2014B AMC 10 真题（P版）

1 答案 C

设一美分硬币有 \(p\) 枚，则五美分硬币有 \(13-p\) 枚；按题意，五美分硬币数也等于 \(p-1\)，所以 \(13-p=p-1\)。因此有 \(7\) 枚一美分硬币和 \(6\) 枚五美分硬币。

硬币的总价值为 \(7+6\cdot 5=37\) 美分。

所以正确答案是 C。

English

Let the number of pennies be \(p.\) Then, the number of nickels is \(13-p\) and \(p-1,\) so \(13-p=p-1.\) This means we have \(7\) pennies and \(6\) nickels.

Therefore, the number of cents is \(7+6\cdot 5=37.\)

Thus, the correct answer is C .

2 答案 E

\[\begin{aligned} \dfrac{2^3 + 2^3}{2^{-3} + 2^{-3}} &= \dfrac{2\cdot 2^3}{2\cdot 2^{-3}} \\&=2^6 \\&= 64 \end{aligned}\]

所以正确答案是 E。

English

\[\begin{aligned} \dfrac{2^3 + 2^3}{2^{-3} + 2^{-3}} &= \dfrac{2\cdot 2^3}{2\cdot 2^{-3}} \\&=2^6 \\&= 64 \end{aligned}\]

Thus, the correct answer is E .

3 答案 E

设全程为 \(t\) 英里，则 \[\begin{aligned}t &= \dfrac t5 + 20 + \dfrac t3 \\ t &= 20 + \dfrac {8t}{15} \\ \dfrac 7{15} t &= 20\\ t &= \dfrac{300}7\end{aligned}\text{。}\]

所以正确答案是 E。

English

Let the path distance be \(t.\) Then, we get: \[\begin{aligned}t &= \dfrac t5 + 20 + \dfrac t3 \\ t &= 20 + \dfrac {8t}{15} \\ \dfrac 7{15} t &= 20\\ t &= \dfrac{300}7.\end{aligned}\]

Thus, the correct answer is E .

4 答案 B

设一个松饼价格为 \(m\)，一根香蕉价格为 \(b\)。\[\begin{aligned} 2(4m+3b) &= 16b + 2m\\ 8m+6b &= 16b + 2m \\ 10b &= 6m\\ m &= \dfrac 53 b\end{aligned}\text{。}\]

所以正确答案是 B。

English

Let the price for a muffin be \(m\) and let the price for a banana be \(b.\) Then, \[\begin{aligned} 2(4m+3b) &= 16b + 2m\\ 8m+6b &= 16b + 2m \\ 10b &= 6m\\ m &= \dfrac 53 b.\end{aligned}\]

Thus, the correct answer is B .

5 答案 A

设每块玻璃的高为 \(5x\)，宽为 \(2x\)。竖直方向上，窗户包含两块玻璃的高度和三条横框，所以高度为 \(10x+6\)。水平方向上，它包含四块玻璃的宽度和五条竖框，所以宽度为 \(8x+10\)。

因为窗户是正方形，\(10x+6=8x+10\)，所以 \(x=2\)。

因此边长为 \(10(2)+6=26\) 英寸。

所以正确答案是 A。

English

Let each pane have height \(5x\) and width \(2x\). Vertically, the window contains two pane heights and three horizontal borders, so its height is \(10x+6\). Horizontally, it contains four pane widths and five vertical borders, so its width is \(8x+10\).

Because the window is square, \(10x+6=8x+10\), so \(x=2\).

Therefore, the side length is \(10(2)+6=26\) inches.

Thus, the correct answer is A .

6 答案 C

每一对气球的价格是一个气球原价的 \(1+\frac23=\frac53\) 倍。

因此，带着刚好够买 \(30\) 个原价气球的钱，Orvin 可以买 \(30\cdot\frac65=36\) 个气球。

所以正确答案是 C。

English

Each pair of balloons costs \(1+\frac23=\frac53\) times the regular price of one balloon.

Thus, with the money for \(30\) regular-price balloons, Orvin can buy \(30\cdot\frac65=36\) balloons.

Thus, the correct answer is C .

