2018B AMC 10 真题（P版）

1 答案 A

整盘面积为 \(20 \times 18 = 360\) 平方英寸。每小块面积为 \(2 \times 2 = 4\) 平方英寸。因此小块数为 \(\frac{360}{4} = 90\)。正确答案是 A。

English

The whole pan has area \(20 \times 18 = 360\) square inches. Each piece is \(2 \times 2 = 4\) square inches. So the number of pieces is \(\frac{360}{4} = 90.\) Thus, A is the correct answer.

2 答案 D

每段都是半小时。第一段行驶 \(60 \cdot \tfrac12 = 30\) 英里，第二段行驶 \(65 \cdot \tfrac12 = 32.5\) 英里。前两段共 \(62.5\) 英里，还剩 \(96 - 62.5 = 33.5\) 英里。最后半小时的速度为 \(\frac{33.5}{\tfrac12} = 67\) 英里每小时。因此正确答案是 D。

English

Each leg is half an hour. In the first, Sam drove \(60 \cdot \tfrac12 = 30\) miles; in the second, \(65 \cdot \tfrac12 = 32.5\) miles. That’s \(62.5\) miles so far. That leaves \(96 - 62.5 = 33.5\) miles for the last half hour, which is a speed of \(\frac{33.5}{\tfrac12} = 67\) mph. Therefore, the answer is D.

3 答案 B

乘积中数字的顺序以及两个乘积相加的顺序都不影响结果，所以只需考虑把四个数字分成两对。有三种分法：\(1\cdot2 + 3\cdot4 = 14\)、\(1\cdot3 + 2\cdot4 = 11\) 以及 \(1\cdot4 + 2\cdot3 = 10\)。因此可以得到 \(3\) 个不同的值。正确答案是 B。

English

Order inside a product doesn’t matter, and neither does the order we add the two products. So all that matters is how the four digits split into two pairs. There are three splits: \(1\cdot2 + 3\cdot4 = 14,\) \(1\cdot3 + 2\cdot4 = 11,\) and \(1\cdot4 + 2\cdot3 = 10.\) That’s \(3\) different values. Thus, B is the correct answer.

4 答案 B

三种不同的面面积是两两乘积 \(XY = 24\)、\(XZ = 48\)、\(YZ = 72\)，次序可以互换。三式相乘得 \((XYZ)^2 = 24 \cdot 48 \cdot 72 = 82944\)，所以 \(XYZ = 288\)。分别除以三个面面积，得到 \(Z = \frac{288}{24} = 12\)、\(Y = \frac{288}{48} = 6\) 和 \(X = \frac{288}{72} = 4\)，因此 \(X + Y + Z = 22\)。正确答案是 B。

English

The three distinct face areas are the pairwise products \(XY = 24,\) \(XZ = 48,\) \(YZ = 72\) in some order. Multiply all three: \((XYZ)^2 = 24 \cdot 48 \cdot 72 = 82944,\) so \(XYZ = 288.\) Now divide by each face area. We get \(Z = \frac{288}{24} = 12,\) \(Y = \frac{288}{48} = 6,\) and \(X = \frac{288}{72} = 4,\) so \(X + Y + Z = 22.\) Therefore, the answer is B.

5 答案 D

反过来计数。这个集合共有 \(2^8 = 256\) 个子集。不含任何质数的子集只能从非质数 \(\{4, 6, 8, 9\}\) 中选出，共 \(2^4 = 16\) 个。所以至少含一个质数的子集有 \(256 - 16 = 240\) 个。正确答案是 D。

English

Count the complement. The set has \(2^8 = 256\) subsets total. A subset avoids every prime exactly when it sticks to the non-primes \(\{4, 6, 8, 9\},\) and there are \(2^4 = 16\) of those. So \(256 - 16 = 240\) subsets contain at least one prime. Thus, D is the correct answer.

6 答案 D

恰好需要第三次抽取，当且仅当前两个筹码的和仍不超过 \(4\)。这样的无序对只有 \(\{1,2\}\) 和 \(\{1,3\}\)。每一对都可以按两种顺序抽出，所以有利的有序前两次抽取共有 \(4\) 种。

设想事先就把五个筹码的完整随机顺序定好。这样前两个筹码的 \(5\cdot4=20\) 种有序结果都是等可能的，即使实际过程会在抽出第一个筹码后就停止也不影响。因此概率为 \(\frac{4}{20}=\frac{1}{5}\)。正确答案是 D。

English

We need a third draw exactly when the first two chips still sum to \(4\) or less. The only such unordered pairs are \(\{1,2\}\) and \(\{1,3\}.\) Each can be drawn in either order, giving \(4\) favorable ordered prefixes.

