2016A AMC 12 真题（P版）

1 答案 B

分解分子可得 \[ \begin{gathered} \dfrac{11!-10!}{9!}\\ =\dfrac{10!(11-1)}{9!}\\ =\dfrac{10\cdot 9!\cdot 10}{9!}\\ =100 \end{gathered}\text{。}\]

所以正确答案是 B。

English

Factoring the numerator gives \[ \begin{gathered} \dfrac{11!-10!}{9!}\\ =\dfrac{10!(11-1)}{9!}\\ =\dfrac{10\cdot 9!\cdot 10}{9!}\\ =100. \end{gathered} \]

Thus, the correct answer is B.

2 答案 C

因为 \(100=10^2\)，\(1000=10^3\)，方程变为 \[ 10^x\cdot 10^{4x}=10^{15}\text{，} \] 所以 \(10^{5x}=10^{15}\)。于是 \(5x=15\)，得到 \(x=3\)。

所以正确答案是 C。

English

Since \(100=10^2\) and \(1000=10^3,\) the equation becomes \[ 10^x\cdot 10^{4x}=10^{15}, \] so \(10^{5x}=10^{15}.\) Then \(5x=15,\) giving \(x=3.\)

Thus, the correct answer is C.

3 答案 B

首先， \[ \begin{gathered} \dfrac{x}{y}=\dfrac{\frac{3}{8}}{-\frac{2}{5}}\\ =\dfrac{3}{8}\cdot\left(-\dfrac{5}{2}\right)\\ =-\dfrac{15}{16} \end{gathered}\text{，}\] 所以 \(\left\lfloor -\dfrac{15}{16}\right\rfloor=-1\)。

因此 \[ \begin{gathered} \text{rem}\left(\dfrac{3}{8},-\dfrac{2}{5}\right)\\ =\dfrac{3}{8}-\left(-\dfrac{2}{5}\right)(-1)\\ =\dfrac{3}{8}-\dfrac{2}{5}\\ =\dfrac{15-16}{40}\\ =-\dfrac{1}{40} \end{gathered}\text{。}\]

所以正确答案是 B。

English

First, \[ \begin{gathered} \dfrac{x}{y}=\dfrac{\frac{3}{8}}{-\frac{2}{5}}\\ =\dfrac{3}{8}\cdot\left(-\dfrac{5}{2}\right)\\ =-\dfrac{15}{16}, \end{gathered} \] and \(\left\lfloor -\dfrac{15}{16}\right\rfloor=-1.\)

Therefore \[ \begin{gathered} \text{rem}\left(\dfrac{3}{8},-\dfrac{2}{5}\right)\\ =\dfrac{3}{8}-\left(-\dfrac{2}{5}\right)(-1)\\ =\dfrac{3}{8}-\dfrac{2}{5}\\ =\dfrac{15-16}{40}\\ =-\dfrac{1}{40}. \end{gathered} \]

Thus, the correct answer is B.

4 答案 D

由平均数条件， \[ \begin{gathered} \small \dfrac{60+100+x+40+50+200+90}{7}\\ =\dfrac{540+x}{7}\\ =x\text{，} \end{gathered} \] 所以 \(540+x=7x\)，得到 \(x=90\)。

按非降序排列，数据为 \(40, 50, 60, 90, 90, 100, 200\)，所以中位数为 \(90\)，众数也为 \(90\)，符合要求。

所以正确答案是 D。

English

The mean condition gives \[ \begin{gathered} \small \dfrac{60+100+x+40+50+200+90}{7}\\ =\dfrac{540+x}{7}\\ =x, \end{gathered} \] so \(540+x=7x\) and \(x=90.\)

In nondecreasing order the data are \(40, 50, 60, 90, 90, 100, 200,\) so the median is \(90\) and the mode is \(90,\) as required.

Thus, the correct answer is D.

5 答案 E

反例必须满足“大于 \(2\) 的偶整数”这个前提，同时不满足“可以写成两个质数之和”这个结论。

所以正确答案是 E。

English

A counterexample must satisfy the hypothesis of being an even integer greater than \(2\) while failing the conclusion that it can be written as the sum of two prime numbers.

Thus, the correct answer is E.

6 答案 D

硬币总数为 \[ \begin{gathered} 1+2+\cdots+N\\ =\dfrac{N(N+1)}{2}\\ =2016\text{，} \end{gathered} \] 所以 \(N(N+1)=4032\)。因为 \(63\cdot 64=4032\)，所以 \(N=63\)，其各位数字之和为 \(6+3=9\)。

所以正确答案是 D。

English

The total number of coins is \[ \begin{gathered} 1+2+\cdots+N\\ =\dfrac{N(N+1)}{2}\\ =2016, \end{gathered} \] so \(N(N+1)=4032.\) Since \(63\cdot 64=4032,\) we have \(N=63,\) and the sum of its digits is \(6+3=9.\)

Thus, the correct answer is D.