7 答案 A

由定义可知 \[A = \dfrac {x+ 100}{100}B \]\[= B + \dfrac x{100}B\text{。}\]

由此可得 \[\begin{aligned} (A-B) &= \dfrac x{100}B\\ 100\left(\dfrac{A-B} B\right) &=x\end{aligned}\text{。}\]

所以正确答案是 A。

English

By definition, we know \[A = \dfrac {x+ 100}{100}B \]\[= B + \dfrac x{100}B.\]

This implies, \[\begin{aligned} (A-B) &= \dfrac x{100}B\\ 100\left(\dfrac{A-B} B\right) &=x.\end{aligned}\]

Thus, the correct answer is A .

8 答案 E

卡车每 \(t\) 秒行驶 \(\dfrac{b}{18}\) 码，因为 \(3\) 英尺等于一码；所以卡车每秒行驶 \(\dfrac{b}{18t}\) 码。

\(3\) 分钟为 \(180\) 秒，因此行驶 \[\dfrac{b}{18t}\cdot 180 = \dfrac {10b}t\text{。}\]

所以正确答案是 E。

English

This means it travels \(\dfrac{b}{18}\) yards in \(t\) seconds since a yard is \(3\) feet. Then, in one second, it travels \(\dfrac{b}{18t}\) yards.

Therefore, in \(3\) minutes which is \(180\) seconds, it travels \[\dfrac{b}{18t}\cdot 180 = \dfrac {10b}t.\]

Thus, the correct answer is E .

9 答案 A

分子和分母同乘后，得到 \[\begin{aligned} \dfrac{wz}{wz}\cdot \dfrac{\frac{1}{w} + \frac{1}{z}}{\frac{1}{w} - \frac{1}{z}} &= 2014\\ \dfrac{w+z}{z-w} &= 2014\\ \dfrac{w+z}{w-z} &= -1\cdot 2014\\ \dfrac{w+z}{w-z}&=-2014\end{aligned}\text{。}\]

所以正确答案是 A。

English

Observe that: \[\begin{aligned} \dfrac{wz}{wz}\cdot \dfrac{\frac{1}{w} + \frac{1}{z}}{\frac{1}{w} - \frac{1}{z}} &= 2014\\ \dfrac{w+z}{z-w} &= 2014\\ \dfrac{w+z}{w-z} &= -1\cdot 2014\\ \dfrac{w+z}{w-z}&=-2014.\end{aligned}\]

Thus, the correct answer is A .

10 答案 C

最左列没有产生第六位进位，所以 \(A+B=D\le 9\)。

个位列也是 \(B+A=D\)，所以个位列也没有进位。十位列给出 \(C+D=D\)，因此 \(C=0\)。

因为 \(A\) 和 \(B\) 是不同的非零数字，\(D=A+B\) 可以是从 \(3\) 到 \(9\) 的任意数字。例如，\((A,B)=(1,2)\)、\((1,3)\)、\((2,3)\)、\((2,4)\)、\((2,5)\)、\((2,6)\)、\((2,7)\) 分别给出 \(D=3,4,5,6,7,8,9\)。

所以 \(D\) 有 \(7\) 个可能值，正确答案是 C。

English

From the leftmost column, there is no carry into a sixth digit, so \(A+B=D\le 9\).

The units column is \(B+A=D\), so it also has no carry. The tens column then gives \(C+D=D\), hence \(C=0\).

Since \(A\) and \(B\) are distinct nonzero digits, \(D=A+B\) can be any digit from \(3\) through \(9\). For example, \((A,B)=(1,2),\) \((1,3),\) \((2,3),\) \((2,4),\) \((2,5),\) \((2,6),\) \((2,7)\) give \(D=3,4,5,6,7,8,9\).

Thus there are \(7\) possible values of \(D\), and the correct answer is C .

11 答案 C

三种连续折扣方案分别使顾客支付标价的 \[ \begin{aligned} 0.85^2&=0.7225,\\ 0.9^3&=0.729,\\ 0.75(0.95)&=0.7125 \end{aligned} \] 倍。因此它们的实际折扣分别为 \(27.75\%\)、\(27.1\%\) 和 \(28.75\%\)。

单次 \(n\%\) 折扣必须大于这三个折扣，所以 \(n>28.75\)。符合条件的最小正整数是 \(29\)。

所以正确答案是 C。

English

The three successive-discount plans leave the customer paying \[ \begin{aligned} 0.85^2&=0.7225,\\ 0.9^3&=0.729,\\ 0.75(0.95)&=0.7125 \end{aligned} \] times the listed price. Their effective discounts are therefore \(27.75\%\), \(27.1\%\), and \(28.75\%\).

