2016B AMC 12 真题（P版）

1 答案 D

当 \(a=\tfrac12\) 时，\(a^{-1}=2\)。分子为 \(2\cdot2+\dfrac{2}{2}=4+1=5\)，再除以 \(a=\tfrac12\)，得到 \(\dfrac{5}{\frac{1}{2}}=10\)。

所以正确答案是 D。

English

With \(a=\tfrac12,\) we have \(a^{-1}=2.\) The numerator is \(2\cdot2+\dfrac{2}{2}=4+1=5,\) and dividing by \(a=\tfrac12\) gives \(\dfrac{5}{\frac{1}{2}}=10.\)

Thus, the correct answer is D.

2 答案 A

调和平均数为 \(\dfrac{2\cdot1\cdot2016}{1+2016}=\dfrac{4032}{2017}\)。因为 \(\dfrac{2016}{2017}\) 非常接近 \(1\)，所以这个数略小于 \(2\)，最接近的整数是 \(2\)。

所以正确答案是 A。

English

The harmonic mean is \(\dfrac{2\cdot1\cdot2016}{1+2016}=\dfrac{4032}{2017}.\) Since \(\dfrac{2016}{2017}\) is very close to \(1,\) this is just under \(2,\) so the closest integer is \(2.\)

Thus, the correct answer is A.

3 答案 D

因为 \(x=-2016\)，所以 \(|x|=2016\)。最内层为 \(|x|-x=2016+2016=4032\)。接着 \(\big|4032\big|-|x|=4032-2016\) \(=2016\)，外层绝对值仍为 \(2016\)。最后减去 \(x\)，得到 \(2016-(-2016)=4032\)。

所以正确答案是 D。

English

Since \(x=-2016,\) \(|x|=2016.\) The innermost expression is \(|x|-x=2016+2016=4032.\) Then \(\big|4032\big|-|x|=4032-2016\) \(=2016,\) and the outer absolute value leaves \(2016.\) Finally subtracting \(x\) gives \(2016-(-2016)=4032.\)

Thus, the correct answer is D.

4 答案 C

设两角为 \(\alpha\lt\beta\)，其中 \(\beta=\tfrac54\alpha\)。较小角的余角较大，所以 \(90-\alpha=2(90-\beta)=180-\tfrac52\alpha\)。由此得到 \(\tfrac32\alpha=90\)，所以 \(\alpha=60^\circ\)，\(\beta=75^\circ\)。两角之和为 \(135^\circ\)。

所以正确答案是 C。

English

Let the angles be \(\alpha\lt\beta\) with \(\beta=\tfrac54\alpha.\) The larger complement belongs to the smaller angle, so \(90-\alpha=2(90-\beta)=180-\tfrac52\alpha.\) This gives \(\tfrac32\alpha=90,\) so \(\alpha=60^\circ\) and \(\beta=75^\circ.\) The sum is \(135^\circ.\)

Thus, the correct answer is C.

5 答案 B

因为 \(919=7\cdot131+2\)，条约签署日比星期四晚 \(131\) 个整周再加 \(2\) 天。

星期四后两天是星期六。所以正确答案是 B。

English

Because \(919=7\cdot131+2,\) the treaty was signed \(131\) full weeks plus \(2\) days after Thursday. Two days beyond Thursday is Saturday.

Thus, the correct answer is B.

6 答案 C

设第一象限中的顶点为 \((x,x^2)\)。由对称性，底边长为 \(BC=2x\)，高为 \(x^2\)，所以面积满足 \(\tfrac12\cdot2x\cdot x^2=x^3=64\)。因此 \(x=4\)，从而 \(BC=2x=8\)。

所以正确答案是 C。

English

Let the vertex in the first quadrant be \((x,x^2).\) By symmetry the base is \(BC=2x\) and the height is \(x^2,\) so \(\tfrac12\cdot2x\cdot x^2=x^3=64.\) Thus \(x=4\) and \(BC=2x=8.\)

Thus, the correct answer is C.

