2013B AMC 12 真题（P版）

1 答案 C

最高气温比最低气温高 \(16\)，所以最低气温比平均值低 \(8\)。平均值为 \(3^\circ\)，因此最低气温为 \(3 - 8 = -5^\circ\)。所以正确答案是 C。

English

The high exceeds the low by \(16,\) so the low is \(8\) below the average. Since the average is \(3^\circ,\) the low temperature is \(3 - 8 = -5^\circ.\) Thus, the correct answer is C.

2 答案 A

花园的长和宽分别为 \(2\cdot 15 = 30\) 英尺和 \(2\cdot 20 = 40\) 英尺，面积为 \(1200\) 平方英尺。每平方英尺能产半磅土豆，所以预计产量为 \(\tfrac12\cdot 1200 = 600\) 磅。正确答案是 A。

English

The garden is \(2\cdot 15 = 30\) feet by \(2\cdot 20 = 40\) feet, an area of \(1200\) square feet. At half a pound per square foot, Mr. Green expects \(\tfrac12\cdot 1200 = 600\) pounds. Thus, the correct answer is A.

3 答案 D

从 \(201\) 往下数时，数值 \(x\) 是第 \((202-x)\) 个数。因此 \(53\) 是第 \((202-53) = 149\) 个数。正确答案是 D。

English

Counting down from \(201,\) the value \(x\) is the \((202-x)\)th number. So \(53\) is the \((202-53) = 149\)th number. Thus, the correct answer is D.

4 答案 B

若每人行驶 \(D\) 英里，两人共行驶 \(2D\) 英里，耗油量为 \(\dfrac{D}{40}+\dfrac{D}{10} = \dfrac{D}{8}\) 加仑。因此合并油耗率为 \(\dfrac{2D}{\frac{D}{8}} = 16\) 英里每加仑。正确答案是 B。

English

If each drives \(D\) miles, together they cover \(2D\) miles using \(\dfrac{D}{40}+\dfrac{D}{10} = \dfrac{D}{8}\) gallons. The combined rate is \(\dfrac{2D}{\frac{D}{8}} = 16\) miles per gallon. Thus, the correct answer is B.

5 答案 C

家长年龄总和为 \(55\cdot 33\)，五年级学生年龄总和为 \(33\cdot 11\)，合计 \(33\cdot 66\)。总人数为 \(88\)，所以平均年龄为 \(\dfrac{33\cdot 66}{88} = 24.75\)。正确答案是 C。

English

The parents’ ages sum to \(55\cdot 33\) and the fifth-graders’ to \(33\cdot 11,\) a total of \(33\cdot 66.\) Dividing by \(88\) people gives \(\dfrac{33\cdot 66}{88} = 24.75.\) Thus, the correct answer is C.

6 答案 B

整理并配方得 \(x^2 - 10x + 25 + y^2 + 6y + 9\) \(= 0\)，即 \((x-5)^2 + (y+3)^2 = 0\)。因此 \(x = 5\)、\(y = -3\)，所以 \(x + y = 2\)。正确答案是 B。

English

Rearranging gives \(x^2 - 10x + 25 + y^2 + 6y + 9\) \(= 0,\) that is \((x-5)^2 + (y+3)^2 = 0.\) Hence \(x = 5\) and \(y = -3,\) so \(x + y = 2.\) Thus, the correct answer is B.

7 答案 E

到数到 \(n\) 的那一轮结束时，共说了 \(1 + 2 + \cdots + n = \tfrac12 n(n+1)\) 个数。当 \(n = 9\) 时，这个总数为 \(45\)。下一轮从 \(1, 2, \ldots\) 开始，所以第 \(53\) 个数是这一轮的第 \(8\) 个数，即 \(8\)。正确答案是 E。

English

After the turn that counts up to \(n,\) exactly \(1 + 2 + \cdots + n = \tfrac12 n(n+1)\) numbers have been said. For \(n = 9\) that is \(45.\) The next turn starts \(1, 2, \ldots,\) so the \(53\)rd number is the \(8\)th number of that turn, namely \(8.\) Thus, the correct answer is E.