A single \(n\%\) discount must exceed all three, so \(n>28.75\). The smallest positive integer that works is \(29\).

Thus, the correct answer is C .

12 答案 C

第五大因数等于 \(2,014,000,000\) 除以第五小因数。

\(2,014,000,000\) 的素因数分解为 \[2^7\cdot 5^6 \cdot 19\cdot 53\text{。}\] 最小的 \(5\) 个因数是 \(1,2,4,5,8\)。所以第五小因数为 \(8\)，第五大因数为 \[\dfrac{2014000000}8 =251750000\text{。}\]

所以正确答案是 C。

English

The fifth-largest divisor is \(2,014,000,000\) divided by the fifth smallest divisor.

The prime factorization of \(2,014,000,000\) is: \[2^7\cdot 5^6 \cdot 19\cdot 53.\] This makes the first \(5\) smallest divisors \(1,2,4,5,8.\) Therefore, the fifth smallest divisor is \(8,\) and the fifth largest divisor must be: \[\dfrac{2014000000}8 =251750000.\]

Thus, the correct answer is C .

13 答案 B

沿着单位正六边形构成的 \(60^\circ\) 网格建立坐标，取 \(A=(0,0)\)。于是图中标出的另外两个顶点可以写成 \(B=(3,\sqrt3)\) 和 \(C=(3,-\sqrt3)\)。

因此 \(BC=2\sqrt3\)，且 \[ \begin{aligned} AB=AC&=\sqrt{3^2+(\sqrt3)^2}\\ &=2\sqrt3 \end{aligned}\text{。}\] 于是 \(\triangle ABC\) 是等边三角形，面积为 \[ \frac{(2\sqrt3)^2\sqrt3}{4}=3\sqrt3\text{。}\]

所以正确答案是 B。

English

Follow the \(60^\circ\) grid formed by the unit hexagons and place \(A=(0,0)\). The marked vertices may then be written as \(B=(3,\sqrt3)\) and \(C=(3,-\sqrt3)\).

Thus \(BC=2\sqrt3\), and \[ \begin{aligned} AB=AC&=\sqrt{3^2+(\sqrt3)^2}\\ &=2\sqrt3. \end{aligned} \] Hence \(\triangle ABC\) is equilateral, with area \[ \frac{(2\sqrt3)^2\sqrt3}{4}=3\sqrt3. \]

Thus, the correct answer is B .

14 答案 D

里程表读数 \(cba\) 与 \(abc\) 之差为 \[100c + 10b+a - 100a - 10b-c \] \[= 99(c-a)\text{，}\]而这个差必须是 \(55\) 的倍数。因为 \(\gcd(55,99)\) 为 \(11\)，所以 \(c-a\) 是 \(5\) 的倍数，且 \(c > a\)。

在 \(a+ b+c \leq 7\) 的限制下，唯一可能是 \(a = 1, b = 0, c = 6\)，其他组合都会有 \(a+b+c > 7\)。因此 \(a^2+b^2+c^2 = 37\)。

所以正确答案是 D。

English

We know that the difference of the numbers \(cba\) and \(abc\) is equal to: \[100c + 10b+a - 100a - 10b-c \]\[= 99(c-a)\] We know that this number also must be a multiple of \(55.\) As \(\gcd(55,99)\) is \(11,\) we know that \(c-a\) is a multiple of \(5,\) and \(c > a.\)

This makes \(a = 1, b = 0, c = 6\) the only possible value with \(a+ b+c \leq 7\) as every other combination has \(a+b+c > 7.\) As such, \(a^2+b^2+c^2 = 37.\)

Thus, the correct answer is D .