7 答案 D

将所有项移到一边，得到 \[ (x^2-y^2)(x+y+1)=0\text{，} \] 再分解为 \[ (x-y)(x+y)(x+y+1)=0\text{。} \] 因此图像是直线 \(x=y\)、\(x=-y\) 和 \(x+y+1=0\) 的并集。

前两条直线交于原点，但第三条直线 \(x+y=-1\) 与 \(x+y=0\) 平行，不经过原点。因此，图像由三条不全经过同一个公共点的直线组成。

所以正确答案是 D。

English

Moving all terms to one side gives \[ (x^2-y^2)(x+y+1)=0, \] which factors as \[ (x-y)(x+y)(x+y+1)=0. \] The graph is therefore the union of the lines \(x=y,\) \(x=-y,\) and \(x+y+1=0.\)

The first two lines intersect at the origin, but the third line \(x+y=-1\) is parallel to \(x+y=0\) and does not pass through the origin. So the graph consists of three lines that do not all pass through a common point.

Thus, the correct answer is D.

8 答案 D

从长方形左上角到右下角的对角线将阴影区域分成四个三角形，它们都在长方形中心相遇。

其中两个三角形的水平底边长为 \(1\)，高为 \(\frac12\cdot 5=\frac52\)；另外两个三角形的竖直底边长为 \(1\)，高为 \(\frac12\cdot 8=4\)。总面积为 \[ \begin{gathered} 2\cdot\dfrac12\cdot 1\cdot\dfrac52\\ {}+2\cdot\dfrac12\cdot 1\cdot 4\\ =\dfrac52+4\\ =\dfrac{13}{2} \end{gathered}\text{。}\]

所以正确答案是 D。

English

The diagonal of the rectangle from the upper-left corner to the lower-right corner divides the shaded region into four triangles, all meeting at the center of the rectangle.

Two of these triangles have a horizontal base of length \(1\) and altitude \(\frac12\cdot 5=\frac52,\) and the other two have a vertical base of length \(1\) and altitude \(\frac12\cdot 8=4.\) The total area is \[ \begin{gathered} 2\cdot\dfrac12\cdot 1\cdot\dfrac52\\ {}+2\cdot\dfrac12\cdot 1\cdot 4\\ =\dfrac52+4\\ =\dfrac{13}{2}. \end{gathered} \]

Thus, the correct answer is D.

9 答案 E

设公共边长为 \(x\)。单位正方形的对角线长为 \(\sqrt{2}\)，由两条各长 \(x\sqrt2\) 的小正方形对角线和一条长为 \(x\) 的小正方形边组成，所以 \[ 2x\sqrt2+x=\sqrt2\text{。} \]

解得 \[ \begin{gathered} x=\dfrac{\sqrt2}{2\sqrt2+1}\\ =\dfrac{\sqrt2\,(2\sqrt2-1)}{(2\sqrt2+1)(2\sqrt2-1)}\\ =\dfrac{4-\sqrt2}{7}\text{。} \end{gathered} \] 因此 \(a=4\)、\(b=7\)，所以 \(a+b=11\)。

所以正确答案是 E。

English

Let \(x\) be the common side length. The diagonal of the unit square has length \(\sqrt{2}\) and consists of two small-square diagonals (each \(x\sqrt2\)) plus one small-square side length \(x,\) so \[ 2x\sqrt2+x=\sqrt2. \]

Solving, \[ \begin{gathered} x=\dfrac{\sqrt2}{2\sqrt2+1}\\ =\dfrac{\sqrt2\,(2\sqrt2-1)}{(2\sqrt2+1)(2\sqrt2-1)}\\ =\dfrac{4-\sqrt2}{7}. \end{gathered} \] Thus \(a=4,\) \(b=7,\) and \(a+b=11.\)

Thus, the correct answer is E.

10 答案 B

五位朋友的总位移为零。Dee 和 Edie 交换座位，所以两人的位移相互抵消。Bea 移动 \(+2\) 个座位，Ceci 移动 \(-1\) 个座位，净位移为 \(+1\)，所以 Ada 必须移动 \(-1\) 个座位才能抵消。

Ada 回来后坐在端座；既然她向左移动了一个座位，该端座只能是 \(1\) 号座，所以她原来坐在 \(2\) 号座。

所以正确答案是 B。

English

The net displacement of all five friends is zero. Dee and Edie swapped seats, so their movements cancel. Bea moved \(+2\) and Ceci moved \(-1,\) a net of \(+1,\) so Ada must move \(-1\) to balance.

Ada returns to an end seat; since she moved one seat to the left, that seat must be seat \(1,\) so she had been sitting in seat \(2.\)

Thus, the correct answer is B.