7 答案 D

第一轮划掉奇数，留下 \(2\) 的倍数。第二轮划掉 \(2,6,10,\ldots\) 留下 \(4\) 的倍数。一般地，第 \(n\) 轮后只剩 \(2^n\) 的倍数。最后幸存的是不超过 \(100\) 的最大 \(2\) 的幂，即 \(2^6=64\)。

所以正确答案是 D。

English

The first pass removes the odd numbers, leaving the multiples of \(2.\) The second pass removes \(2,6,10,\ldots,\) leaving the multiples of \(4.\) In general, after the \(n\)th pass only the multiples of \(2^n\) remain. The surviving number is the highest power of \(2\) not exceeding \(100,\) which is \(2^6=64.\)

Thus, the correct answer is D.

8 答案 D

重量与面积成正比，面积随边长平方缩放。第二块边长是第一块的 \(\tfrac53\) 倍，所以重量为 \(12\cdot\left(\dfrac53\right)^2=\dfrac{100}{3}\approx33.3\) 盎司。

所以正确答案是 D。

English

Weight is proportional to area, and area scales with the square of the side length. The second side is \(\tfrac53\) times the first, so its weight is \(12\cdot\left(\dfrac53\right)^2=\dfrac{100}{3}\approx33.3\) ounces.

Thus, the correct answer is D.

9 答案 B

设较短边有 \(x\) 根桩，则较长边有 \(2x\) 根。四个角都被重复计算一次，所以 \(2x+2(2x)-4=20\)，得 \(x=4\)。较短边有 \(4\) 根桩，长度为 \((4-1)\cdot4=12\) 码；较长边有 \(8\) 根桩，长度为 \((8-1)\cdot4=28\) 码。面积为 \(12\cdot28=336\)。

所以正确答案是 B。

English

Let the shorter side have \(x\) posts, so the longer side has \(2x.\) Counting all posts and subtracting the four corners counted twice, \(2x+2(2x)-4=20,\) giving \(x=4.\) The shorter side has \(4\) posts, or \((4-1)\cdot4=12\) yards, and the longer side has \(8\) posts, or \((8-1)\cdot4=28\) yards. The area is \(12\cdot28=336.\)

Thus, the correct answer is B.

10 答案 A

边 \(\overline{PQ}\) 与 \(\overline{RS}\) 的斜率为 \(-1\)，\(\overline{QR}\) 与 \(\overline{PS}\) 的斜率为 \(1\)，所以 \(PQRS\) 是长方形，边长为 \((a-b)\sqrt2\) 与 \((a+b)\sqrt2\)。它的面积为 \(2(a-b)(a+b)=2(a^2-b^2)\) \(=16\)，所以 \(a^2-b^2=8\)。唯一相差 \(8\) 的两个完全平方数是 \(9\) 和 \(1\)，因此 \(a=3\)、\(b=1\)，从而 \(a+b=4\)。

所以正确答案是 A。

English

The sides \(\overline{PQ}\) and \(\overline{RS}\) have slope \(-1,\) and \(\overline{QR}\) and \(\overline{PS}\) have slope \(1,\) so \(PQRS\) is a rectangle with sides \((a-b)\sqrt2\) and \((a+b)\sqrt2.\) Its area is \(2(a-b)(a+b)=2(a^2-b^2)\) \(=16,\) so \(a^2-b^2=8.\) The only perfect squares differing by \(8\) are \(9\) and \(1,\) giving \(a=3,\) \(b=1,\) and \(a+b=4.\)

Thus, the correct answer is A.