15 答案 A

设 \(AD=h\)。因为 \(DC=AB=2h\)，所以长方形 \(ABCD\) 的面积为 \(2h^2\)。

射线 \(DE\) 和 \(DF\) 与 \(DC\) 的夹角分别为 \(60^\circ\) 和 \(30^\circ\)。

因此 \[AE=h\tan30^\circ=\dfrac{h}{\sqrt3}\] 且 \[AF=h\tan60^\circ=h\sqrt3\text{。}\]

于是 \[\begin{aligned}EF&=AF-AE\\&=h\sqrt3-\dfrac{h}{\sqrt3}\\&=\dfrac{2h\sqrt3}{3}\end{aligned}\text{。}\] \(\triangle DEF\) 的面积为 \[\dfrac12\cdot EF\cdot h=\dfrac{h^2\sqrt3}{3}\text{。}\]

所求面积比为 \[\dfrac{\frac{h^2\sqrt3}{3}}{2h^2}=\dfrac{\sqrt3}{6}\text{。}\] 所以正确答案是 A。

English

Let \(AD=h.\) Since \(DC=AB=2h,\) the area of rectangle \(ABCD\) is \(2h^2.\)

The rays \(DE\) and \(DF\) make angles of \(60^\circ\) and \(30^\circ,\) respectively, with \(DC.\)

Therefore, \[AE=h\tan30^\circ=\dfrac{h}{\sqrt3}\] and \[AF=h\tan60^\circ=h\sqrt3.\]

Hence \[\begin{aligned}EF&=AF-AE\\&=h\sqrt3-\dfrac{h}{\sqrt3}\\&=\dfrac{2h\sqrt3}{3}.\end{aligned}\] The area of \(\triangle DEF\) is \[\dfrac12\cdot EF\cdot h=\dfrac{h^2\sqrt3}{3}.\]

The desired ratio is \[\dfrac{\frac{h^2\sqrt3}{3}}{2h^2}=\dfrac{\sqrt3}{6}.\] Thus, the correct answer is A.

16 答案 B

共有 \(6^4\) 个等可能有序结果。

恰好三枚相同时，重复点数有 \(6\) 种，不同点数有 \(5\) 种，不同骰子的位置有 \(4\) 种，共 \(6\cdot5\cdot4=120\) 种。

四枚全相同有 \(6\) 种。

所求概率为 \(\frac{120+6}{6^4}=\frac{126}{1296}=\frac7{72}\)。

所以正确答案是 B。

English

There are \(6^4\) equally likely ordered outcomes.

If exactly three dice show the same value, choose the repeated value in \(6\) ways, the different value in \(5\) ways, and the position of the different die in \(4\) ways. This gives \(6\cdot5\cdot4=120\) outcomes.

If all four dice match, there are \(6\) outcomes.

The probability is \(\frac{120+6}{6^4}=\frac{126}{1296}=\frac7{72}\).

Thus, the correct answer is B .

17 答案 D

先提取明显的 \(2\) 的幂：\(10^{1002}-4^{501}=2^{1002}(5^{1002}-1)\)。

因为 \(5^{1002}-1=(5^{501}-1)(5^{501}+1)\)，且 \(501\) 为奇数，所以 \(5^{501}-1\) 可被 \(4\) 整除但不能被 \(8\) 整除，而 \(5^{501}+1\) 可被 \(2\) 整除但不能被 \(4\) 整除。

因此 \(5^{1002}-1\) 正好贡献 \(2^3\)，整个表达式可被 \(2^{1005}\) 整除，但不能被 \(2^{1006}\) 整除。

所以正确答案是 D。

English

Factor out the obvious power of \(2\): \(10^{1002}-4^{501}=2^{1002}(5^{1002}-1)\).

Since \(5^{1002}-1=(5^{501}-1)(5^{501}+1)\), and \(501\) is odd, \(5^{501}-1\) is divisible by \(4\) but not by \(8\), while \(5^{501}+1\) is divisible by \(2\) but not by \(4\).

Thus \(5^{1002}-1\) contributes exactly \(2^3\), so the whole expression is divisible by \(2^{1005}\) but not \(2^{1006}\).

Thus, the correct answer is D .

18 答案 E

总和为 \(11\cdot10=110\)。

要最大化最大项，应最小化其余十项之和。按非递减顺序排列时，第六项为 \(9\)，且 \(8\) 必须是唯一众数。若 \(8\) 出现两次，前十项最小和为 \(1+2+3+8+8+9\) \(+10+11+12+13\) \(=77\)，最大项为 \(33\)。

若 \(8\) 出现三次，前十项可取 \(1,1,8,8,8,9,9,10,10,11\)，其和为 \(75\)，最大项可为 \(35\)。

若 \(8\) 出现四次或五次，前十项最小和至少为 \(80\)，最大项至多为 \(30\)。

因此最大可能值为 \(35\)，正确答案是 E。

English

The list has total sum \(11\cdot10=110\). To maximize the largest entry, minimize the sum of the other ten entries.