11 答案 E

会唱歌、跳舞、表演的人数分别为 \(100-42=58\)、\(100-65=35\)、\(100-29=71\)，总和为 \(58+35+71=164\)。

因为没有人三种才艺都会，每个学生有一种或两种才艺，所以只有一种才艺的学生被数一次，有两种才艺的学生被数两次。因此，被重复计算的人数为 \(164-100=64\)。

所以正确答案是 E。

English

The numbers who can sing, dance, and act are \(100-42=58,\) \(100-65=35,\) and \(100-29=71,\) respectively, for a total of \(58+35+71=164.\)

Since no student has all three talents, each student has one or two talents, so single-talent students are counted once and two-talent students are counted twice. The number counted twice is \(164-100=64.\)

Thus, the correct answer is E.

12 答案 C

在 \(\triangle ABC\) 中应用角平分线定理，得到 \(BD:DC=AB:AC=6:8\)，所以 \(BD=\dfrac{6}{6+8}\cdot 7=3\)。

在 \(\triangle ABD\) 中，\(BF\) 平分 \(\angle ABD\)，再次应用角平分线定理可得 \[ \begin{gathered} AF:FD=AB:BD\\ =6:3=2:1 \end{gathered}\text{。}\]

所以正确答案是 C。

English

Applying the Angle Bisector Theorem to \(\triangle ABC\) gives \(BD:DC=AB:AC=6:8,\) so \(BD=\dfrac{6}{6+8}\cdot 7=3.\)

Now \(BF\) lies along the bisector of \(\angle ABD\) in \(\triangle ABD,\) so by the Angle Bisector Theorem again, \[ \begin{gathered} AF:FD=AB:BD\\ =6:3=2:1. \end{gathered} \]

Thus, the correct answer is C.

13 答案 A

写 \(N=5k\)。从一端开始，将红球的可能位置编号为 \(0,1,\ldots,5k\)，共有 \(5k+1\) 个等可能位置。

红球两侧的绿球都少于 \(\dfrac{3}{5}\) 的情形，恰好对应红球位于 \(2k+1,2k+2,\ldots,3k-1\) 中的某个位置，共有 \(k-1\) 个。因此 \[ P(N)=1-\dfrac{k-1}{5k+1}=\dfrac{4k+2}{5k+1}\text{。}\]

解不等式 \(\dfrac{4k+2}{5k+1}\lt\dfrac{321}{400}\)，得到 \(400(4k+2)\lt 321(5k+1)\)，即 \(1600k+800\lt 1605k+321\)，所以 \(5k\gt 479\)，也就是 \(k\gt 95.8\)。因此 \(k=96\)、\(N=480\)，其各位数字之和为 \(4+8+0=12\)。

所以正确答案是 A。

English

Write \(N=5k.\) Number the positions of the red ball \(0,1,\ldots,5k\) from one end; there are \(5k+1\) equally likely positions.

Fewer than \(\dfrac{3}{5}\) of the green balls lie on each side exactly when the red ball is in one of the positions \(2k+1,2k+2,\ldots,3k-1,\) which is \(k-1\) positions. Hence \[ P(N)=1-\dfrac{k-1}{5k+1}=\dfrac{4k+2}{5k+1}. \]

Solving \(\dfrac{4k+2}{5k+1}\lt\dfrac{321}{400}\) gives \(400(4k+2)\lt 321(5k+1),\) so \(1600k+800\lt 1605k+321\) and \(5k\gt 479,\) meaning \(k\gt 95.8.\) Thus \(k=96\) and \(N=480,\) whose digit sum is \(4+8+0=12.\)

Thus, the correct answer is A.

14 答案 C

每个顶点属于 \(3\) 个面，所以 \(6S=3(1+2+\cdots+8)=108\)，从而每个面的和为 \(S=18\)。

包含 \(1\) 且和为 \(18\) 的四元子集是 \(\{1,2,7,8\}\)，\(\{1,3,6,8\}\)，\(\{1,4,5,8\}\)，和 \(\{1,4,6,7\}\)。其中只有一个不含 \(8\)，所以经过 \(1\) 的三个不同面中至少有两个包含 \(8\)。立方体的两个顶点恰好在相邻时才共同位于两个面上；因此 \(1\) 和 \(8\) 相邻。

旋转立方体，使 \(1\) 位于左下前顶点，\(8\) 位于右下前顶点。经过 \(1\) 而不含 \(8\) 的唯一一个面必须放置 \(4,6,7\)，它们有 \(3!=6\) 种排列。每种排列都会迫使 \(5,3,2\) 位于三个相对顶点，其余各面的和也都为 \(18\)。因此共有 \(6\) 种标法。

因此，正确答案是 C。

English

Each vertex belongs to \(3\) faces, so \(6S=3(1+2+\cdots+8)=108,\) giving each face-sum \(S=18.\)

The four-element subsets containing \(1\) with sum \(18\) are \(\{1,2,7,8\},\) \(\{1,3,6,8\},\) \(\{1,4,5,8\},\) and \(\{1,4,6,7\}.\) Only one omits \(8,\) so at least two of the three distinct faces through \(1\) contain \(8.\) Two vertices of a cube lie on two common faces exactly when they are adjacent; hence \(1\) and \(8\) are adjacent.