8 答案 B

联立 \(3x - 2y = 1\) 与 \(y = 1\)，得 \(B = (1, 1)\)。点 \(A = (-1, -2)\) 到直线 \(y = 1\) 的距离为 \(3\)，所以 \(\tfrac12\cdot BC\cdot 3 = 3\)，从而 \(BC = 2\)。于是 \(C = (3, 1)\) 或 \(C = (-1, 1)\)；后者会使 \(\ell_3\) 竖直，所以 \(C = (3, 1)\)。因此斜率为 \(\dfrac{1 - (-2)}{3 - (-1)} = \dfrac34\)。正确答案是 B。

English

Solving \(3x - 2y = 1\) with \(y = 1\) gives \(B = (1, 1).\) The distance from \(A = (-1, -2)\) to the line \(y = 1\) is \(3,\) so \(\tfrac12\cdot BC\cdot 3 = 3\) gives \(BC = 2.\) Then \(C = (3, 1)\) or \(C = (-1, 1);\) the latter makes \(\ell_3\) vertical, so \(C = (3, 1)\) and the slope is \(\dfrac{1 - (-2)}{3 - (-1)} = \dfrac34.\) Thus, the correct answer is B.

9 答案 C

因为 \(12! = 2^{10}\cdot 3^5\cdot 5^2\cdot 7\cdot 11\)，能整除它的最大完全平方数为 \(2^{10}\cdot 3^4\cdot 5^2\)，其平方根为 \(2^5\cdot 3^2\cdot 5\)。指数和为 \(5 + 2 + 1 = 8\)。正确答案是 C。

English

Since \(12! = 2^{10}\cdot 3^5\cdot 5^2\cdot 7\cdot 11,\) the largest perfect square dividing it is \(2^{10}\cdot 3^4\cdot 5^2,\) whose square root is \(2^5\cdot 3^2\cdot 5.\) The exponents sum to \(5 + 2 + 1 = 8.\) Thus, the correct answer is C.

10 答案 E

经过 \(m\) 次红色摊位交换和 \(n\) 次蓝色摊位交换后，Alex 有 \(75 - (2m - n)\) 个红色代币、\(75-(3n-m)\) 个蓝色代币，以及 \(m+n\) 个银色代币。停止时，红色代币数为 \(0\) 或 \(1\)，蓝色代币数为 \(0,1\)，或 \(2\)。对这六种情形求解代币方程，只得到 \((m,n)=(59,44)\)，终态为 \((1,2)\)，或 \((m,n)=(60,45)\)，终态为 \((0,0)\)。后者不可能到达：最后一次交换前的状态只能是 \((-1,3)\) 或 \((2,-1)\)。因此 Alex 最终有 \(59+44=103\) 个银色代币。所以正确答案是 E。

English

After \(m\) red-booth and \(n\) blue-booth exchanges, Alex has \(75 - (2m - n)\) red tokens, \(75-(3n-m)\) blue tokens, and \(m+n\) silver tokens. At termination the red count is \(0\) or \(1,\) and the blue count is \(0,1,\) or \(2.\) Solving the token equations over these six cases leaves only \((m,n)=(59,44),\) ending at \((1,2),\) or \((m,n)=(60,45),\) ending at \((0,0).\) The latter is unreachable: its final exchange would have to start at either \((-1,3)\) or \((2,-1).\) Hence Alex finishes with \(59+44=103\) silver tokens. Thus, the correct answer is E.

11 答案 A

取东、北、上为 \(x, y, z\) 方向。飞行 \(7\) 英尺后，蜜蜂 \(A\) 在 \((2, 3, 2)\)，蜜蜂 \(B\) 在 \((-3, -4, 0)\)，距离为 \(\sqrt{78} \lt 10\)。下一英尺中，\(A\) 向东到 \((3, 3, 2)\)，\(B\) 向西到 \((-4, -4, 0)\)，距离为 \(\sqrt{102} \gt 10\)。所以它们相距 \(10\) 英尺时，\(A\) 正向东，\(B\) 正向西。正确答案是 A。

English

Take east, north, up as \(x, y, z.\) After \(7\) feet bee \(A\) is at \((2, 3, 2)\) and bee \(B\) is at \((-3, -4, 0),\) a distance \(\sqrt{78} \lt 10.\) On the next foot bee \(A\) moves east to \((3, 3, 2)\) and bee \(B\) moves west to \((-4, -4, 0),\) a distance \(\sqrt{102} \gt 10.\) So they pass through \(10\) feet apart while \(A\) heads east and \(B\) heads west. Thus, the correct answer is A.