In nondecreasing order, the sixth entry is \(9\), and \(8\) must be the unique mode. If \(8\) appears twice, the least possible first ten entries sum to \(1+2+3+8+8+9\) \(+10+11+12+13\) \(=77\), giving largest entry \(33\).

If \(8\) appears three times, the least possible first ten entries are \(1,1,8,8,8,9,9,10,10,11\), with sum \(75\), giving largest entry \(35\).

If \(8\) appears four or five times, the least possible sum of the first ten entries is at least \(80\), so the largest entry is at most \(30\).

Therefore the largest possible entry is \(35\), and the correct answer is E .

19 答案 D

在外圆上固定第一个端点 \(A\)。从 \(A\) 作两条与内圆相切的弦，并把它们的另一个端点记为 \(B\) 和 \(C\)。从 \(A\) 出发的弦与内圆相交，当且仅当它的另一个端点位于小弧 \(BC\) 上。

设 \(O\) 为两圆的共同圆心，\(D\) 为一个切点，则 \(\triangle AOD\) 是直角三角形，其中 \(OA=2\)，\(OD=1\)。所以 \(\angle OAD=30^\circ\)。两条切线弦在 \(A\) 处所成角为 \(60^\circ\)，因此所截小弧 \(BC\) 的度数为 \(120^\circ\)。

所以所求概率为 \[\dfrac {120^\circ}{360^\circ} = \dfrac 13\text{。}\]

所以正确答案是 D。

English

Fix the first endpoint \(A\) on the outer circle. Draw the two chords from \(A\) that are tangent to the inner circle, and call their other endpoints \(B\) and \(C\). A chord from \(A\) meets the inner circle exactly when its second endpoint lies on the minor arc \(BC\).

If \(O\) is the common center and \(D\) is a tangency point, then \(\triangle AOD\) is right, with \(OA=2\) and \(OD=1\). Hence \(\angle OAD=30^\circ\). The two tangent chords therefore make a \(60^\circ\) angle at \(A\), so the intercepted minor arc \(BC\) measures \(120^\circ\).

Therefore, the probability is \[\dfrac {120^\circ}{360^\circ} = \dfrac 13 .\]

Thus, the correct answer is D .

20 答案 C

首先注意到 \[x^4-51x^2+50 \]\[= (x^2-50)(x^2-1)\text{。}\]

乘积为负，当且仅当两个因子异号。由于 \(x^2-50<x^2-1\)，必须有 \(x^2-50<0<x^2-1\)。所以 \(1<x^2<50\)，即 \(2\le |x|\le7\)。正整数解有 \(6\) 个，负整数解也有 \(6\) 个，共有 \(12\) 个。

所以正确答案是 C。

English

First, note that \[x^4-51x^2+50 \]\[= (x^2-50)(x^2-1).\]

The product is negative exactly when its two factors have opposite signs. Since \(x^2-50<x^2-1\), this requires \(x^2-50<0<x^2-1\). Thus \(1<x^2<50\), or \(2\le |x|\le7\). There are \(6\) positive and \(6\) negative integer solutions, for a total of \(12\).

Thus, the correct answer is C .

21 答案 B

从 \(D\) 和 \(C\) 向 \(AB\) 作垂线，垂足分别为 \(E\) 和 \(F\)。如图，不妨取 \(AD=10\)、\(BC=14\)；交换两腰只会把梯形翻转。

设 \(AE=x\)，高为 \(h\)。由于 \(EF=CD=21\)，且 \(AB=33\)，所以 \(FB=12-x\)。

由两个直角三角形可得 \[10^2=x^2+h^2\] 和 \[14^2=(12-x)^2+h^2\text{。}\]两式相减得 \(96=144-24x\)，所以 \(x=2\)，且 \(h^2=96\)。

较短的对角线是 \(AC\)，它的水平位移为 \(AE+EF=2+21=23\)。因此 \[AC=\sqrt{23^2+96}=\sqrt{625}=25\text{。}\]

所以正确答案是 B。

English

Let the feet of the perpendiculars from \(D\) and \(C\) to \(AB\) be \(E\) and \(F\), respectively. As drawn, take \(AD=10\) and \(BC=14\); interchanging the two legs only reflects the trapezoid.