Rotate the cube so that \(1\) is at the lower-left-front vertex and \(8\) at the lower-right-front vertex. The unique face through \(1\) that does not contain \(8\) must use \(4,6,7,\) and these can be placed in \(3!=6\) orders. Each order forces \(5,3,2\) at the three opposite vertices, and the remaining face sums are then \(18.\) Hence there are \(6\) arrangements.

Thus, the correct answer is C.

15 答案 D

三个圆心到直线 \(l\) 的高度分别为 \(1\)、\(2\)、\(3\)。由于圆 \(Q\) 与圆 \(P\) 外切，\(PQ=3\)，所以水平距离为 \(P'Q'=\sqrt{3^2-1^2}=\sqrt{8}\)。同理，圆 \(Q\) 与圆 \(R\) 外切，\(QR=5\)，所以 \(Q'R'=\sqrt{5^2-1^2}=\sqrt{24}\)。

取 \(P=(0,1)\)、\(Q=(\sqrt8,2)\)、\(R=(\sqrt8+\sqrt{24},3)\)。由鞋带公式，面积为 \[ \begin{gathered} \small \dfrac12\left|\sqrt8(3-1)+(\sqrt8+\sqrt{24})(1-2)\right|\\ =\dfrac12\left(\sqrt{24}-\sqrt8\right)\\ =\sqrt6-\sqrt2 \end{gathered}\text{。}\]

所以正确答案是 D。

English

The centers lie at heights \(1,\) \(2,\) and \(3\) above line \(l.\) Since circle \(Q\) is externally tangent to circle \(P,\) we have \(PQ=3,\) so the horizontal distance is \(P'Q'=\sqrt{3^2-1^2}=\sqrt{8}.\) Since circle \(Q\) is tangent to circle \(R,\) we have \(QR=5,\) so \(Q'R'=\sqrt{5^2-1^2}=\sqrt{24}.\)

Place \(P=(0,1),\) \(Q=(\sqrt8,2),\) and \(R=(\sqrt8+\sqrt{24},3).\) By the shoelace formula, the area is \[ \begin{gathered} \small \dfrac12\left|\sqrt8(3-1)+(\sqrt8+\sqrt{24})(1-2)\right|\\ =\dfrac12\left(\sqrt{24}-\sqrt8\right)\\ =\sqrt6-\sqrt2. \end{gathered} \]

Thus, the correct answer is D.

16 答案 D

令 \(u=\log_3 x\)，则 \(\log_x 3=\dfrac1u\)，\(\log_{\frac{1}{3}}x=-u\)，且 \(\log_x\dfrac13=-\dfrac1u\)。两条图像相交时，\(u,\dfrac1u,-u,-\dfrac1u\) 中有两个相等，并且对应的 \(x\gt 0\) 有效。

令 \(u=\dfrac1u\) 得 \(u=\pm1\)，所以 \(x=3\) 或 \(x=\dfrac13\)；令 \(-u=-\dfrac1u\) 得到同样的值。令 \(u=-u\) 得 \(u=0\)，即 \(x=1\)，此时 \(\log_3 x\) 与 \(\log_{\frac{1}{3}}x\) 都等于 \(0\)。其余配对没有实数解。

不同交点为 \((1,0)\)、\((3,1)\)、\(\left(\dfrac13,-1\right)\)、\((3,-1)\) 和 \(\left(\dfrac13,1\right)\)，所以共有 \(5\) 个。

所以正确答案是 D。

English

Let \(u=\log_3 x.\) Then \(\log_x 3=\dfrac1u,\) \(\log_{\frac{1}{3}}x=-u,\) and \(\log_x\dfrac13=-\dfrac1u.\) Two graphs meet where two of \(u,\dfrac1u,-u,-\dfrac1u\) are equal for some valid \(x\gt 0.\)

Setting \(u=\dfrac1u\) gives \(u=\pm1,\) so \(x=3\) or \(x=\dfrac13;\) setting \(-u=-\dfrac1u\) gives the same values. Setting \(u=-u\) gives \(u=0,\) i.e. \(x=1,\) where \(\log_3 x\) and \(\log_{\frac{1}{3}}x\) are both \(0.\) The remaining pairings have no real solution.

The distinct intersection points are \((1,0),\) \((3,1),\) \(\left(\dfrac13,-1\right),\) \((3,-1),\) and \(\left(\dfrac13,1\right),\) so there are \(5.\)

Thus, the correct answer is D.