12 答案 D

城市 \(E\)（道路 \(AE, DE\)）是在 \(A\)-\(D\) 旅程上的绕行，城市 \(C\)（道路 \(BC, CD\)）是在 \(B\)-\(D\) 旅程上的绕行。把它们化简后，得到只含 \(A, B, D\) 的图，其中有两条 \(A\)-\(D\) 连接、两条 \(B\)-\(D\) 连接和一条 \(A\)-\(B\) 道路。恰好各用一次的 \(A\) 到 \(B\) 通路有 \(4\) 类：\(ABDADB\)、\(ADABDB\)、\(ADBADB\)、\(ADBDAB\)。每个绕行（经 \(E\)、经 \(C\)）都可放在两次经过中的任一次，所以每类给出 \(4\) 条实际路线，共 \(4\cdot 4 = 16\) 条。正确答案是 D。

English

City \(E\) (roads \(AE, DE\)) is a detour on an \(A\)–\(D\) trip, and city \(C\) (roads \(BC, CD\)) is a detour on a \(B\)–\(D\) trip. Replace them to get a graph on \(A, B, D\) with two \(A\)–\(D\) connections, two \(B\)–\(D\) connections, and one \(A\)–\(B\) road. The trails from \(A\) to \(B\) using each once are of \(4\) types: \(ABDADB,\) \(ADABDB,\) \(ADBADB,\) and \(ADBDAB.\) Each detour (through \(E,\) through \(C\)) can be taken on either passage, so each type gives \(4\) actual routes, for \(4\cdot 4 = 16\) routes. Thus, the correct answer is D.

13 答案 D

三角形的三个角成等差数列，当且仅当中间的角为 \(60^\circ\)。令 \(\angle DBA = x\)、\(\angle ADB = y\)，则四边形 \(ABCD\) 的四个角为 \(x, y, 180 - y, 180-x\)，它们本身也必须成等差数列。按递增顺序排列，它们只能是 \(x,y,180-y,180-x\) 或 \(x,180-y,y,180-x\)，于是分别得到 \(3y=x+180\) 或 \(3y=360-x\)。而三角形的三个角 \(x,y,180-x-y\) 中有一个等于 \(60^\circ\)。代入后只剩下两组角 \((60,80,100,120)\) 和 \((45,75,105,135)\)。最大的两个角之和至多为 \(105 + 135 = 240\)。所以正确答案是 D。

English

The angles of a triangle form an arithmetic progression exactly when the middle one is \(60^\circ.\) With \(\angle DBA = x\) and \(\angle ADB = y,\) the four angles of \(ABCD\) are \(x, y, 180 - y, 180-x,\) which must itself be an arithmetic progression. In increasing order they are either \(x,y,180-y,180-x\) or \(x,180-y,y,180-x,\) giving \(3y=x+180\) or \(3y=360-x.\) One of the triangle angles \(x,y,180-x-y\) is \(60^\circ.\) Substitution leaves the angle sets \((60,80,100,120)\) and \((45,75,105,135).\) The two largest angles sum to at most \(105 + 135 = 240.\) Thus, the correct answer is D.

14 答案 C

以 \(a_1, a_2\) 开始的序列第七项为 \(5a_1 + 8a_2\)。对两个序列，\(5a_1 + 8a_2 = 5b_1 + 8b_2\)，所以 \(5(b_1 - a_1) = 8(a_2 - b_2)\)。因为 \(\gcd(5, 8) = 1\)，需要 \(b_1 - a_1\) 是 \(8\) 的倍数，且 \(a_2 - b_2\) 是 \(5\) 的倍数。设 \(a_1 \lt b_1\)，并利用不降条件，有 \(a_1 \le b_1 - 8 \le b_2 - 8 \le a_2 - 13\)。取 \(a_1 = 0\)、\(b_1 = b_2 = 8\)、\(a_2 = 13\)，得到 \(N = 5\cdot 0 + 8\cdot 13 = 104\)。正确答案是 C。