Imagine that a complete random ordering of all five chips is chosen in advance. Then all \(5\cdot4=20\) ordered first-two-chip prefixes are equally likely, even when the actual process would stop after the first chip. Thus the probability is \(\frac{4}{20}=\frac{1}{5}.\) Therefore, the answer is D.

7 答案 D

设每个小半圆半径为 \(r\)。因为 \(N\) 个小直径覆盖大直径，所以大半圆半径为 \(Nr\)。小半圆面积之和为 \(A = N \cdot \tfrac12 \pi r^2\)，大半圆面积为 \(\tfrac12 \pi (Nr)^2\)，所以剩余区域面积为 \(B = \tfrac12 \pi r^2(N^2 - N)\)。因此 \(A : B = N : N(N-1)\) \(= 1 : (N-1)\)。令 \(N - 1 = 18\)，得 \(N = 19\)。正确答案是 D。

English

Let each small semicircle have radius \(r.\) The \(N\) diameters cover the big diameter, so the large radius is \(Nr.\) Then \(A = N \cdot \tfrac12 \pi r^2,\) and the large semicircle has area \(\tfrac12 \pi (Nr)^2,\) so the leftover region is \(B = \tfrac12 \pi r^2(N^2 - N).\) This gives \(A : B = N : N(N-1)\) \(= 1 : (N-1).\) Set \(N - 1 = 18,\) and \(N = 19.\) Thus, D is the correct answer.

8 答案 C

\(n\) 级楼梯中，竖直牙签数为 \((1 + 2 + \cdots + n) + n\) \(= \tfrac{n(n+1)}{2} + n\)，水平牙签数与其相同，因此总数为 \(n(n+1) + 2n = n(n+3)\)。检验可知 \(n = 3\) 时有 \(18\) 根牙签。令 \(n(n+3) = 180\)。因式分解得 \((n-12)(n+15) = 0\)，所以 \(n = 12\)。正确答案是 C。

English

In an \(n\)-step staircase the vertical toothpicks number \((1 + 2 + \cdots + n) + n\) \(= \tfrac{n(n+1)}{2} + n,\) and there are just as many horizontal ones. That’s a total of \(n(n+1) + 2n = n(n+3).\) Check: \(n = 3\) gives \(18,\) as it should. Now solve \(n(n+3) = 180.\) This factors as \((n-12)(n+15) = 0,\) so \(n = 12.\) Therefore, the answer is C.

9 答案 D

对每个骰子，把点数 \(k\) 替换为 \(7 - k\)。这会把所有掷骰结果一一配对，且概率不变；原总和为 \(s\) 时，新总和为 \(7 \cdot 7 - s = 49 - s\)。因此总和 \(s\) 与 \(49 - s\) 出现的概率相同。与 \(10\) 配对的是 \(49 - 10 = 39\)。正确答案是 D。

English

Replace each die’s value \(k\) by \(7 - k.\) This pairs up outcomes one-to-one and keeps their probabilities, and it sends a total of \(s\) to \(7 \cdot 7 - s = 49 - s.\) So the sums \(s\) and \(49 - s\) are equally likely. The partner of \(10\) is \(49 - 10 = 39.\) Thus, D is the correct answer.

10 答案 E

将 \(A\) 放在原点，并令三条棱沿坐标轴：\(A = (0,0,0)\)、\(B = (3,0,0)\)、\(C = (3,1,0)\)、\(E = (0,0,2)\)、\(H = (0,1,2)\)、\(F = (3,0,2)\)、\(G = (3,1,2)\)，所以 \(M = (3, \tfrac12, 2)\)。底面 \(BCHE\) 是矩形，且 \(BC = 1\)，\(BE = \sqrt{3^2 + 2^2} = \sqrt{13}\)，所以底面积为 \(\sqrt{13}\)。底面所在平面为 \(2x + 3z = 6\)，点 \(M\) 到该平面的距离为 \(\frac{|2\cdot3 + 3\cdot2 - 6|}{\sqrt{13}} = \frac{6}{\sqrt{13}}\)。体积为 \(\tfrac13 \cdot \sqrt{13} \cdot \frac{6}{\sqrt{13}} = 2\)。正确答案是 E。

English

Put \(A\) at the origin with edges along the axes: \(A = (0,0,0),\) \(B = (3,0,0),\) \(C = (3,1,0),\) \(E = (0,0,2),\) \(H = (0,1,2),\) \(F = (3,0,2),\) \(G = (3,1,2),\) so \(M = (3, \tfrac12, 2).\) The base \(BCHE\) is a rectangle with \(BC = 1\) and \(BE = \sqrt{3^2 + 2^2} = \sqrt{13},\) hence area \(\sqrt{13}.\) Its plane is \(2x + 3z = 6,\) and \(M\) sits at distance \(\frac{|2\cdot3 + 3\cdot2 - 6|}{\sqrt{13}} = \frac{6}{\sqrt{13}}\) from it. The volume is \(\tfrac13 \cdot \sqrt{13} \cdot \frac{6}{\sqrt{13}} = 2.\) Therefore, the answer is E.