17 答案 B

设正方形 \(ABCD\) 的边长为 \(6\)。每个等边三角形的高为 \(3\sqrt3\)，其中心到正方形相应边的距离是该高的 \(\frac13\)，即 \(\sqrt3\)。

正方形 \(ABCD\) 的对角线长为 \(6\sqrt2\)。正方形 \(EFGH\) 的对角线等于 \(ABCD\) 的边长加上两段长为 \(\sqrt3\) 的距离，即 \(6+2\sqrt3\)。面积比等于对角线长度之比的平方：\[ \begin{gathered} \left(\dfrac{6+2\sqrt3}{6\sqrt2}\right)^2\\ =\left(\dfrac{3+\sqrt3}{3\sqrt2}\right)^2\\ =\dfrac{12+6\sqrt3}{18}\\ =\dfrac{2+\sqrt3}{3} \end{gathered}\text{。}\]

所以正确答案是 B。

English

Let square \(ABCD\) have side length \(6.\) Each equilateral triangle has height \(3\sqrt3,\) and its center lies \(\frac13\) of that height, namely \(\sqrt3,\) from the square’s side.

Square \(ABCD\) has diagonal \(6\sqrt2.\) Square \(EFGH\) has diagonal equal to the side of \(ABCD\) plus twice \(\sqrt3,\) namely \(6+2\sqrt3.\) The area ratio is the square of the ratio of diagonals: \[ \begin{gathered} \left(\dfrac{6+2\sqrt3}{6\sqrt2}\right)^2\\ =\left(\dfrac{3+\sqrt3}{3\sqrt2}\right)^2\\ =\dfrac{12+6\sqrt3}{18}\\ =\dfrac{2+\sqrt3}{3}. \end{gathered} \]

Thus, the correct answer is B.

18 答案 D

数 \(110n^3\) 能被三个不同的质数 \(2,5,11\) 整除。若其质因数指数为 \(r_1,r_2,\ldots\)，则 \((r_1+1)(r_2+1)\cdots=110\)，其中 \(110=2\cdot5\cdot11\)。因为已经至少有三个大于 \(1\) 的因数，所以恰好只有三个，这说明没有其他质数整除 \(n\)。

\(110n^3\) 中的每个指数都满足 \(1\pmod3\)，因此每个因数个数中的因子 \(r_i+1\) 都满足 \(2\pmod3\)。因数 \(2,5,11\) 都具有这种形式，所以指数按某种顺序为 \(1,4,10\)。从 \(110\) 中已有的指数减去 \(1\)，再除以 \(3\)，得到 \(n\) 中的指数按某种顺序为 \(0,1,3\)。因此 \(n=pq^3\)，其中不同质数 \(p,q\) 选自 \(2,5,11\)。

所以 \(n^4\) 中的指数为 \(4\) 和 \(12\)，而 \(81=3^4\) 引入第三个质数，因为 \(n\) 不能被 \(3\) 整除。因此 \(81n^4\) 的因数个数为 \[ \begin{gathered} (4+1)(4+1)(12+1)\\ =5\cdot5\cdot13\\ =325 \end{gathered}\text{。}\]

因此，正确答案是 D。

English

The number \(110n^3\) is divisible by the three distinct primes \(2,5,11.\) If its prime exponents are \(r_1,r_2,\ldots,\) then \((r_1+1)(r_2+1)\cdots=110,\) where \(110=2\cdot5\cdot11.\) Because there are already at least three factors greater than \(1,\) there are exactly three, so no other prime divides \(n.\)

Each exponent in \(110n^3\) is \(1\pmod3,\) so each divisor-count factor \(r_i+1\) is \(2\pmod3.\) The factors \(2,5,11\) all have that form, so the exponents are \(1,4,10\) in some order. After subtracting the exponent \(1\) from \(110\) and dividing by \(3,\) the exponents in \(n\) are \(0,1,3\) in some order. Thus \(n=pq^3\) for two distinct primes \(p,q\) chosen from \(2,5,11.\)

Consequently \(n^4\) has exponents \(4\) and \(12,\) while \(81=3^4\) introduces a third prime because \(n\) is not divisible by \(3.\) Hence the number of divisors of \(81n^4\) is \[ \begin{gathered} (4+1)(4+1)(12+1)\\ =5\cdot5\cdot13\\ =325. \end{gathered} \]

Thus, the correct answer is D.

19 答案 B

统计 \(8\) 次投掷中累计总和达到过 \(4\) 的序列数。如果反面至多出现 \(2\) 次，他一定能达到 \(4\)，贡献 \[ \begin{gathered} \binom80+\binom81+\binom82\\ =1+8+28\\ =37 \end{gathered} \] 个序列。

恰有 \(3\) 次反面时，他只能在第 \(4\) 次或第 \(6\) 次投掷首次达到 \(4\)。要在第 \(4\) 次达到，前四次必须全为正面，之后剩下的那个正面有 \(4\) 个可能位置。否则第 \(5\) 和第 \(6\) 次是正面，前四次中有一次反面，最后两次是反面，同样有 \(4\) 种可能。因此这一情形贡献 \(8\) 个序列。恰有 \(4\) 次反面时，只有 HHHHTTTT 可行，贡献 \(1\) 个序列。正面少于 \(4\) 次时不可能达到 \(4\)。

所以有利序列共有 \(37+8+1=46\) 个，全部序列有 \(2^8=256\) 个，概率为 \(\dfrac{46}{256}=\dfrac{23}{128}\)。因此 \(a+b=23+128=151\)。

因此，正确答案是 B。

English

Count the sequences of \(8\) tosses whose running total reaches \(4.\) With at most \(2\) tails he certainly reaches \(4,\) contributing \[ \begin{gathered} \binom80+\binom81+\binom82\\ =1+8+28\\ =37 \end{gathered} \] sequences.