English

A sequence starting \(a_1, a_2\) has seventh term \(5a_1 + 8a_2.\) For the two sequences, \(5a_1 + 8a_2 = 5b_1 + 8b_2,\) so \(5(b_1 - a_1) = 8(a_2 - b_2).\) Since \(\gcd(5, 8) = 1,\) we need \(b_1 - a_1\) to be a multiple of \(8\) and \(a_2 - b_2\) to be a multiple of \(5.\) Taking \(a_1 \lt b_1\) with nondecreasing terms gives \(a_1 \le b_1 - 8 \le b_2 - 8 \le a_2 - 13.\) Choosing \(a_1 = 0,\) \(b_1 = b_2 = 8,\) \(a_2 = 13\) yields \(N = 5\cdot 0 + 8\cdot 13 = 104.\) Thus, the correct answer is C.

15 答案 B

因为 \(2013 = 3\cdot 11\cdot 61\)，分子中至少需要一个 \(61!\) 来提供质因数 \(61\)，所以 \(a_1 \ge 61\)。但是 \(61!\) 还含有 \(2013\) 所不含的质因数 \(59\)，所以分母中必须有 \(b_1 \ge 59\)。因此 \(a_1 + b_1 \ge 120\)。取 \(a_1 = 61\)、\(b_1 = 59\)，并使用 \(2013 = \dfrac{61!\,11!\,3!}{59!\,10!\,5!}\)，可以达到这个下界。于是 \(|a_1 - b_1| = 2\)。所以正确答案是 B。

English

Since \(2013 = 3\cdot 11\cdot 61,\) the numerator needs a factorial at least \(61!\) to supply the prime \(61,\) so \(a_1 \ge 61.\) But \(61!\) also has a factor of \(59,\) which \(2013\) does not, so the denominator needs \(b_1 \ge 59.\) Thus \(a_1 + b_1 \ge 120,\) attained by \(a_1 = 61,\) \(b_1 = 59\) via \(2013 = \dfrac{61!\,11!\,3!}{59!\,10!\,5!}.\) Then \(|a_1 - b_1| = 2.\) Thus, the correct answer is B.

16 答案 A

等角五边形的内角全为 \(108^\circ\)，所以星形的每个尖角都是底角为 \(72^\circ\)、顶角为 \(36^\circ\) 的等腰三角形。由于底角相等，每个尖角贡献的两条边都是其所对应五边形边长的同一固定倍数 \(c\)。把五个尖角相加，星形周长为 \(2c\cdot(\text{五边形周长}) = 2c\)，与各边的具体长度无关。因此 \(s\) 是常数，其最大值与最小值之差为 \(0\)。所以正确答案是 A。

English

An equiangular pentagon has all interior angles \(108^\circ,\) so each point of the star is an isosceles triangle with base angles \(72^\circ\) and apex \(36^\circ.\) By the equal base angles, each point contributes two sides that are the same fixed multiple \(c\) of the pentagon side it rests on. Summing over the five points, the star perimeter equals \(2c\cdot(\text{pentagon perimeter}) = 2c,\) independent of the individual side lengths. So \(s\) is constant, and the difference between its maximum and minimum values is \(0.\) Thus, the correct answer is A.

17 答案 D

由方程可得 \(a + b = 2 - c\) 和 \(a^2 + b^2 = 12 - c^2\)。具有给定的和与平方和的实数 \(a, b\) 存在，当且仅当 \((a + b)^2 \le 2(a^2 + b^2)\)，也就是 \((2 - c)^2 \le 2(12 - c^2)\)。化简得 \((3c - 10)(c + 2) \le 0\)，所以 \(-2 \le c \le \tfrac{10}{3}\)。最大值与最小值之差为 \(\tfrac{10}{3} - (-2) = \tfrac{16}{3}\)。所以正确答案是 D。

English

From the equations, \(a + b = 2 - c\) and \(a^2 + b^2 = 12 - c^2.\) Real numbers \(a, b\) with a given sum and sum of squares exist iff \((a + b)^2 \le 2(a^2 + b^2),\) i.e. \((2 - c)^2 \le 2(12 - c^2).\) This simplifies to \((3c - 10)(c + 2) \le 0,\) so \(-2 \le c \le \tfrac{10}{3}.\) The difference is \(\tfrac{10}{3} - (-2) = \tfrac{16}{3}.\) Thus, the correct answer is D.