11 答案 D

左边位于 \(x=k\) 的正方形，必须低于其宽度范围内 \(y=\pi x\) 的最低点，即 \(y=\pi k\)。对边长 \(1\)，可能的左边位置 \(k=1,2,3,4\) 分别贡献 \(3+6+9+12=30\) 个正方形。对边长 \(2\)，左边位置 \(k=1,2,3\) 贡献 \(2+5+8=15\) 个；对边长 \(3\)，左边位置 \(k=1,2\) 贡献 \(1+4=5\) 个。边长至少为 \(4\) 的正方形无法放入，所以总数是 \(30+15+5=50\)。

因此，正确答案是 D。

English

A square whose left edge is \(x=k\) must fit below the lowest point of \(y=\pi x\) over its width, namely \(y=\pi k.\) For side length \(1,\) the possible left edges \(k=1,2,3,4\) contribute \(3+6+9+12=30\) squares. For side length \(2,\) the left edges \(k=1,2,3\) contribute \(2+5+8=15,\) and for side length \(3,\) the left edges \(k=1,2\) contribute \(1+4=5.\) A square of side at least \(4\) cannot fit, so the total is \(30+15+5=50.\)

Thus, the correct answer is D.

12 答案 C

像棋盘一样染色，使四个角和中心同色。由于连续数字位于相邻的异色格中，沿数字链颜色交替，所以这五个同色格中是五个奇数 \(1,3,5,7,9\)，其和为 \(25\)。四个角的和为 \(18\)，所以中心格中的数为 \(25-18=7\)。

所以正确答案是 C。

English

Color the grid like a checkerboard so the four corners and the center share one color. Since consecutive numbers occupy adjacent (opposite colored) squares, the numbers alternate parity along the chain, so the five same-colored cells contain the five odd numbers \(1,3,5,7,9,\) which sum to \(25.\) The four corners add to \(18,\) so the center is \(25-18=7.\)

Thus, the correct answer is C.

13 答案 E

设飞机位于点 \(C\)，其正下方的地面点为 \(D\)，高度为 \(h\)。三角形 \(ACD\) 和 \(BCD\) 都是 \(30\)-\(60\)-\(90\) 直角三角形，所以 \(AD=\sqrt3\,h\)，\(BD=\dfrac{h}{\sqrt3}\)。Alice 向北看而 Bob 向西看，因此 \(\angle ADB=90^\circ\)，从而 \(AD^2+BD^2=AB^2=100\)。于是 \(3h^2+\dfrac{h^2}{3}=\dfrac{10h^2}{3}=100\)，解得 \(h=\sqrt{30}\approx5.48\)，最接近 \(5.5\)。

所以正确答案是 E。

English

Let the airplane be at \(C,\) directly above point \(D\) on the ground at altitude \(h.\) Triangles \(ACD\) and \(BCD\) are \(30\)-\(60\)-\(90\) right triangles, so \(AD=\sqrt3\,h\) and \(BD=\dfrac{h}{\sqrt3}.\) Since Alice looks north and Bob looks west, \(\angle ADB=90^\circ,\) so \(AD^2+BD^2=AB^2=100.\) Then \(3h^2+\dfrac{h^2}{3}=\dfrac{10h^2}{3}=100,\) giving \(h=\sqrt{30}\approx5.48,\) closest to \(5.5.\)

Thus, the correct answer is E.

14 答案 E

设公比为 \(r\)。因为第二项为 \(1\)，第一项为 \(\dfrac1r\)，所以 \(S=\dfrac{\frac{1}{r}}{1-r}=\dfrac{1}{r-r^2}\)。收敛要求 \(|r|\lt1\)，而 \(S\gt0\) 又迫使 \(0\lt r\lt1\)。因此，要使 \(S\) 最小，就要使 \(r-r^2\) 最大。抛物线 \(r-r^2\) 在 \(r=\tfrac12\) 处达到最大值 \(\tfrac14\)，所以 \(S\) 的最小值为 \(\dfrac{1}{\frac{1}{4}}=4\)。

因此，正确答案是 E。

English

Let \(r\) be the common ratio. Since the second term is \(1,\) the first term is \(\dfrac1r,\) so \(S=\dfrac{\frac{1}{r}}{1-r}=\dfrac{1}{r-r^2}.\) Convergence requires \(|r|\lt1,\) and \(S\gt0\) then forces \(0\lt r\lt1.\) Therefore \(S\) is smallest when \(r-r^2\) is largest. The parabola \(r-r^2\) peaks at \(r=\tfrac12,\) where it equals \(\tfrac14,\) so the smallest value of \(S\) is \(\dfrac{1}{\frac{1}{4}}=4.\)

Thus, the correct answer is E.