With exactly \(3\) tails, he can first reach \(4\) on toss \(4\) or toss \(6.\) Reaching it on toss \(4\) requires four initial heads, after which the remaining head has \(4\) possible positions. Otherwise, tosses \(5\) and \(6\) are heads, one of the first four tosses is a tail, and the last two are tails, again giving \(4\) possibilities. Thus this case contributes \(8.\) With exactly \(4\) tails, only HHHHTTTT works, giving \(1.\) He cannot reach \(4\) with fewer than \(4\) heads.

So there are \(37+8+1=46\) favorable sequences out of \(2^8=256,\) a probability of \(\dfrac{46}{256}=\dfrac{23}{128}.\) Then \(a+b=23+128=151.\)

Thus, the correct answer is B.

20 答案 A

令 \(b=c=a\)，得到 \(a\diamond 1=a\diamond(a\diamond a)\) \(=(a\diamond a)\cdot a=a\)。再令 \(c=b\)，得到 \(a=a\diamond 1\) \(=a\diamond(b\diamond b)=(a\diamond b)\cdot b\)，所以 \(a\diamond b=\dfrac{a}{b}\)。

因此 \[ \begin{gathered} 2016\diamond(6\diamond x)\\ =2016\diamond\dfrac{6}{x}\\ =\dfrac{2016}{\frac{6}{x}}\\ =336x=100 \end{gathered}\text{，}\] 所以 \(x=\dfrac{100}{336}=\dfrac{25}{84}\)，且 \(p+q=25+84=109\)。

所以正确答案是 A。

English

Setting \(b=c=a\) gives \(a\diamond 1=a\diamond(a\diamond a)\) \(=(a\diamond a)\cdot a=a.\) Then setting \(c=b\) gives \(a=a\diamond 1\) \(=a\diamond(b\diamond b)=(a\diamond b)\cdot b,\) so \(a\diamond b=\dfrac{a}{b}.\)

Therefore \[ \begin{gathered} 2016\diamond(6\diamond x)\\ =2016\diamond\dfrac{6}{x}\\ =\dfrac{2016}{\frac{6}{x}}\\ =336x=100, \end{gathered} \] so \(x=\dfrac{100}{336}=\dfrac{25}{84}\) and \(p+q=25+84=109.\)

Thus, the correct answer is A.

21 答案 E

设长为 \(200\) 的边所对的圆心角为 \(\theta\)，圆的半径为 \(R=200\sqrt2\)。对以圆心为顶点的等腰三角形应用余弦定理，得到 \[ \begin{gathered} 200^2=2R^2(1-\cos\theta)\\ =160000(1-\cos\theta) \end{gathered}\text{，}\] 所以 \(\cos\theta=\dfrac34\)。

三条相等的边占据三段圆心角均为 \(\theta\) 的连续弧，所以第四段弧的圆心角为 \(2\pi-3\theta\)。它所对的弦与圆心角为 \(3\theta\) 的弦等长，并且 \[ \begin{gathered} \cos 3\theta=4\cos^3\theta-3\cos\theta\\ =4\cdot\dfrac{27}{64}-\dfrac94\\ =-\dfrac{9}{16} \end{gathered}\text{。}\] 它的长度平方为 \[ \begin{gathered} 2R^2(1-\cos 3\theta)\\ =160000\left(1+\dfrac{9}{16}\right)\\ =160000\cdot\dfrac{25}{16}\\ =250000 \end{gathered}\text{，}\] 所以第四条边长为 \(500\)。

所以正确答案是 E。

English

Let \(\theta\) be the central angle subtending a side of length \(200,\) with radius \(R=200\sqrt2.\) By the law of cosines on the isosceles triangle from the center, \[ \begin{gathered} 200^2=2R^2(1-\cos\theta)\\ =160000(1-\cos\theta), \end{gathered} \] so \(\cos\theta=\dfrac34.\)

The three equal sides use three consecutive arcs of angle \(\theta,\) so the fourth arc has angle \(2\pi-3\theta.\) Its chord has the same length as a chord with central angle \(3\theta,\) and \[ \begin{gathered} \cos 3\theta=4\cos^3\theta-3\cos\theta\\ =4\cdot\dfrac{27}{64}-\dfrac94\\ =-\dfrac{9}{16}. \end{gathered} \] Its length squared is \[ \begin{gathered} 2R^2(1-\cos 3\theta)\\ =160000\left(1+\dfrac{9}{16}\right)\\ =160000\cdot\dfrac{25}{16}\\ =250000, \end{gathered} \] so the fourth side is \(500.\)

Thus, the correct answer is E.