18 答案 B

按模 \(5\) 分析。因为 \(2013 \equiv 3\)，无论谁先走，Jenna 都能获胜。如果 Jenna 先走，她先拿走 \(3\) 枚，使剩余数量成为 \(5\) 的倍数；此后 Barbara 拿走 \(2\) 枚时，她就拿走 \(3\) 枚，Barbara 拿走 \(4\) 枚时，她就拿走 \(1\) 枚，从而始终留下 \(5\) 的倍数，并最终拿走最后一枚。如果 Jenna 后走，她可以使每轮结束后的硬币数保持 \(\equiv 3 \pmod 5\)，直到 Barbara 面对 \(3\) 枚硬币，只能拿走 \(2\) 枚，把最后一枚留给 Jenna。因为 \(2014 \equiv 4\)，这时先手获胜：Jenna 先走可把局面化为 \(2013\) 枚的情形；Barbara 先走则可先拿走 \(4\) 枚，此后维持 \(5\) 的倍数。因此选择 B。所以正确答案是 B。

English

Work modulo \(5.\) With \(2013 \equiv 3\) coins, Jenna wins either way: going first she takes \(3\) to leave a multiple of \(5,\) then answers Barbara’s \(2\) with \(3\) and \(4\) with \(1\) to keep multiples of \(5,\) eventually taking the last coin; going second she keeps the count \(\equiv 3 \pmod 5\) until Barbara is stuck at \(3\) coins, must remove \(2,\) and leaves Jenna the last coin. With \(2014 \equiv 4\) coins, whoever goes first wins: Jenna first reduces to the \(2013\) case, while Barbara first takes \(4\) and then keeps multiples of \(5.\) This is choice B. Thus, the correct answer is B.

19 答案 B

从 \(A\) 到 \(BC\) 的高可得 \(BD = 5\)、\(CD = 9\)、\(AD = 12\)。因为 \(DE \perp AC\)，所以 \(AED \sim ADC\)，从而 \(DE = \tfrac{36}{5}\)、\(AE = \tfrac{48}{5}\)。又因为 \(\angle AFB = \angle ADB = 90^\circ\)，四边形 \(ABDF\) 是圆内接四边形，所以 \(\angle ABD = \angle AFE\)。于是直角三角形 \(ABD\) 与 \(AFE\) 相似，故 \(\dfrac{FE}{5} = \dfrac{\frac{48}{5}}{12}\)，即 \(FE = 4\)。因此 \(DF = DE - FE\) \(= \tfrac{36}{5} - 4\) \(= \tfrac{16}{5}\)，所以 \(m + n = 21\)。所以正确答案是 B。

English

The altitude from \(A\) to \(BC\) gives \(BD = 5,\) \(CD = 9,\) \(AD = 12.\) Because \(DE \perp AC,\) triangle \(AED \sim ADC,\) giving \(DE = \tfrac{36}{5}\) and \(AE = \tfrac{48}{5}.\) Since \(\angle AFB = \angle ADB = 90^\circ,\) quadrilateral \(ABDF\) is cyclic, so \(\angle ABD = \angle AFE,\) making right triangles \(ABD\) and \(AFE\) similar: \(\dfrac{FE}{5} = \dfrac{\frac{48}{5}}{12},\) so \(FE = 4.\) Hence \(DF = DE - FE\) \(= \tfrac{36}{5} - 4\) \(= \tfrac{16}{5},\) and \(m + n = 21.\) Thus, the correct answer is B.