15 答案 D

将相对面配成 \((a,b),(c,d),(e,f)\)。每个顶点乘积从每对相对面中取一个数，所以八个乘积之和可分解为 \((a+b)(c+d)(e+f)\)。三个因子的总和固定为 \(2+3+4+5+6+7=27\)，乘积在三者相等、各为 \(9\) 时最大。这可由 \((2,7),(3,6),(4,5)\) 达到，得到 \(9\cdot9\cdot9=729\)。

所以正确答案是 D。

English

Pair the opposite faces as \((a,b),(c,d),(e,f).\) Each vertex product uses one face from each pair, so the sum of all eight products factors as \((a+b)(c+d)(e+f).\) The three factors have fixed total \(2+3+4+5+6+7=27,\) and a product with fixed sum is largest when the factors are equal, at \(9\) each. This balance is achievable with \((2,7),(3,6),(4,5),\) giving \(9\cdot9\cdot9=729.\)

Thus, the correct answer is D.

16 答案 E

连续整数之和等于项数乘以中位数。若项数为奇数，中位数是 \(345\) 的整数因数，得到项数为 \(3\)（中位数 \(115\)）、\(5\)（中位数 \(69\)）、\(15\)（中位数 \(23\)）和 \(23\)（中位数 \(15\)）的序列。若项数为偶数，中位数为半整数；符合要求的项数为 \(2,6,10\)，对应的中位数依次为 \(172.5,57.5,34.5\)。更长的序列会含有非正项。因此共有 \(4+3=7\) 种。

所以正确答案是 E。

English

A sum of consecutive integers equals the count times the median. For an odd number of terms, the median is an integer divisor of \(345,\) giving runs of \(3\) (median \(115\)), \(5\) (median \(69\)), \(15\) (median \(23\)), and \(23\) (median \(15\)) terms. For an even number of terms the median is a half-integer. The positive possibilities have lengths \(2,6,10,\) with respective medians \(172.5,57.5,34.5.\) Longer divisor-based runs would force a nonpositive first term. This gives \(4+3=7\) ways.

Thus, the correct answer is E.

17 答案 D

令 \(x=BH\)，则 \(CH=8-x\)。由两个直角三角形可得 \(AH^2=7^2-x^2=9^2-(8-x)^2\)，所以 \(x=2\)，且 \(AH=\sqrt{45}\)。在 \(\triangle ACH\) 中，由角平分线定理，\(\dfrac{AP}{PH}=\dfrac{CA}{CH}=\dfrac96\)，所以 \(AP=\dfrac35 AH\)。同理，在 \(\triangle ABH\) 中，\(\dfrac{AQ}{QH}=\dfrac{BA}{BH}=\dfrac72\)，所以 \(AQ=\dfrac79 AH\)。于是 \[ \begin{aligned} PQ &= AQ-AP \\ &= \left(\dfrac79-\dfrac35\right)AH \\ &= \dfrac{8}{45}\sqrt{45} \\ &= \dfrac{8}{15}\sqrt5 \end{aligned}\text{。}\]

所以正确答案是 D。

English

Let \(x=BH.\) Then \(CH=8-x,\) and from the two right triangles \(AH^2=7^2-x^2=9^2-(8-x)^2.\) This gives \(x=2\) and \(AH=\sqrt{45}.\) By the angle bisector theorem in \(\triangle ACH,\) \(\dfrac{AP}{PH}=\dfrac{CA}{CH}=\dfrac96,\) so \(AP=\dfrac35 AH.\) Similarly in \(\triangle ABH,\) \(\dfrac{AQ}{QH}=\dfrac{BA}{BH}=\dfrac72,\) so \(AQ=\dfrac79 AH.\) Then \[ \begin{aligned} PQ &= AQ-AP \\ &= \left(\dfrac79-\dfrac35\right)AH \\ &= \dfrac{8}{45}\sqrt{45} \\ &= \dfrac{8}{15}\sqrt5. \end{aligned} \]

Thus, the correct answer is D.