Let \(AE=x\) and let the altitude be \(h\). Since \(EF=CD=21\) and \(AB=33\), we have \(FB=12-x\).

The two right triangles give \[10^2=x^2+h^2\] and \[14^2=(12-x)^2+h^2.\] Subtracting yields \(96=144-24x\), so \(x=2\) and \(h^2=96\).

The shorter diagonal is \(AC\), whose horizontal displacement is \(AE+EF=2+21=23\). Therefore \[AC=\sqrt{23^2+96}=\sqrt{625}=25.\]

Thus, the correct answer is B .

22 答案 B

从正方形中心到一个半圆圆心的距离可由直角三角形的斜边求出。

一条直角边是从正方形中心到边中点的距离，长度为 \(1\)。

另一条直角边是从边中点到半圆圆心的距离，长度为 \(\dfrac 12\)。这也说明半圆半径为 \(\dfrac 12\)。

因此正方形中心到半圆圆心的距离为 \[\sqrt{1^2 + \left(\dfrac 12\right)^2} = \dfrac {\sqrt 5}2\text{。}\] 再减去半圆半径 \(\dfrac 12\)，得到小圆半径 \[\dfrac{\sqrt 5 -1}2\text{。}\]

所以正确答案是 B。

English

The distance from the center of the square to the center of the semicircles can be found as a hypotenuse of a right triangle.

One of the legs is from the center of the square to the center of one of the sides which is of distance \(1.\)

The other leg is from the center of the side to the center of one of the semicircles which is of distance \(\dfrac 12.\) This also shows that the radius of the semicircles is \(\dfrac 12.\)

Therefore, the distance from the center of the square to the center of the semicircle is \[\sqrt{1^2 + \left(\dfrac 12\right)^2} = \dfrac {\sqrt 5}2.\] Then we subtract \(\dfrac 12\) for the radius of the semicircle. This makes the radius of the circle \[\dfrac{\sqrt 5 -1}2 .\]

Thus, the correct answer is B .

23 答案 E

设上底半径为 \(1\)，下底半径为 \(R\)，内切球半径为 \(a\)。

在截面中，球与两个底面相切，所以截锥的高为 \(2a\)。若以球心为原点，一条斜边连接 \((1,a)\) 与 \((R,-a)\)。它的方程是 \[2ax+(R-1)y-a(R+1)=0\text{。}\]因为这条直线与半径为 \(a\) 的圆相切，它到原点的距离为 \(a\)。所以 \[\frac{a(R+1)}{\sqrt{4a^2+(R-1)^2}}=a\text{，}\]化简得 \(R=a^2\)。

截锥的体积为 \(\frac13\pi(R^2+R+1)(2a)\) \(=\frac{2a\pi}{3}(a^4+a^2+1)\)。

它等于球体积 \(\frac{8a^3\pi}{3}\) 的两倍。约去公因子可得 \(a^4-3a^2+1=0\)，所以 \(R^2-3R+1=0\)。

由于下底半径大于上底半径，\(R>1\)。因此 \(R=\frac{3+\sqrt5}{2}\)，正确答案是 E。

English

Let the top radius be \(1\), the bottom radius be \(R\), and the inscribed sphere radius be \(a\).

In the cross-section, the sphere is tangent to the two bases, so the frustum height is \(2a\). A slanted side joins \((1,a)\) to \((R,-a)\), if the sphere’s center is the origin. Its equation is \[2ax+(R-1)y-a(R+1)=0.\] Because this line is tangent to the circle of radius \(a\), its distance from the origin is \(a\). Thus \[\frac{a(R+1)}{\sqrt{4a^2+(R-1)^2}}=a,\] which simplifies to \(R=a^2\).

The frustum volume is \(\frac13\pi(R^2+R+1)(2a)\) \(=\frac{2a\pi}{3}(a^4+a^2+1)\).

This is twice the sphere volume, \(\frac{8a^3\pi}{3}\). Cancelling gives \(a^4-3a^2+1=0\), so \(R^2-3R+1=0\).

Since the bottom radius is larger than the top radius, \(R>1\). Thus \(R=\frac{3+\sqrt5}{2}\), and the correct answer is E .