20 答案 A

每个点 \((t, t^2)\) 都在 \(y = t^2\) 上，而连接参数为 \(t_1, t_2\) 的两点所得弦的斜率为 \(t_1 + t_2\)。当 \(135^\circ \lt x \lt 180^\circ\) 时，\(\cos x\) 和 \(\tan x\) 都介于 \(\cot x\) 与 \(\sin x\) 之间，所以 \(P\)、\(S\) 位于 \(Q\)、\(R\) 之间，平行边为 \(QR\) 和 \(PS\)。由斜率相等可得 \(\cot x + \sin x = \tan x + \cos x\)。两边乘以 \(\sin x\cos x\) 并化简，得 \(\cos x + \sin x - \sin x\cos x = 0\)。两边平方，再利用 \(2\sin x\cos x = \sin 2x\)，可得 \(1 + \sin 2x = \tfrac14\sin^2 2x\)。它在 \((-1, 1)\) 中唯一的根为 \(\sin 2x = 2 - 2\sqrt2\)。所以正确答案是 A。

English

Each point \((t, t^2)\) lies on \(y = t^2,\) and the chord through parameters \(t_1, t_2\) has slope \(t_1 + t_2.\) For \(135^\circ \lt x \lt 180^\circ,\) both \(\cos x\) and \(\tan x\) lie between \(\cot x\) and \(\sin x,\) so \(P\) and \(S\) sit between \(Q\) and \(R\) and the parallel sides are \(QR\) and \(PS.\) Equal slopes give \(\cot x + \sin x = \tan x + \cos x.\) Multiplying by \(\sin x\cos x\) and simplifying yields \(\cos x + \sin x - \sin x\cos x = 0.\) Squaring and using \(2\sin x\cos x = \sin 2x\) gives \(1 + \sin 2x = \tfrac14\sin^2 2x,\) whose only root in \((-1, 1)\) is \(\sin 2x = 2 - 2\sqrt2.\) Thus, the correct answer is A.

21 答案 C

两条共同焦点为 \(O\) 的抛物线恰有 \(2\) 个交点，除非它们的准线平行且 \(O\) 在两准线之间的带状区域外，此时不相交。不相交的对具有相同斜率，且 \(y\)-截距同号。共有 \(5\) 个斜率；对每个斜率，同号截距对有 \(2\binom{3}{2} = 6\) 对。每个相交对贡献 \(2\) 个点，且没有三条共点，所以总数为 \(2\left(\binom{30}{2} - 5\cdot 6\right)\) \(= 2(435 - 30)\) \(= 810\)。所以正确答案是 C。

English

Two parabolas with common focus \(O\) meet in exactly \(2\) points, except when their directrices are parallel and \(O\) lies outside the strip between them, in which case they do not meet. The non-intersecting pairs have directrices of equal slope and \(y\)-intercepts of the same sign. There are \(5\) slopes, and for each, \(2\binom{3}{2} = 6\) same-sign intercept pairs. Since every intersecting pair meets in \(2\) points and no point lies on three parabolas, the total is \(2\left(\binom{30}{2} - 5\cdot 6\right)\) \(= 2(435 - 30)\) \(= 810.\) Thus, the correct answer is C.

22 答案 A

写作 \(\log_n x = \tfrac{\log x}{\log n}\) 和 \(\log_m x = \tfrac{\log x}{\log m}\)，方程成为关于 \(\log x\) 的二次方程，其两根之和为 \(\log(x_1 x_2)\) \(= \tfrac18(7\log m + 6\log n)\)。于是 \(N^8=m^7n^6\)，其中 \(N=x_1x_2\)。对每个整除 \(mn\) 的质数，设它在 \(m,n\) 中的指数分别为 \(a,b\)。那么 \(7a+6b\equiv0\pmod8\)。\(a\) 为奇数是不可能的。若 \(a=0\)，则 \(b\) 是 \(4\) 的倍数，这个质数至少贡献 \(p^3\)，但这时还必须有另一个质数整除 \(m\)。若 \(a\equiv0\pmod8\) 且为正，则它对 \(N\) 的贡献至少是 \(p^7\)；若 \(a\equiv2\pmod8\)，则 \(b\equiv3\pmod4\)，此时最小的贡献是 \(p^4\)。其余任何正偶数 \(a\) 的贡献都更大。因此最小值只在 \(p=2\) 且 \((a,b)=(2,3)\) 时取到，此时 \(N=16\)、\(m=4\)、\(n=8\)。从而 \(m+n=12\)。所以正确答案是 A。

