2024B AMC 10 真题（P版）

1 答案 B

这个位置左边有 \(1012\) 个人，右边有 \(1009\) 个人。再加上这个人自己，总人数为 \(1012 + 1009 + 1 = 2022\)。也可以看作两个位置计数在这个人处重合一次，所以 \(1013 + 1010 - 1 = 2022\)。因此正确答案是 B。

English

There are \(1012\) people to the left of this spot and \(1009\) to the right. Add those two groups plus the person themselves: \(1012 + 1009 + 1 = 2022.\) Or, just as fast, the two positions overlap on one person, so \(1013 + 1010 - 1 = 2022.\) Thus, B is the correct answer.

2 答案 B

有 \(10! = 10 \cdot 9 \cdot 8 \cdot 7! = 720 \cdot 7!\)。而 \(720 = 6!\)，所以 \(10! = 6! \cdot 7! = 7! \cdot 6!\)。两项相同，因此 \(10! - 7! \cdot 6! = 0\)。正确答案是 B。

English

Write \(10! = 10 \cdot 9 \cdot 8 \cdot 7! = 720 \cdot 7!.\) But \(720 = 6!\) too, so \(10! = 6! \cdot 7! = 7! \cdot 6!.\) The two terms are the same. That makes \(10! - 7! \cdot 6! = 0.\) Therefore, the answer is B.

3 答案 E

两边除以 \(2\)，得 \(|x| \le 3.5\pi \approx 10.996\)。满足条件的整数从 \(-10\) 到 \(10\)，共有 \(21\) 个。因此正确答案是 E。

English

Divide by \(2\) to get \(|x| \le 3.5\pi \approx 10.996.\) The integers that fit run from \(-10\) up to \(10,\) and there are \(21\) of them. Thus, E is the correct answer.

4 答案 D

第 \(g\) 组含有 \(g\) 个球，因此前 \(g\) 组一共用掉 \(\tfrac{g(g+1)}{2}\) 个球。现在 \(\tfrac{63 \cdot 64}{2} = 2016\)，而 \(\tfrac{64 \cdot 65}{2} = 2080\)，所以球 \(2024\) 在第 \(64\) 组中（球 \(2017\) 到 \(2080\)）。箱子按 \(A, B, C, D, E\) 循环，所以第 \(g\) 组进入编号为 \((g - 1) \bmod 5\) 的箱子。对 \(g = 64\)，有 \(63 \bmod 5 = 3\)，即箱子 \(D\)。因此正确答案是 D。

English

Group \(g\) holds \(g\) balls, so the first \(g\) groups swallow \(\tfrac{g(g+1)}{2}\) of them. Now \(\tfrac{63 \cdot 64}{2} = 2016\) and \(\tfrac{64 \cdot 65}{2} = 2080,\) which puts ball \(2024\) in group \(64\) (balls \(2017\) through \(2080\)). The bins cycle \(A, B, C, D, E,\) so group \(g\) lands in bin number \((g - 1) \bmod 5.\) For \(g = 64\) that’s \(63 \bmod 5 = 3,\) bin \(D.\) Therefore, the answer is D.

5 答案 B

完整和为 \(1 + 3 + \cdots + 99 = 50^2 = 2500\)。把一项 \(t\) 从加号改为减号，会使总和减少 \(2t\)，所以被变号的项之和必须大于 \(1250\)。为了使用最少项，应选择最大的奇数。最大的 \(k\) 个奇数之和为 \(99 + 97 + \cdots = k(100 - k)\)。需要 \(k(100 - k) \gt 1250\)。当 \(k = 14\) 时只有 \(1204\)，当 \(k = 15\) 时为 \(1275\)。所以 \(15\) 次变号可以做到。因此正确答案是 B。

English

The full sum is \(1 + 3 + \cdots + 99 = 50^2 = 2500.\) Flipping a term \(t\) drops the total by \(2t,\) so to go negative the flipped terms have to add up to more than \(1250.\) The greedy move is to flip the biggest odd numbers: flipping the top \(k\) gives \(99 + 97 + \cdots = k(100 - k).\) We want \(k(100 - k) \gt 1250.\) At \(k = 14\) it’s only \(1204,\) but at \(k = 15\) it jumps to \(1275.\) So \(15\) flips do it. Thus, B is the correct answer.

