is \(d,\) the \(y\)-intercept, which is less than \(6.\)

Let \(R\) be the product of the real zeros, so \(R \gt 4.5.\) The product of the non-real zeros is \(\dfrac dR,\) which is less than \(\dfrac{6}{4.5} \lt 2.\)

This is smaller than every other listed quantity.

Thus, the correct answer is C.

对数之和为整数 \(k\)，当且仅当六个数的乘积为 \(10^k\)。因为 \(10 = 2 \cdot 5\)，每个选中的数都必须形如 \(2^a 5^b\)，所以只能来自 \[ 1, 2, 4, 5, 8, 10, 16, 20, 25, 32, 40\text{。}\]

对每个数记录因数 \(2\) 的个数比因数 \(5\) 的个数多多少：\(0, 1, 2, -1, 3, 0, 4, 1, -2, 5, 2\)。乘积为 \(10\) 的幂，要求六个所选数中 \(2\) 与 \(5\) 的总指数相等，也就是这些差值之和为 \(0\)。

负差值只有 \(5\) 的 \(-1\) 和 \(25\) 的 \(-2\)。总差值为 \(0\) 的六数彩票必须同时包含两者：缺少任意一个，都没有足够的零和小正差值凑满六个数。因此另外四个数的差值之和必须为 \(3\)。差值为 \(0\) 的数有两个（\(1,10\)），差值为 \(1\) 的数有两个（\(2,20\)），差值为 \(2\) 的数有两个（\(4,40\)）；若使用差值至少为 \(3\) 的数，就没有足够的零差值数凑满四个。因此必须取两个差值为 \(0\) 的数、一个差值为 \(1\) 的数和一个差值为 \(2\) 的数。恰有四张有效彩票：\(\{1, 5, 10, 20, 25, 40\}\)，\(\{1, 2, 5, 10, 25, 40\}\)，\(\{1, 2, 4, 5, 10, 25\}\)，和 \(\{1, 4, 5, 10, 20, 25\}\)。

Gamble 教授持有其中一张，而只有一张与中奖彩票相同，所以概率为 \(\dfrac14\)。

所以正确答案是 B。

English

The sum of the logarithms is an integer \(k\) exactly when the product of the six numbers is \(10^k.\) Since \(10 = 2 \cdot 5,\) each chosen number must be of the form \(2^a 5^b,\) so it comes from \[ 1, 2, 4, 5, 8, 10, 16, 20, 25, 32, 40. \]

For each, record the excess of factors of \(2\) over factors of \(5\): \(0, 1, 2, -1, 3, 0, 4, 1, -2, 5, 2.\) The product is a power of \(10\) only if the six chosen values have equal totals of \(2\)s and \(5\)s, i.e. their excesses sum to \(0.\)

The only negative excesses are \(-1\) for \(5\) and \(-2\) for \(25.\) A six-number ticket with total excess \(0\) must contain both: omitting either leaves too few zero and small positive excesses to reach six numbers. The other four numbers must therefore have total excess \(3.\) There are two numbers of excess \(0\) (\(1,10\)), two of excess \(1\) (\(2,20\)), and two of excess \(2\) (\(4,40\)); any number of excess at least \(3\) would leave too few zeros to complete a four-number selection. Thus we must take both excess-\(0\) numbers, one excess-\(1\) number, and one excess-\(2\) number. This gives exactly four valid tickets: \(\{1, 5, 10, 20, 25, 40\},\) \(\{1, 2, 5, 10, 25, 40\},\) \(\{1, 2, 4, 5, 10, 25\},\) and \(\{1, 4, 5, 10, 20, 25\}.\)

Professor Gamble holds one of these four, and only one matches the winning ticket, so the probability is \(\dfrac14.\)

Thus, the correct answer is B.

每条圆弧的半径都是 \(AB\)，且 \(C\) 到 \(A\) 和 \(B\) 的距离都为 \(AB\)，所以 \(\triangle ABC\) 是等边三角形。因此圆弧 \(BC\) 对应半径为 \(AB\) 的圆上的 \(60^\circ\) 圆心角，该大圆的整周长为 \(6 \cdot 12 = 72\)。

设小圆半径为 \(r\)，并在线段 \(\overline{AB}\) 的中点 \(D\) 处相切，其中 \(AD = \tfrac12 AB\)。由点幂定理，\(AD^2 = AB(AB - 2r)\)，所以 \[ \frac{AB^2}{4} = AB^2 - 2r\,AB\text{，} \] 从而 \(2r = \tfrac34 AB\)，因而 \(r = \tfrac38 AB\)。

周长之比等于半径之比，所以小圆的周长为 \(\tfrac38 \cdot 72 = 27\)。

因此，正确答案是 D。

English

Each arc has radius \(AB,\) and \(C\) is at distance \(AB\) from both \(A\) and \(B,\) so \(\triangle ABC\) is equilateral. Thus arc \(BC\) subtends \(60^\circ\) of a circle of radius \(AB,\) whose full circumference is \(6 \cdot 12 = 72.\)

Let the small circle have radius \(r\) and touch \(\overline{AB}\) at its midpoint \(D,\) where \(AD = \tfrac12 AB.\) By Power of a Point, \(AD^2 = AB(AB - 2r),\) so \[ \frac{AB^2}{4} = AB^2 - 2r\,AB, \] giving \(2r = \tfrac34 AB,\) hence \(r = \tfrac38 AB.\)

The circumferences are in the ratio of the radii, so the small circle’s circumference is \(\tfrac38 \cdot 72 = 27.\)

Thus, the correct answer is D.

将八种不同颜色分配给八个面共有 \(8!\) 种方式。两个分配表示同一个八面体，当且仅当其中一个可以旋转成另一个。

正八面体的旋转群有 \(24\) 个元素。因为八种颜色全都不同，没有非平凡旋转会固定一种着色，所以每个可区分的八面体恰好对应 \(24\) 种分配。

因此可区分的八面体数量为 \[ \frac{8!}{24} = \frac{40320}{24} = 1680\text{。}\]

因此，正确答案是 E。

English

There are \(8!\) ways to assign the eight distinct colors to the eight faces. Two assignments give the same octahedron exactly when one is a rotation of the other.

The rotation group of a regular octahedron has \(24\) elements. Because all eight colors are different, no nontrivial rotation fixes a coloring, so each distinguishable octahedron corresponds to exactly \(24\) assignments.

Therefore the number of distinguishable octahedrons is \[ \frac{8!}{24} = \frac{40320}{24} = 1680. \]

Thus, the correct answer is E.