18 答案 B

由对称性，只看第一象限，此时方程为 \(x^2+y^2=x+y\)，即 \(\left(x-\tfrac12\right)^2+\left(y-\tfrac12\right)^2=\tfrac12\)。这是一个圆心在 \(\left(\tfrac12,\tfrac12\right)\)、经过 \((1,0)\) 和 \((0,1)\) 的圆。由于圆心是连接 \((1,0)\) 与 \((0,1)\) 的弦的中点，第一象限内围成的区域由面积为 \(\tfrac12\) 的直角三角形和半径为 \(\dfrac{\sqrt2}{2}\)、面积为 \(\dfrac\pi4\) 的半圆组成。乘以 \(4\) 个象限，总面积为 \(4\left(\tfrac12+\tfrac\pi4\right)=\pi+2\)。

所以正确答案是 B。

English

By symmetry, consider the first quadrant, where the equation is \(x^2+y^2=x+y,\) or \(\left(x-\tfrac12\right)^2+\left(y-\tfrac12\right)^2=\tfrac12.\) This is a circle centered at \(\left(\tfrac12,\tfrac12\right)\) passing through \((1,0)\) and \((0,1);\) since the center is the midpoint of that chord, the enclosed first-quadrant region is the right triangle with legs to \((1,0)\) and \((0,1)\) (area \(\tfrac12\)) plus a semicircle of radius \(\dfrac{\sqrt2}{2}\) (area \(\dfrac\pi4\)). Multiplying by \(4\) for all quadrants gives \(4\left(\tfrac12+\tfrac\pi4\right)=\pi+2.\)

Thus, the correct answer is B.

19 答案 B

单个玩家第一次正面出现在第 \(n\) 次的概率为 \(\left(\tfrac12\right)^n\)。三人都在第 \(n\) 次停止的概率为 \(\left(\left(\tfrac12\right)^n\right)^3=\left(\tfrac18\right)^n\)。对 \(n\ge1\) 求和，得 \(\displaystyle\sum_{n=1}^\infty\left(\tfrac18\right)^n =\frac{\frac{1}{8}}{1-\frac{1}{8}}=\frac17\)。

所以正确答案是 B。

English

A player’s first head comes on flip \(n\) with probability \(\left(\tfrac12\right)^n.\) All three stopping on the same flip \(n\) has probability \(\left(\left(\tfrac12\right)^n\right)^3=\left(\tfrac18\right)^n.\) Summing over \(n\ge1,\) \(\displaystyle\sum_{n=1}^\infty\left(\tfrac18\right)^n =\frac{\frac{1}{8}}{1-\frac{1}{8}}=\frac17.\)

Thus, the correct answer is B.

20 答案 A

因为每队赢 \(10\) 场、输 \(10\) 场，所以共有 \(21\) 支队伍，三队组总数为 \(\binom{21}{3}=1330\)。一个三队组不是循环关系，当且仅当其中某一队击败另外两队。选定这支胜队有 \(21\) 种方式，再从被它击败的 \(10\) 支队伍中选出 \(2\) 支，所以非循环三队组共有 \(21\cdot\binom{10}{2}=21\cdot45=945\) 个。因此循环三队组共有 \(1330-945=385\) 个。

所以正确答案是 A。

English

Since each team won \(10\) and lost \(10,\) there are \(21\) teams and \(\binom{21}{3}=1330\) triples. A triple is not cyclic exactly when one team beats both others. Choosing that team (\(21\) ways) and \(2\) of the \(10\) teams it beat gives \(21\cdot\binom{10}{2}=21\cdot45=945\) non-cyclic triples. Thus the cyclic triples number \(1330-945=385.\)

Thus, the correct answer is A.