6 答案 B

若长和宽为 \(\ell\) 和 \(w\)，周长为 \(2(\ell + w)\)，在 \(\ell w = 2024\) 且两边尽量接近时最小。分解 \(2024 = 2^3 \cdot 11 \cdot 23\)。最接近 \(\sqrt{2024} \approx 45\) 的因数对是 \(44 \times 46\)，周长为 \(2(44 + 46) = 180\)。因此正确答案是 B。

English

The perimeter \(2(\ell + w)\) with \(\ell w = 2024\) is smallest when \(\ell\) and \(w\) are as close together as possible. Factor \(2024 = 2^3 \cdot 11 \cdot 23.\) The divisor pair nearest \(\sqrt{2024} \approx 45\) is \(44 \times 46,\) which gives perimeter \(2(44 + 46) = 180.\) Therefore, the answer is B.

7 答案 A

提出公共幂：\(7^{2024} + 7^{2025} + 7^{2026}\) \(= 7^{2024}(1 + 7 + 49)\) \(= 7^{2024} \cdot 57\)。而 \(57 = 3 \cdot 19\)，所以整个乘积是 \(19\) 的倍数。余数为 \(0\)。因此正确答案是 A。

English

Pull out the common power: \(7^{2024} + 7^{2025} + 7^{2026}\) \(= 7^{2024}(1 + 7 + 49)\) \(= 7^{2024} \cdot 57.\) And \(57 = 3 \cdot 19,\) so the product is a multiple of \(19.\) The remainder is \(0.\) Thus, A is the correct answer.

8 答案 D

因为 \(42 = 2 \cdot 3 \cdot 7\) 有 \((1+1)^3 = 8\) 个因数，可以把每个因数和它的互补因数配对，所以 \(N = 42^{\frac{8}{2}} = 42^4\)。只看个位数字，\(2^4 = 16\) 的个位是 \(6\)，所以 \(42^4\) 也是如此。因此正确答案是 D。

English

Since \(42 = 2 \cdot 3 \cdot 7\) has \((1+1)^3 = 8\) divisors, we can pair each divisor with its complement, so \(N = 42^{\frac{8}{2}} = 42^4.\) Only the units digit matters, and \(2^4 = 16\) ends in \(6,\) so \(42^4\) does too. Therefore, the answer is D.

9 答案 A

由平均数可得 \(a + b + c = 0\) 且 \(a^2 + b^2 + c^2 = 30\)。平方第一个等式：\((a+b+c)^2\) \(= a^2 + b^2 + c^2\) \(+ 2(ab + bc + ca)\)。于是 \(0 = 30 + 2(ab + bc + ca)\)，所以 \(ab + bc + ca = -15\)。它们的平均数为 \(-\frac{15}{3} = -5\)。因此正确答案是 A。

English

The means tell us \(a + b + c = 0\) and \(a^2 + b^2 + c^2 = 30.\) Square the first: \((a+b+c)^2\) \(= a^2 + b^2 + c^2\) \(+ 2(ab + bc + ca),\) so \(0 = 30 + 2(ab + bc + ca)\) and \(ab + bc + ca = -15.\) Their mean is \(-\frac{15}{3} = -5.\) Thus, A is the correct answer.

10 答案 A

面积比在仿射变换下不变，所以取方便坐标：\(A = (0,0)\)、\(B = (1,0)\)、\(C = (1,1)\)、\(D = (0,1)\)，此时 \(E = (0, \tfrac12)\)。直线 \(AC\) 为 \(y = x\)，直线 \(EB\) 从 \((0, \tfrac12)\) 到 \((1, 0)\)，它们相交于 \(F = (\tfrac13, \tfrac13)\)。用鞋带公式可得四边形 \(CDEF\) 的面积为 \(\tfrac{5}{12}\)，三角形 \(CFB\) 的面积为 \(\tfrac13\)。所以比值为 \(\tfrac{5}{12} : \tfrac13 = 5 : 4\)。因此正确答案是 A。