24 答案 B

单个数字给出 \(1\) 到 \(5\)，它们的补集给出 \(10\) 到 \(14\)，全部五个数给出 \(15\)。因此只需要检查能否得到和 \(6\) 与 \(7\)。

若无法得到和 \(6\)，则 \(1\) 不与 \(5\) 相邻。通过旋转和翻转，可写成 \(1bc5e\)。相邻块 \(bc\) 不能是 \(\{2,3\}\) 或 \(\{2,4\}\)，因为 \(1+2+3=6\) 且 \(2+4=6\)。于是 \(e=2\)，再避免连续块 \(2,1,3\)，得到坏排列 \(14352\)。

若无法得到 \(7\)，则 \(2\) 不与 \(5\) 相邻，可写成 \(2bc5e\)。此时 \(bc\) 不能是 \(\{3,4\}\) 或 \(\{1,4\}\)，所以 \(e=4\)。再避免连续块 \(4,2,1\)，得到 \(b=3,\ c=1\)，即坏排列 \(23154\)。

这两个排列确实都是坏排列，分别无法得到和 \(6\) 与和 \(7\)。因此共有 \(2\) 种坏排列。

所以正确答案是 B。

English

Single numbers give sums \(1\) through \(5\), complements give sums \(10\) through \(14\), and all five numbers give \(15\). So an arrangement is good exactly when consecutive blocks can make sums \(6\) and \(7\).

If sum \(6\) is impossible, then \(1\) is not adjacent to \(5\). By rotating and reflecting, write the arrangement as \(1bc5e\). The adjacent pair \(bc\) cannot be \(\{2,3\}\) or \(\{2,4\}\), since \(1+2+3=6\) and \(2+4=6\). Thus \(e=2\), and avoiding the consecutive block \(2,1,3\) forces the bad arrangement \(14352\).

If sum \(7\) is impossible, then \(2\) is not adjacent to \(5\). Similarly write the arrangement as \(2bc5e\). Now \(bc\) cannot be \(\{3,4\}\) or \(\{1,4\}\), so \(e=4\). To avoid the consecutive block \(4,2,1\), the remaining order must be \(b=3,\ c=1\), giving \(23154\).

These two arrangements are indeed bad, one missing sum \(6\) and the other missing sum \(7\). Hence there are \(2\) bad arrangements.

Thus, the correct answer is B .

25 答案 C

设 \(p_i\) 为从第 \(i\) 片叶开始最终逃脱的概率。边界条件为 \(p_0=0\)、\(p_{10}=1\)，由对称性 \(p_5=\frac12\)。

对 \(1\le i\le 4\)，有递推式 \(p_i=\frac{i}{10}p_{i-1}+\frac{10-i}{10}p_{i+1}\)。

从 \(p_5=\frac12\) 向下推，得到 \(p_4=\frac25p_3+\frac3{10}\) 和 \(p_3=\frac3{10}p_2+\frac7{10}p_4=\frac5{12}p_2+\frac7{24}\)。

接着可得 \(p_2=\frac15p_1+\frac45p_3=\frac3{10}p_1+\frac7{20}\)，而 \(p_1=\frac9{10}p_2\)。

代入 \(p_2\) 的表达式，得到 \(p_1=\frac9{10}\left(\frac3{10}p_1+\frac7{20}\right)\)，所以 \(p_1=\frac{63}{146}\)。

所以正确答案是 C。

English

Let \(p_i\) be the probability that the frog eventually escapes starting from pad \(i\). Then \(p_0=0\), \(p_{10}=1\), and by symmetry \(p_5=\frac12\).

For \(1\le i\le 4\), \(p_i=\frac{i}{10}p_{i-1}+\frac{10-i}{10}p_{i+1}\).

Working downward from \(p_5=\frac12\), we get \(p_4=\frac25p_3+\frac3{10}\), then \(p_3=\frac3{10}p_2+\frac7{10}p_4=\frac5{12}p_2+\frac7{24}\).

Next \(p_2=\frac15p_1+\frac45p_3=\frac3{10}p_1+\frac7{20}\). Finally \(p_1=\frac9{10}p_2\).

Substituting the expression for \(p_2\) gives \(p_1=\frac9{10}\left(\frac3{10}p_1+\frac7{20}\right)\), so \(p_1=\frac{63}{146}\).

Thus, the correct answer is C .