21 答案 B

设 \(D=(0,0)\)、\(C=(1,0)\)、\(B=(1,1)\)、\(A=(0,1)\)，并令 \(q_i=DQ_i\)。直线 \(AQ_i\) 与 \(\overline{BD}\)（即直线 \(y=x\)）相交于 \(P_i\)，交点的两个坐标都为 \(\dfrac{q_i}{1+q_i}\)，所以 \(q_{i+1}=\dfrac{q_i}{1+q_i}\)。由 \(q_1=\tfrac12\) 可得 \(q_i=\dfrac{1}{i+1}\)。三角形 \(\triangle DQ_iP_i\) 的底为 \(DQ_i=\dfrac{1}{i+1}\)，高为 \(P_i\) 的 \(y\) 坐标，即 \(q_{i+1}=\dfrac{1}{i+2}\)。于是 \[ \begin{gathered} \text{面积 }\triangle DQ_iP_i=\tfrac12\cdot\dfrac{1}{i+1} \\ \quad{}\cdot\dfrac{1}{i+2} \\ {}=\tfrac12\left(\dfrac{1}{i+1}-\dfrac{1}{i+2}\right)\text{。} \end{gathered} \] 求和后中间项相消，得到 \(\tfrac12\cdot\tfrac12=\tfrac14\)。

所以正确答案是 B。

English

Place \(D=(0,0),\) \(C=(1,0),\) \(B=(1,1),\) \(A=(0,1),\) and let \(q_i=DQ_i.\) Intersecting line \(AQ_i\) with \(\overline{BD}\) (the line \(y=x\)) gives \(P_i\) with both coordinates \(\dfrac{q_i}{1+q_i},\) so \(q_{i+1}=\dfrac{q_i}{1+q_i}.\) From \(q_1=\tfrac12\) this yields \(q_i=\dfrac{1}{i+1}.\) The base of \(\triangle DQ_iP_i\) is \(DQ_i=\dfrac{1}{i+1}\) and its height is the \(y\)-coordinate of \(P_i,\) which is \(q_{i+1}=\dfrac{1}{i+2}.\) Then \[ \begin{gathered} \text{Area of }\triangle DQ_iP_i=\tfrac12\cdot\dfrac{1}{i+1} \\ \quad{}\cdot\dfrac{1}{i+2} \\ {}=\tfrac12\left(\dfrac{1}{i+1}-\dfrac{1}{i+2}\right). \end{gathered} \] Summing telescopes to \(\tfrac12\cdot\tfrac12=\tfrac14.\)

Thus, the correct answer is B.

22 答案 B

周期为 \(6\) 要求 \(10^6-1=3^3\cdot7\cdot11\cdot13\cdot37\) 是 \(n\) 的倍数。周期为 \(4\) 要求 \(10^4-1=3^2\cdot11\cdot101\) 是 \(n+6\) 的倍数，而 \(10^2-1=3^2\cdot11\) 不是 \(n+6\) 的倍数（否则周期会是 \(1\) 或 \(2\)）。因此 \(n+6\) 是 \(101\) 的倍数。又因为 \(n+6\) 整除 \(3^2\cdot11\cdot101\) 且小于 \(1006\)，所以只有 \(n+6=101,303,909\) 三种可能，对应 \(n=95,297,903\)。其中只有 \(297=3^3\cdot11\) 整除 \(10^6-1\)，所以 \(n=297\)。

最后，\(10^6\equiv1\pmod{297}\)，而 \(10^2\not\equiv1\) 且 \(10^3\not\equiv1\pmod{297}\)，所以它的周期恰为 \(6\)。此外，\(303\) 整除 \(10^4-1\) 但不整除 \(10^2-1\)，所以 \(\frac{1}{303}\) 的周期恰为 \(4\)。因此 \(n=297\) 位于 \([201,400]\) 中。