English

Area ratios don’t change under an affine map, so drop in convenient coordinates: \(A = (0,0),\) \(B = (1,0),\) \(C = (1,1),\) \(D = (0,1),\) which makes \(E = (0, \tfrac12).\) Line \(AC\) is \(y = x,\) and line \(EB\) runs from \((0, \tfrac12)\) to \((1, 0);\) they cross at \(F = (\tfrac13, \tfrac13).\) The shoelace formula gives quadrilateral \(CDEF\) area \(\tfrac{5}{12}\) and triangle \(CFB\) area \(\tfrac13.\) So the ratio is \(\tfrac{5}{12} : \tfrac13 = 5 : 4.\) Therefore, the answer is A.

11 答案 C

设 \(X = (0,0)\)、\(Y = (8,0)\)、\(W = (0,4)\)、\(Z = (8,4)\)，于是 \(M = (m, 0)\)、\(A = (8, a)\)。直角条件表示 \(\overrightarrow{MW} \cdot \overrightarrow{MA} = 0\)，给出 \(-m(8 - m) + 4a = 0\)，即 \(m(8 - m) = 4a\)。面积相等 \([WXM] = 2m\) 和 \([WAZ] = 4(4 - a)\) 迫使 \(m = 8 - 2a\)，所以 \(a = \tfrac{8 - m}{2}\)。代回得 \((8 - m)(2 - m) = 0\)。取 \(M \ne Y\)，得到 \(m = 2\)、\(a = 3\)。于是 \([WMA] = 32\) \(- [WXM]\) \(- [MYA]\) \(- [AZW]\) \(= 32 - 4 - 9 - 4\) \(= 15\)。因此正确答案是 C。

English

Set \(X = (0,0),\) \(Y = (8,0),\) \(W = (0,4),\) \(Z = (8,4),\) so \(M = (m, 0)\) and \(A = (8, a).\) The right angle means \(\overrightarrow{MW} \cdot \overrightarrow{MA} = 0,\) which gives \(-m(8 - m) + 4a = 0,\) that is \(m(8 - m) = 4a.\) Equal areas \([WXM] = 2m\) and \([WAZ] = 4(4 - a)\) force \(m = 8 - 2a,\) so \(a = \tfrac{8 - m}{2}.\) Substitute back and \((8 - m)(2 - m) = 0.\) Taking \(M \ne Y\) leaves \(m = 2\) and \(a = 3.\) Then \([WMA] = 32\) \(- [WXM]\) \(- [MYA]\) \(- [AZW]\) \(= 32 - 4 - 9 - 4\) \(= 15.\) Thus, C is the correct answer.

12 答案 A

把每名学生会说的语言看作一个集合。条件表示没有人的集合包含另一个人的集合。每个人会说同样数量 \(k\) 种语言，而两个不同的 \(k\) 元集合不可能互相包含，所以只需要有 \(100\) 个不同的 \(k\) 元子集。设语言总数为 \(n\)，需要 \(\binom{n}{k} \ge 100\)。当 \(n = 8\) 时最多为 \(\binom{8}{4} = 70\)，不足 \(100\)。但 \(\binom{9}{4} = 126 \ge 100\)。因此 \(9\) 种语言既足够也必要。正确答案是 A。

English

Give each student the set of languages they speak. The condition says no one’s set sits inside another’s. Everyone speaks the same number \(k\) of languages, and two distinct \(k\)-element sets can never contain each other, so all we need is \(100\) different \(k\)-subsets of the \(n\) languages, i.e. \(\binom{n}{k} \ge 100.\) With \(n = 8\) the best we can manage is \(\binom{8}{4} = 70,\) short of \(100.\) But \(\binom{9}{4} = 126 \ge 100.\) So \(9\) languages are both enough and necessary. Therefore, the answer is A.