English

Writing \(\log_n x = \tfrac{\log x}{\log n}\) and \(\log_m x = \tfrac{\log x}{\log m},\) the equation becomes a quadratic in \(\log x\) whose roots sum to \(\log(x_1 x_2)\) \(= \tfrac18(7\log m + 6\log n).\) Hence \(N^8=m^7n^6,\) where \(N=x_1x_2.\) For each prime dividing \(mn,\) let its exponents in \(m,n\) be \(a,b.\) Then \(7a+6b\equiv0\pmod8.\) An odd \(a\) is impossible. If \(a=0,\) then \(b\) is a multiple of \(4\) and this prime contributes at least \(p^3,\) but some other prime must divide \(m.\) If \(a\equiv0\pmod8\) is positive, its contribution to \(N\) is at least \(p^7;\) if \(a\equiv2\pmod8,\) then \(b\equiv3\pmod4\) and the least contribution is \(p^4.\) Every other positive even \(a\) gives more. Thus the minimum uses only \(p=2\) with \((a,b)=(2,3),\) giving \(N=16,\) \(m=4,\) and \(n=8.\) So \(m+n=12.\) Thus, the correct answer is A.

23 答案 E

因为 \(\mathrm{lcm}(25, 36, 100) = 900\)，关于 \(N\) 的条件只取决于 \(N \bmod 900\)，所以考虑 \(0 \le N \le 899\)。设以 \(5\) 为底的表示的末两位为 \(a_1, a_0\)，以 \(6\) 为底的表示的末两位为 \(b_1,b_0\)。模 \(10\) 时，所要求的相等以及 \(N\equiv a_0\pmod5\) 迫使 \(a_0=b_0\)。再对模 \(25\) 和 \(36\) 应用中国剩余定理，得到 \[ \begin{aligned} N&\equiv180a_1+150b_1 \\ &\quad {}+a_0\pmod{900} \end{aligned}\text{。}\] 把这个余数的两倍模 \(100\) 与十进制数 \(10(a_1+b_1)+a_0+b_0\) 比较，可化简为 \(5a_1\equiv b_1\pmod{10}\)。因此有效的数对恰好是 \((0,0),(2,0),(4,0),(1,5),(3,5)\)。每一对都可与 \(5\) 种 \(a_0\) 的取法 \((0 \le a_0 \le 4)\) 结合，共给出 \(25\) 个 \(N\)。所以正确答案是 E。

English

Because \(\mathrm{lcm}(25, 36, 100) = 900,\) the condition on \(N\) depends only on \(N \bmod 900,\) so consider \(0 \le N \le 899.\) Let the last two base-\(5\) digits be \(a_1, a_0\) and the last two base-\(6\) digits be \(b_1,b_0.\) Modulo \(10,\) the desired equality and \(N\equiv a_0\pmod5\) force \(a_0=b_0.\) The Chinese Remainder Theorem applied modulo \(25\) and \(36\) then gives \[ \begin{aligned} N&\equiv180a_1+150b_1 \\ &\quad {}+a_0\pmod{900}. \end{aligned} \] Comparing twice this residue modulo \(100\) with the decimal number \(10(a_1+b_1)+a_0+b_0\) reduces to \(5a_1\equiv b_1\pmod{10}.\) Hence the valid pairs are exactly \((0,0),(2,0),(4,0),(1,5),(3,5).\) Each combines with \(5\) choices of \(a_0\) \((0 \le a_0 \le 4),\) giving \(25\) values of \(N.\) Thus, the correct answer is E.

24 答案 A

令 \(\alpha = \angle ACN = \angle NCB\)、\(x = BN\)。由于 \(\triangle BXN\) 是等边三角形，所以 \(\angle BXC = \angle CNA = 120^\circ\)，由此得到 \(\triangle ABC \sim \triangle BMC\) 和 \(\triangle ANC \sim \triangle BXC\)。由第一组相似关系，并利用 \(MC = \tfrac12 AC = 1\)，可得 \(\dfrac{BC}{2} = \dfrac{MC}{BC}\)，所以 \(BC = \sqrt2\)。由第二组相似关系可得 \(CX = (\sqrt2 + 1)x\)。在 \(\triangle BCX\) 中，对 \(\angle BXC = 120^\circ\) 应用余弦定理，得到 \(2 = x^2\) \(+ (\sqrt2 + 1)^2 x^2\) \(+ (\sqrt2 + 1)x^2\) \(= (5 + 3\sqrt2)x^2\)。因此 \(BN^2 = x^2\) \(= \dfrac{2}{5 + 3\sqrt2}\) \(= \dfrac{10 - 6\sqrt2}{7}\)。所以正确答案是 A。