22 答案 A

因为 \(\text{lcm}(x,y)=2^3\cdot3^2\)，且 \(\text{lcm}(x,z)=2^3\cdot3\cdot5^2\)，所以 \(5^2\) 整除 \(z\)，而 \(x\) 和 \(y\) 都不能被 \(5\) 整除。同理，\(3^2\) 整除 \(y\)，而 \(x\) 和 \(z\) 都不能被 \(3^2\) 整除；此外，\(x\) 必须含有因子 \(2^3\)。

写 \(x=2^3\cdot3^{\,j}\)、\(y=2^{\,k}\cdot3^2\)、\(z=2^{\,m}\cdot3^{\,n}\cdot5^2\)，最小公倍数条件要求 \(\max(j,n)=1\)，且 \(\max(k,m)=2\)。第一对可以取 \((1,0),(0,1),(1,1)\)，第二对可以取 \((2,0),(2,1),(2,2),(0,2),(1,2)\)，因此共有 \(3\cdot5=15\) 个有序三元组。

所以正确答案是 A。

English

Because \(\text{lcm}(x,y)=2^3\cdot3^2\) and \(\text{lcm}(x,z)=2^3\cdot3\cdot5^2,\) the factor \(5^2\) divides \(z\) while neither \(x\) nor \(y\) is divisible by \(5.\) Also \(3^2\) divides \(y,\) while neither \(x\) nor \(z\) is divisible by \(3^2,\) and \(x\) must have the factor \(2^3.\)

Writing \(x=2^3\cdot3^{\,j},\) \(y=2^{\,k}\cdot3^2,\) and \(z=2^{\,m}\cdot3^{\,n}\cdot5^2,\) the lcm conditions require \(\max(j,n)=1\) and \(\max(k,m)=2.\) The first pair can be \((1,0),(0,1),(1,1),\) and the second can be \((2,0),(2,1),(2,2),(0,2),(1,2).\) Thus there are \(3\cdot5=15\) ordered triples.

Thus, the correct answer is A.

23 答案 C

有序三元组 \((x,y,z)\) 填满体积为 \(1\) 的单位立方体。不能构成三角形，恰好意味着某个数大于或等于另外两个数之和。

区域 \(z\ge x+y\) 是一个四面体，顶点为 \((0,0,0),(0,0,1),(0,1,1),(1,0,1)\)，体积为 \(\frac16\)。类似地，区域 \(x\ge y+z\) 和 \(y\ge x+z\) 的体积也各为 \(\frac16\)，而且内部互不相交。因此不能构成三角形的概率为 \(3\cdot\frac16=\frac12\)，能构成三角形的概率为 \(1-\frac12=\frac12\)。

所以正确答案是 C。

English

The ordered triples \((x,y,z)\) fill the unit cube of volume \(1.\) They fail to form a triangle exactly when one value is at least the sum of the other two.

The region \(z\ge x+y\) is a tetrahedron with vertices \((0,0,0),(0,0,1),(0,1,1),(1,0,1)\) of volume \(\frac16.\) The analogous regions \(x\ge y+z\) and \(y\ge x+z\) also have volume \(\frac16\) and have disjoint interiors. So the failure probability is \(3\cdot\frac16=\frac12,\) and the triangle probability is \(1-\frac12=\frac12.\)

Thus, the correct answer is C.

24 答案 B

由韦达定理，实根 \(r,s,t\) 满足 \(r+s+t=a\)，\(rs+st+tr=b\)，且 \(rst=a\)，所以 \(r+s+t=rst\)。

乘积为正，说明三个根全为正，或有两个根为负。在后一种情形中，将根写成 \(-u,-v,w\)，其中 \(u,v,w\gt0\)。和为正迫使 \(w\gt u+v\)，于是 \(b=uv-w(u+v)\lt0\)，矛盾。因此三个根都为正。

由算术平均数与几何平均数不等式，\(27rst\le(r+s+t)^3=(rst)^3\)，所以 \(a=rst\ge 3\sqrt3\)，等号成立当且仅当 \(r=s=t=\sqrt3\)。在这个最小的 \(a\) 下，\[ \begin{gathered} b=rs+st+tr\\ =3r^2=3\cdot 3=9 \end{gathered}\text{。}\]

因此，正确答案是 B。

English

By Vieta’s formulas, the real roots \(r,s,t\) satisfy \(r+s+t=a,\) \(rs+st+tr=b,\) and \(rst=a,\) so \(r+s+t=rst.\)

The positive product means either all three roots are positive or two are negative. In the latter case write the roots as \(-u,-v,w\) with \(u,v,w\gt0.\) The positive sum forces \(w\gt u+v,\) and then \(b=uv-w(u+v)\lt0,\) a contradiction. Hence all three roots are positive.