13 答案 B

因为 \(1183 = 7 \cdot 13^2\)，所以 \(\sqrt{1183} = 13\sqrt7\)。把所给等式两边平方，可知 \(\sqrt{xy}\) 是有理数。因此 \(x\) 与 \(y\) 有相同的无平方因子部分：记 \(x=da^2\) 和 \(y=db^2\)，其中 \(d\) 无平方因子，而 \(a,b\) 为正整数。于是 \((a+b)\sqrt d=13\sqrt7\)，所以 \(d=7\)，\(a+b=13\)。因此 \(x+y=7(a^2+b^2)\)，它在 \(a\) 与 \(b\) 尽量接近时取到最小值。取 \(a=6\) 和 \(b=7\)，得 \(x+y=7(36+49)=595\)。因此正确答案是 B。

English

Since \(1183 = 7 \cdot 13^2,\) we have \(\sqrt{1183} = 13\sqrt7.\) Squaring the given equation shows that \(\sqrt{xy}\) is rational. Thus \(x\) and \(y\) have the same squarefree part: write \(x=da^2\) and \(y=db^2,\) where \(d\) is squarefree and \(a,b\) are positive integers. Then \((a+b)\sqrt d=13\sqrt7,\) so \(d=7\) and \(a+b=13.\) Therefore \(x+y=7(a^2+b^2),\) which is smallest when \(a\) and \(b\) are as close as possible. Taking \(a=6\) and \(b=7\) gives \(x+y=7(36+49)=595.\) Thus, B is the correct answer.

14 答案 B

\(B\) 是正方形 \(|x| + |y| \le 8\)，面积为 \(2 \cdot 8^2 = 128\)。目标条件 \((x^2 + y^2 - 25)^2 \le 49\) 等价于 \(|x^2 + y^2 - 25| \le 7\)，也就是 \(18 \le x^2 + y^2 \le 32\)，这是面积为 \(\pi(32 - 18) = 14\pi\) 的圆环。它是否在 \(B\) 中？原点到边 \(x + y = 8\) 的距离是 \(\tfrac{8}{\sqrt2} = 4\sqrt2 = \sqrt{32}\)，正好是外半径，所以圆环在正方形内。概率为 \(\tfrac{14\pi}{128} = \tfrac{7}{64}\pi\)，因此 \(m + n = 71\)。正确答案是 B。

English

\(B\) is the square \(|x| + |y| \le 8,\) with area \(2 \cdot 8^2 = 128.\) The target condition \((x^2 + y^2 - 25)^2 \le 49\) unpacks to \(|x^2 + y^2 - 25| \le 7,\) that is \(18 \le x^2 + y^2 \le 32,\) an annulus of area \(\pi(32 - 18) = 14\pi.\) Does it fit inside \(B?\) The distance from the origin to an edge \(x + y = 8\) is \(\tfrac{8}{\sqrt2} = 4\sqrt2 = \sqrt{32},\) exactly the outer radius, so yes, the annulus sits inside the square. The probability is \(\tfrac{14\pi}{128} = \tfrac{7}{64}\pi,\) giving \(m + n = 71.\) Therefore, the answer is B.

15 答案 C

六个固定数之和为 \(24.8\)。若平均数是整数 \(k\)，则 \(x+y+z=9k-24.8\)。因为固定数已经从 \(1\) 延伸到 \(7\)，极差条件给出三种情况。

若 \(z\le7\)，则 \(x=0\)。对 \(y+z\) 的范围限制迫使 \(k=3\) 或 \(4\)。当 \(k=3\) 时，中位数为 \(2.2\)；当 \(k=4\) 时，\(y+z=11.2\)，所以 \(y\ge4.2\)，而中位数为整数仅当 \(y=5\)，得到 \((x,y,z)=(0,5,6.2)\)。

若 \(x\ge1\)，则 \(z=8\)。此时 \(k=4\) 给出中位数 \(3.2\)。当 \(k=5\) 时，\(x+y=12.2\)；中位数为整数仅当 \(x=6\)，得到 \((6,6.2,8)\)。

剩余情况满足 \(0<x<1\) 且 \(z=x+7\)。总和介于 \(31.8\) 与 \(41.8\) 之间，所以 \(k=4\)，且 \(y=4.2-2x\)。中位数为整数仅当 \(y=4\)，得到 \((0.1,4,7.1)\)。因此恰有 \(3\) 个有序三元组符合条件。所以正确答案是 C。

English

The six fixed numbers total \(24.8.\) If the mean is the integer \(k,\) then \(x+y+z=9k-24.8.\) Because the fixed entries already run from \(1\) to \(7,\) the range condition gives three cases.