因此，正确答案是 B。

English

Period \(6\) requires \(10^6-1=3^3\cdot7\cdot11\cdot13\cdot37\) to be divisible by \(n.\) Period \(4\) requires \(10^4-1=3^2\cdot11\cdot101\) to be divisible by \(n+6,\) while \(10^2-1=3^2\cdot11\) is not divisible by \(n+6\) (else the period would be \(1\) or \(2\)). Hence \(n+6\) is a multiple of \(101.\) Since \(n+6\) also divides \(3^2\cdot11\cdot101\) and is less than \(1006,\) the only possibilities are \(n+6=101,303,909,\) giving \(n=95,297,903.\) Only \(297=3^3\cdot11\) divides \(10^6-1,\) so \(n=297.\)

Finally, \(10^6\equiv1\pmod{297},\) while \(10^2\not\equiv1\) and \(10^3\not\equiv1\pmod{297},\) so its period is exactly \(6.\) Also \(303\) divides \(10^4-1\) but not \(10^2-1,\) so the period of \(\frac{1}{303}\) is exactly \(4.\) Thus \(n=297\) lies in \([201,400].\)

Thus, the correct answer is B.

23 答案 A

区域 \(|x|+|y|+|z|\le1\) 是一个正八面体，顶点为 \((\pm1,0,0),(0,\pm1,0),(0,0,\pm1)\)，体积为 \(2\cdot\tfrac13\cdot(\sqrt2)^2\cdot1=\tfrac43\)。第二个区域是同一个八面体向上平移 \(1\) 后得到的。两者的交集是另一个对角线长为 \(1\) 的正八面体，其线性尺寸是原八面体的一半，所以体积为 \(\left(\tfrac12\right)^3\cdot\tfrac43=\tfrac16\)。

所以正确答案是 A。

English

The region \(|x|+|y|+|z|\le1\) is a regular octahedron with vertices at \((\pm1,0,0),(0,\pm1,0),(0,0,\pm1),\) whose volume is \(2\cdot\tfrac13\cdot(\sqrt2)^2\cdot1=\tfrac43.\) The second region is the same octahedron shifted up by \(1.\) Their intersection is bounded by another regular octahedron with diagonals of length \(1,\) half the linear dimensions of the first, so its volume is \(\left(\tfrac12\right)^3\cdot\tfrac43=\tfrac16.\)

Thus, the correct answer is A.

24 答案 D

将每个数写成 \(77\) 乘以一个约化值，我们需要 \(\gcd=1\) 且 \(\text{lcm}=m=\frac{n}{77}\)。对每个整除 \(m\) 且最大指数为 \(M\) 的质数 \(p\)，有效指数四元组的个数为 \((M+1)^4-2M^4\) \(+(M-1)^4\) \(=2(6M^2+1)\)。所有质数的贡献乘积必须等于 \(77{,}000=2^3\cdot5^3\cdot7\cdot11\)。当 \(M=1,2,3\) 时，\(2(6M^2+1)\) 分别等于 \(14\)、\(50\) 和 \(110\)，并且 \(14\cdot50\cdot110=77{,}000\)，所以指数 \(1,2,3\) 给出一个候选分解。

每个质数恰好贡献一个因数 \(2\)，所以 \(m\) 恰好有三个质因数。它们的奇数因子 \(6M^2+1\) 必须整除 \(\frac{77{,}000}{2^3}=9625\)。检查这些因数得到 \(M=1,2,3,8\)，相应的奇数因子为 \(7,25,55,385\)。若取 \(M=8\)，其余两个奇数因子的乘积只能为 \(25\)，但每个至少为 \(7\)，不可能。因此最大指数恰为 \(1,2,3\)。为使 \(m=\frac{n}{77}\) 最小，把最大指数分配给最小质数：\(m=2^3\cdot3^2\cdot5=360\)，所以 \(n=77\cdot360=27{,}720\)。