English

Let \(\alpha = \angle ACN = \angle NCB\) and \(x = BN.\) Since \(\triangle BXN\) is equilateral, \(\angle BXC = \angle CNA = 120^\circ,\) which gives \(\triangle ABC \sim \triangle BMC\) and \(\triangle ANC \sim \triangle BXC.\) From the first, with \(MC = \tfrac12 AC = 1,\) we get \(\dfrac{BC}{2} = \dfrac{MC}{BC},\) so \(BC = \sqrt2.\) From the second, \(CX = (\sqrt2 + 1)x.\) The Law of Cosines in \(\triangle BCX\) with \(\angle BXC = 120^\circ\) gives \(2 = x^2\) \(+ (\sqrt2 + 1)^2 x^2\) \(+ (\sqrt2 + 1)x^2\) \(= (5 + 3\sqrt2)x^2.\) Hence \(BN^2 = x^2\) \(= \dfrac{2}{5 + 3\sqrt2}\) \(= \dfrac{10 - 6\sqrt2}{7}.\) Thus, the correct answer is A.

25 答案 B

因为系数为实数，非实根成共轭对，所以 \(P(z)\) 可分解为互不相同的线性因子 \((z - c)\)（其中 \(c \in \mathbb{Z}\)）以及二次因子 \((z - (a+ib))(z - (a-ib))\) \(= z^2 - 2az + (a^2 + b^2)\)。每个因子的常数项都整除 \(50\)。对 \(d=1,2,5,10,25,50\)，满足 \(a^2+b^2=d\) 且 \(b\ne0\) 的共轭对个数依次为 \(1,2,4,4,5,6\)。再加上两个线性选择 \(z-d,z+d\)，就得到 \(|B_1|=3\)、\(|B_2|=4\)、\(|B_5|=6\)、\(|B_{10}|=6\)、\(|B_{25}|=7\) 和 \(|B_{50}|=8\)。把 \(50\) 分解成若干个大于 \(1\) 的因子大小之积，方式有 \(50,25\cdot2,10\cdot5\) 和 \(5\cdot5\cdot2\)。在最后一种方式中，各根互不相同要求选取两个不同的 \(B_5\) 因子。最后，再计入 \(z+1\) 和 \(z^2+1\) 可以自由出现（而 \(z-1\) 是否出现由其余因子乘积的符号决定），总数为 \[ \begin{aligned} &2^2\left(8 + 7\cdot 4 + 6\cdot 6 + 4\binom{6}{2}\right) \\ &\quad = 4(8 + 28 + 36 + 60) = 528 \end{aligned}\text{。}\] 所以正确答案是 B。

English

Since the coefficients are real, nonreal roots occur in conjugate pairs, so \(P(z)\) factors into distinct linear factors \((z - c)\) with \(c \in \mathbb{Z}\) and quadratics \((z - (a+ib))(z - (a-ib))\) \(= z^2 - 2az + (a^2 + b^2).\) Each factor’s constant term divides \(50.\) For \(d=1,2,5,10,25,50,\) the numbers of conjugate pairs with \(a^2+b^2=d\) and \(b\ne0\) are \(1,2,4,4,5,6,\) respectively. Adding the two linear choices \(z-d,z+d\) gives \(|B_1|=3,\) \(|B_2|=4,\) \(|B_5|=6,\) \(|B_{10}|=6,\) \(|B_{25}|=7,\) and \(|B_{50}|=8.\) The factor-magnitude partitions of \(50\) using values greater than \(1\) are \(50,25\cdot2,10\cdot5,\) and \(5\cdot5\cdot2.\) Distinct roots require choosing two different \(B_5\) factors in the last case. Finally, account for the free presence of \(z+1\) and \(z^2+1\) (with \(z-1\) forced by the sign of the remaining product), gives \[ \begin{aligned} &2^2\left(8 + 7\cdot 4 + 6\cdot 6 + 4\binom{6}{2}\right) \\ &\quad = 4(8 + 28 + 36 + 60) = 528. \end{aligned} \] Thus, the correct answer is B.