If \(z\le7,\) then \(x=0.\) The bounds on \(y+z\) force \(k=3\) or \(4.\) For \(k=3,\) the median is \(2.2;\) for \(k=4,\) we have \(y+z=11.2,\) so \(y\ge4.2\) and the median is integral only when \(y=5,\) giving \((x,y,z)=(0,5,6.2).\)

If \(x\ge1,\) then \(z=8.\) Here \(k=4\) gives median \(3.2.\) For \(k=5,\) we have \(x+y=12.2;\) the median is integral only when \(x=6,\) giving \((6,6.2,8).\)

The remaining case has \(0<x<1\) and \(z=x+7.\) The total lies between \(31.8\) and \(41.8,\) so \(k=4\) and \(y=4.2-2x.\) The median can be an integer only when \(y=4,\) giving \((0.1,4,7.1).\) Hence exactly \(3\) ordered triples work. Thus, C is the correct answer.

16 答案 A

在 \(1, \ldots, 2024\) 中有 \(1012\) 个偶数和 \(1012\) 个奇数。每次操作把 \(4\) 个数换成 \(1\) 个数，所以总数减少 \(3\)。如果某次操作用掉 \(e\) 个偶数，那么它写下的结果要么是奇数，使偶数的个数减少 \(e\)；要么是偶数，使偶数的个数减少 \(e-1\)。无论哪种情形，偶数的个数至多减少 \(3\)。因此要消去全部 \(1012\) 个偶数，至少需要 \(\lceil\frac{1012}{3}\rceil=338\) 次操作。这个次数可以达到：先做 \(337\) 次求和，每次取一个奇数和三个偶数，再做一次求和，取三个奇数和最后一个偶数。这样每次写下的都是奇数。因此白板上最多能剩下 \(2024-3\cdot338=1010\) 个数。因此正确答案是 A。

English

Among \(1, \ldots, 2024\) there are \(1012\) even numbers and \(1012\) odd numbers. Each operation replaces \(4\) entries by \(1,\) so the total count falls by \(3.\) If a move consumes \(e\) even entries, its output is either odd, reducing the even count by \(e,\) or even, reducing it by \(e-1.\) In either case the even count falls by at most \(3.\) Therefore eliminating all \(1012\) even entries takes at least \(\lceil\frac{1012}{3}\rceil=338\) moves. This is achievable: use \(337\) sums containing one odd and three evens, then one sum containing three odds and the final even. Every output is odd. Thus the maximum remaining count is \(2024-3\cdot338=1010.\) Therefore, the answer is A.

17 答案 D

如果没有并列，\(5\) 只蜗牛有 \(5! = 120\) 种名次顺序。现在允许恰好一组大小为 \(k\) 的并列，其中 \(2 \le k \le 5\)。先用 \(\binom{5}{k}\) 选出这一组，再把它看作一个块，连同其余单个蜗牛一共 \(6 - k\) 个块，排列方式为 \((6 - k)!\)。对 \(k\) 求和，得 \(\binom{5}{2}4!\) \(+ \binom{5}{3}3!\) \(+ \binom{5}{4}2!\) \(+ \binom{5}{5}1!\) \(= 240 + 60 + 10 + 1\) \(= 311\)。加上无并列的 \(120 + 311 = 431\) 种。因此正确答案是 D。

English

If nobody ties, the \(5\) snails finish in \(5! = 120\) orders. Now allow exactly one tied group of size \(k\) with \(2 \le k \le 5.\) Choose the group in \(\binom{5}{k}\) ways, then treat it as one block, leaving \(6 - k\) blocks to arrange in \((6 - k)!\) ways. Summing over \(k:\) \(\binom{5}{2}4!\) \(+ \binom{5}{3}3!\) \(+ \binom{5}{4}2!\) \(+ \binom{5}{5}1!\) \(= 240 + 60 + 10 + 1\) \(= 311.\) Add the no-tie count: \(120 + 311 = 431.\) Thus, D is the correct answer.