因此，正确答案是 D。

English

Writing each entry as \(77\) times a reduced value, we need \(\gcd=1\) and \(\text{lcm}=m=\frac{n}{77}.\) For each prime \(p\) dividing \(m\) with maximum exponent \(M,\) the number of valid exponent quadruples is \((M+1)^4-2M^4\) \(+(M-1)^4\) \(=2(6M^2+1).\) The total over all primes must equal \(77{,}000=2^3\cdot5^3\cdot7\cdot11.\) Since \(2(6M^2+1)\) equals \(14,\) \(50,\) and \(110\) for \(M=1,2,3,\) and \(14\cdot50\cdot110=77{,}000,\) the exponents \(1,2,3\) give one candidate factorization.

Every prime contributes exactly one factor of \(2,\) so exactly three primes divide \(m.\) Their odd factors \(6M^2+1\) must divide \(\frac{77{,}000}{2^3}=9625.\) Checking the divisors gives \(M=1,2,3,8,\) with odd factors \(7,25,55,385.\) The choice \(M=8\) leaves only \(25\) for the product of the other two odd factors, but each is at least \(7,\) so this is impossible. Therefore the maximum exponents are exactly \(1,2,3.\) To minimize \(m=\frac{n}{77},\) assign the largest exponent to the smallest prime: \(m=2^3\cdot3^2\cdot5=360,\) so \(n=77\cdot360=27{,}720.\)

Thus, the correct answer is D.

25 答案 A

写成 \(a_n=2^{\frac{b_n}{19}}\)。递推变为 \(b_0=0\)，\(b_1=1\)，\(b_n=b_{n-1}+2b_{n-2}\)，解为 \(b_n=\tfrac13\bigl(2^n-(-1)^n\bigr)\)。乘积 \(a_1\cdots a_k\) 为整数，当且仅当 \(b_1+\cdots+b_k\) 是 \(19\) 的倍数。对 \(b_n\) 的公式求和可得：当 \(k\) 为奇数时，\(b_1+\cdots+b_k=\dfrac{2^{k+1}-1}{3}\)；当 \(k\) 为偶数时，\(b_1+\cdots+b_k=\dfrac{2^{k+1}-2}{3}\)。

\(2\) 模 \(19\) 的阶为 \(18\)，因为 \(2^9\equiv-1\pmod{19}\) 且 \(2^6\not\equiv1\pmod{19}\)。对奇数 \(k\)，整除条件要求 \(k+1\) 是 \(18\) 的倍数，最早在 \(k=17\) 时发生。对偶数 \(k\)，条件要求 \(k\) 是 \(18\) 的倍数，最早在 \(k=18\) 时发生。因此最小正整数 \(k\) 是 \(17\)。

因此，正确答案是 A。

English

Write \(a_n=2^{\frac{b_n}{19}}.\) The recursion becomes \(b_0=0,\) \(b_1=1,\) \(b_n=b_{n-1}+2b_{n-2},\) solved by \(b_n=\tfrac13\bigl(2^n-(-1)^n\bigr).\) The product \(a_1\cdots a_k\) is an integer exactly when \(b_1+\cdots+b_k\) is divisible by \(19.\) Summing the formula for \(b_n\) gives \(b_1+\cdots+b_k=\dfrac{2^{k+1}-1}{3}\) when \(k\) is odd, and \(b_1+\cdots+b_k=\dfrac{2^{k+1}-2}{3}\) when \(k\) is even.

The order of \(2\) modulo \(19\) is \(18\) because \(2^9\equiv-1\pmod{19}\) and \(2^6\not\equiv1\pmod{19}.\) For odd \(k,\) divisibility therefore requires \(k+1\) to be divisible by \(18,\) first occurring at \(k=17.\) For even \(k,\) it requires \(k\) to be divisible by \(18,\) first occurring at \(k=18.\) Hence the smallest positive \(k\) is \(17.\)

Thus, the correct answer is A.