By the AM-GM inequality, \(27rst\le(r+s+t)^3=(rst)^3,\) so \(a=rst\ge 3\sqrt3,\) with equality if and only if \(r=s=t=\sqrt3.\) At this smallest \(a,\) \[ \begin{gathered} b=rs+st+tr\\ =3r^2=3\cdot 3=9. \end{gathered} \]

Thus, the correct answer is B.

25 答案 E

取 \(k=2j\)。最小的 \(k+1\) 位完全平方数是 \(10^{k}=(10^{j})^2\)。Silvia 擦除数字后，黑板上显示的数为 \(\left\lfloor \frac{n^2}{10^{k}}\right\rfloor\)，其中 \(n=10^{j}, 10^{j}+1,\ldots\)。

令 \(M=10^k\)。从 \(n\) 到 \(n+1\) 要出现至少为 \(2\) 的跳跃，需要 \((n+1)^2-n^2=2n+1\gt M\)，因此写 \(n=\frac{M}{2}+m\)，其中 \(m\ge0\)。当 \(m=0\) 时跳跃只有 \(1\)，所以第一次更大的跳跃满足 \(m\ge1\)。令 \(A=\left\lfloor \frac{n^2}{M}\right\rfloor\)，\(B=\left\lfloor\frac{(n+1)^2}{M}\right\rfloor\)。因为 \(M\) 能被 \(4\) 整除，\[ \begin{aligned} A&=\dfrac M4+m+\left\lfloor\dfrac{m^2}{M}\right\rfloor,\\ B&=\dfrac M4+m+1\\ &\quad{}+\left\lfloor\dfrac{(m+1)^2}{M}\right\rfloor \end{aligned}\text{。}\]

因此第一次至少为 \(2\) 的跳跃出现在使 \(m^2\lt M\le(m+1)^2\) 成立的最小 \(m\) 处。因为 \(\sqrt M=10^j\)，所以 \(m=10^j-1\)。间隙之前最后显示的数是 \(\frac{M}{4}+10^j-1\)，所以最小的未出现正整数为 \[f(2j)=\dfrac{10^{2j}}4+10^j\text{。}\]

对 \(j=1,\ldots,1008\) 求和，得到 \[ \begin{gathered} \sum_{j=1}^{1008}f(2j)\\ =25\sum_{j=0}^{1007}10^{2j}\\ {}+10\sum_{j=0}^{1007}10^{j}\\ =\underbrace{2525\cdots25}_{2016\text{ 位数字}}\\ {}+\underbrace{111\cdots10}_{1009\text{ 位数字}} \end{gathered}\text{。}\] 相加时没有进位，所以各位数字之和为 \(1008\cdot(2+5)\) \(+1008\cdot 1=1008\cdot 8=8064\)。

所以正确答案是 E。

English

Take \(k=2j.\) The smallest perfect square with \(k+1\) digits is \(10^{k}=(10^{j})^2,\) and after Silvia erases, the numbers shown are \(\left\lfloor \frac{n^2}{10^{k}}\right\rfloor\) for \(n=10^{j}, 10^{j}+1,\ldots\)

Put \(M=10^k.\) A jump of at least \(2\) from \(n\) to \(n+1\) requires \((n+1)^2-n^2=2n+1\gt M,\) so write \(n=\frac{M}{2}+m\) with \(m\ge0.\) The case \(m=0\) gives a jump of only \(1,\) so the first larger jump has \(m\ge1.\) Let \(A=\left\lfloor \frac{n^2}{M}\right\rfloor\) and \(B=\left\lfloor\frac{(n+1)^2}{M}\right\rfloor.\) Because \(M\) is divisible by \(4,\) \[ \begin{aligned} A&=\dfrac M4+m+\left\lfloor\dfrac{m^2}{M}\right\rfloor,\\ B&=\dfrac M4+m+1\\ &\quad{}+\left\lfloor\dfrac{(m+1)^2}{M}\right\rfloor. \end{aligned} \]

Therefore the first jump of at least \(2\) occurs at the first \(m\) for which \(m^2\lt M\le(m+1)^2.\) Since \(\sqrt M=10^j,\) this is \(m=10^j-1.\) The last displayed value before the gap is \(\frac{M}{4}+10^j-1,\) so the smallest missing integer is \[f(2j)=\dfrac{10^{2j}}4+10^j.\]

Summing over \(j=1,\ldots,1008,\) \[ \begin{gathered} \sum_{j=1}^{1008}f(2j)\\ =25\sum_{j=0}^{1007}10^{2j}\\ {}+10\sum_{j=0}^{1007}10^{j}\\ =\underbrace{2525\cdots25}_{2016\text{ digits}}\\ {}+\underbrace{111\cdots10}_{1009\text{ digits}}. \end{gathered} \] There are no carries, so the digit sum is \(1008\cdot(2+5)\) \(+1008\cdot 1=1008\cdot 8=8064.\)

Thus, the correct answer is E.