18 答案 B

这里 \(125 = 5^3\)，且 \(\varphi(125) = 100\)。若 \(\gcd(n, 5) = 1\)，欧拉定理给出 \(n^{100} \equiv 1 \pmod{125}\)。若 \(n\) 是 \(5\) 的倍数，则 \(n^{100}\) 含有因子 \(5^{100}\)，因而含有 \(125\)，所以 \(n^{100} \equiv 0 \pmod{125}\)。因此只可能有 \(0\) 和 \(1\) 两种余数。正确答案是 B。

English

Here \(125 = 5^3\) and \(\varphi(125) = 100.\) If \(\gcd(n, 5) = 1,\) Euler’s theorem gives \(n^{100} \equiv 1 \pmod{125}.\) And if \(n\) is divisible by \(5,\) then \(n^{100}\) carries a factor of \(5^{100},\) hence of \(125,\) so \(n^{100} \equiv 0 \pmod{125}.\) That leaves only two possible remainders, \(0\) and \(1.\) Therefore, the answer is B.

19 答案 C

任意两个格点决定的斜率都是有理数。所以无理斜率的直线至多含一个格点：它可以有 \(0\) 个（例如 \(y = \sqrt2\,x + \tfrac12\)），也可以恰有 \(1\) 个（例如 \(y = \sqrt2\,x\)），但不可能有两个。有理斜率（包括 \(0\)）的直线若经过格点 \((x_0, y_0)\)，则对其最简斜率 \(\tfrac{p}{q}\)，也经过 \((x_0 + q, y_0 + p)\)，所以会经过无穷多个格点；这样的直线要么没有格点（用无理截距平移即可），要么多于两个，不可能恰有一个或两个。因此每一行正好有两个“可能”。对零斜率和非零有理斜率，是“零个”和“多于两个”两列；对无理斜率，是“零个”和“恰好一个”两列。总数为 \(6\)。因此正确答案是 C。

English

Any two lattice points give a rational slope. So a line with irrational slope holds at most one lattice point: it can have \(0\) (say \(y = \sqrt2\,x + \tfrac12\)) or exactly \(1\) (say \(y = \sqrt2\,x\)), never two. A line with rational slope (zero included) through a lattice point \((x_0, y_0)\) also passes through \((x_0 + q, y_0 + p)\) for its reduced slope \(\tfrac{p}{q},\) so it hits infinitely many; such a line has either \(0\) lattice points (shift it by an irrational intercept) or more than two, never exactly one or two. So each row gives exactly two “Possible” entries. For zero and nonzero rational slope those are the “zero” and “more than two” columns; for irrational slope, the “zero” and “exactly one” columns. That’s \(6\) in all. Thus, C is the correct answer.

20 答案 A

只要左右脚模式中一个 \(L\) 与一个 \(R\) 相邻，这两只鞋就必须是一双。因此内部连续段的长度不能为 \(1\)：其中唯一一只鞋会被迫同时成为两侧鞋子的配对。三只 \(L\) 和三只 \(R\) 的唯一可能模式是 \(LLLRRR\)，\(RRRLLL\)，\(LLRRRL\)，\(LRRLLR\)，\(LRRRLL\)，\(RLLLRR\)，\(RLLRRL\)，和 \(RRLLLR\)。对任意一种只切换一次的模式，选择切换处的一双并排列剩余鞋子，得到 \(3\cdot2\cdot2=12\) 种排列。其他六种模式各有 \(6\) 种把三双鞋分配到切换处的方法。因此总数为 \(2\cdot12+6\cdot6=60\)。所以答案是 A。

English

Wherever an \(L\) meets an \(R\) in the side pattern, those two shoes must be mates. Thus an interior run cannot have length \(1:\) its lone shoe would have to be the mate of both neighbors. With three \(L\)’s and three \(R\)’s, the only possible patterns are \(LLLRRR,\) \(RRRLLL,\) \(LLRRRL,\) \(LRRLLR,\) \(LRRRLL,\) \(RLLLRR,\) \(RLLRRL,\) and \(RRLLLR.\) For either one-switch pattern, choose the pair at the switch and order the remaining shoes, giving \(3\cdot2\cdot2=12\) arrangements. Each of the other six patterns has \(6\) assignments of the three pairs to its switches. Hence the total is \(2\cdot12+6\cdot6=60.\) Therefore, the answer is A.

21 答案 C

两个半径为 \(R\) 和 \(r\) 的圆靠在地面上并互相相切时，它们与地面的接触点水平距离为 \(2\sqrt{Rr}\)。所以半径 \(1\) 和 \(\tfrac14\) 的两根管接触地面的点相距 \(2\sqrt{1 \cdot \tfrac14} = 1\)。设第三根管半径为 \(r\)。它到大管接触点的距离为 \(2\sqrt{r}\)，到小管接触点的距离为 \(2\sqrt{\tfrac14 r} = \sqrt{r}\)。若夹在两者之间，\(2\sqrt r + \sqrt r = 1\)，所以 \(\sqrt r = \tfrac13\)，\(r = \tfrac19\)。若在小管外侧，\(2\sqrt r - \sqrt r = 1\)，所以 \(r = 1\)。（在大管外侧不可能。）半径之和为 \(\tfrac19 + 1 = \tfrac{10}{9}\)。因此正确答案是 C。

English

Two circles of radii \(R\) and \(r\) resting on the floor and touching each other have contact points a horizontal distance \(2\sqrt{Rr}\) apart. So the radius-\(1\) and radius-\(\tfrac14\) pipes touch the floor \(2\sqrt{1 \cdot \tfrac14} = 1\) apart. A third pipe of radius \(r\) sits \(2\sqrt{r}\) from the big pipe’s contact point and \(2\sqrt{\tfrac14 r} = \sqrt{r}\) from the small pipe’s. Nestled between them, \(2\sqrt r + \sqrt r = 1,\) so \(\sqrt r = \tfrac13\) and \(r = \tfrac19.\) Sitting past the small pipe, \(2\sqrt r - \sqrt r = 1,\) so \(r = 1.\) (Past the big pipe can’t happen.) The sum is \(\tfrac19 + 1 = \tfrac{10}{9}.\) Thus, C is the correct answer.

22 答案 A

先把 \(16\) 个人分成 \(4\) 个不可区分的 \(4\) 人组，共有 \(\dfrac{16!}{(4!)^4 \, 4!}\) 种方法；每个委员会再选主席和秘书，有 \(4 \cdot 3 = 12\) 种，因而贡献因子 \(12^4\)。现在数因子 \(3\)。在 \(16!\) 中有 \(\lfloor \frac{16}{3} \rfloor + \lfloor \frac{16}{9} \rfloor = 6\)；分母 \((4!)^4 \, 4!\) 贡献 \(4 \cdot 1 + 1 = 5\)；而 \(12^4\) 贡献 \(4\)。指数为 \(6 - 5 + 4 = 5\)，所以 \(r = 5\)。因此正确答案是 A。

English

Split \(16\) people into \(4\) indistinguishable groups of \(4\) in \(\dfrac{16!}{(4!)^4 \, 4!}\) ways, then each committee picks a chairperson and a secretary in \(4 \cdot 3 = 12\) ways, a factor of \(12^4.\) Now count factors of \(3.\) In \(16!\) there are \(\lfloor \frac{16}{3} \rfloor + \lfloor \frac{16}{9} \rfloor = 6;\) the denominator \((4!)^4 \, 4!\) contributes \(4 \cdot 1 + 1 = 5;\) and \(12^4\) contributes \(4.\) The exponent is \(6 - 5 + 4 = 5,\) so \(r = 5.\) Therefore, the answer is A.

23 答案 B

使用 \(F_{2k} = F_k L_k\)，于是每一项 \(\dfrac{F_{2k}}{F_k} = L_k\)，即第 \(k\) 个卢卡斯数。原和变为 \(\sum_{k=1}^{10} L_k\)。由于 \(L_1 = 1\)，\(L_2 = 3\)，\(L_3 = 4\)，\(\ldots\)，\(L_{10} = 123\)，恒等式 \(\sum_{k=1}^{n} L_k = L_{n+2} - 3\) 给出 \(L_{12} - 3 = 322 - 3 = 319\)。因此正确答案是 B。

English

Use \(F_{2k} = F_k L_k,\) so each term \(\dfrac{F_{2k}}{F_k} = L_k,\) the \(k\)th Lucas number. That collapses the sum to \(\sum_{k=1}^{10} L_k.\) With \(L_1 = 1,\) \(L_2 = 3,\) \(L_3 = 4,\) \(\ldots,\) \(L_{10} = 123,\) the identity \(\sum_{k=1}^{n} L_k = L_{n+2} - 3\) gives \(L_{12} - 3 = 322 - 3 = 319.\) Thus, B is the correct answer.

24 答案 E

通分到 \(8\)：\(P(m) = \dfrac{m^8 + m^4 + 2m^2 + 4m}{8}\)。若 \(m\) 为偶数，分子每一项都被 \(8\) 整除。若 \(m\) 为奇数，则 \(m^2 \equiv m^4 \equiv m^8 \equiv 1\) 且 \(4m \equiv 4 \pmod 8\)，所以分子为 \(1 + 1 + 2 + 4 = 8 \equiv 0 \pmod 8\)。两种情况下 \(P(m)\) 都是整数，因此 \(4\) 个值全是整数。正确答案是 E。

English

Put everything over \(8:\) \(P(m) = \dfrac{m^8 + m^4 + 2m^2 + 4m}{8}.\) If \(m\) is even, every term up top is divisible by \(8.\) If \(m\) is odd, then \(m^2 \equiv m^4 \equiv m^8 \equiv 1\) and \(4m \equiv 4 \pmod 8,\) so the numerator is \(1 + 1 + 2 + 4 = 8 \equiv 0 \pmod 8.\) Either way \(P(m)\) is an integer, so all \(4\) values are integers. Therefore, the answer is E.

25 答案 E

重新标记砖块的尺寸，使新块的边长为 \(7a,2b,2c\)。这三条边必定就是旧块的三条边长 \(3a,3b,3c\) 各增加 \(1\) 之后的结果。新边不可能与字母相同的旧边对应：\(7a=3a+1\) 没有正整数解，而 \(2b=3b+1\) 和 \(2c=3c+1\) 会给出负的边长。因此这个对应必定是两个三轮换之一。在其中一种排法下，\[ \begin{aligned} 7a&=3c+1,\\ 2b&=3a+1,\\ 2c&=3b+1 \end{aligned}\text{。}\]后两个方程给出 \(b=\frac{3a+1}{2}\) 和 \(c=\frac{9a+5}{4}\)，代入第一个方程得 \(28a=27a+19\)，所以 \((a,b,c)=(19,29,44)\)。另一个三轮换只是把 \(b\) 和 \(c\) 互换。这三个长度两两互质，且 \(a+b+c=19+29+44=92\)。因此正确答案是 E。

English

Relabel the brick dimensions so the new block has sides \(7a,2b,2c.\) These must be the three old side lengths \(3a,3b,3c,\) each increased by \(1.\) A new side cannot match the old side with the same letter: \(7a=3a+1\) has no positive integer solution, while \(2b=3b+1\) and \(2c=3c+1\) would give negative lengths. Therefore the matching must be one of the two three-cycles. In one orientation, \[ \begin{aligned} 7a&=3c+1,\\ 2b&=3a+1,\\ 2c&=3b+1. \end{aligned} \] The last two equations give \(b=\frac{3a+1}{2}\) and \(c=\frac{9a+5}{4}.\) Substituting into the first gives \(28a=27a+19,\) so \((a,b,c)=(19,29,44).\) The other cycle merely exchanges \(b\) and \(c.\) These lengths are pairwise relatively prime, and \(a+b+c=19+29+44=92.\) Thus, E is the correct answer.