11 答案 C

考虑 \(p^2 + 26\)。若 \(p = 3\)，则它等于 \(35 = 5 \cdot 7\)，不是质数。若 \(p\) 是其他质数，则它不被 \(3\) 整除，所以 \(p^2 \equiv 1 \pmod 3\)，并且 \(p^2 + 26 \equiv 1 + 2 \equiv 0 \pmod 3\)。无论哪种情况，它都是大于 \(3\) 的 \(3\) 的倍数，因此不是质数。正确答案是 C。

English

Look at \(p^2 + 26.\) When \(p = 3,\) it’s \(35 = 5 \cdot 7.\) For any other prime, \(p\) isn’t divisible by \(3,\) so \(p^2 \equiv 1 \pmod 3\) and \(p^2 + 26 \equiv 1 + 2 \equiv 0 \pmod 3.\) Either way it’s a multiple of \(3\) bigger than \(3,\) hence composite. So it’s never prime. Thus, C is the correct answer.

12 答案 C

把圆心 \(O\) 放在原点，则可取 \(A = (-12, 0)\)、\(B = (12, 0)\)，而 \(C\) 在半径为 \(12\) 的圆上运动。因为 \(A + B = 0\)，重心为 \(\tfrac13(A + B + C) = \tfrac13 C\)。当 \(C\) 绕圆运动时，\(\tfrac13 C\) 描出半径为 \(\tfrac{12}{3} = 4\) 的圆，少掉 \(C = A\) 或 \(B\) 时的两个点不影响面积。面积为 \(\pi \cdot 4^2 = 16\pi \approx 50\)。正确答案是 C。

English

Put the center \(O\) at the origin, so \(A = (-12, 0)\) and \(B = (12, 0),\) while \(C\) runs over the circle of radius \(12.\) Then \(A + B = 0,\) so the centroid is \(\tfrac13(A + B + C) = \tfrac13 C.\) As \(C\) circles, \(\tfrac13 C\) traces a circle of radius \(\tfrac{12}{3} = 4\) (minus the two points where \(C = A\) or \(B\)). Its area is \(\pi \cdot 4^2 = 16\pi \approx 50.\) Therefore, the answer is C.

13 答案 C

第 \(k\) 项为 \(10^{k+1} + 1\)，它能被 \(101\) 整除，当且仅当 \(10^{k+1} \equiv -1 \pmod{101}\)。因为 \(10^2 = 100 \equiv -1 \pmod{101}\)，所以 \(10^m \equiv -1\) 当且仅当 \(m \equiv 2 \pmod 4\)。因此需要 \(k + 1 \equiv 2\)，即 \(k \equiv 1 \pmod 4\)。在 \(k = 1, 2, \ldots, 2018\) 中，符合条件的是 \(1, 5, \ldots, 2017\)，共有 \(505\) 个。正确答案是 C。

English

The \(k\)-th term is \(10^{k+1} + 1,\) which \(101\) divides iff \(10^{k+1} \equiv -1 \pmod{101}.\) Notice \(10^2 = 100 \equiv -1 \pmod{101}.\) So \(10^m \equiv -1\) exactly when \(m \equiv 2 \pmod 4,\) meaning \(k + 1 \equiv 2,\) that is \(k \equiv 1 \pmod 4.\) Among \(k = 1, 2, \ldots, 2018,\) the values \(1, 5, \ldots, 2017\) number \(505.\) Thus, C is the correct answer.

14 答案 D

众数出现 \(10\) 次。为了让不同值个数尽量少，其余每个值都应尽量多出现，但最多只能出现 \(9\) 次，否则会与众数并列。若共有 \(d\) 个不同值，最多可有 \(10 + 9(d-1)\) 项。需要 \(10 + 9(d-1) \ge 2018\)，因此 \(d - 1 \ge 223.1\)，也就是 \(d \ge 225\)。正确答案是 D。

English

The mode shows up \(10\) times. To keep the number of distinct values small, let every other value repeat as much as the rules allow, which is \(9\) times each (any more would tie the mode). With \(d\) distinct values the list holds at most \(10 + 9(d-1)\) entries. We need \(10 + 9(d-1) \ge 2018,\) so \(d - 1 \ge 223.1,\) giving \(d \ge 225.\) Therefore, the answer is D.

15 答案 A

设包装纸边长为 \(s\)，盒子底面边长为 \(w\)。包装纸中心到一个角的距离为 \(\tfrac{s}{\sqrt2}\)。盒子底面相对于包装纸旋转了 \(45^\circ\)，所以从底面中心到一条边的距离为 \(\tfrac{w}{2}\)。一个角折到盒子顶部中心时，依次经过 \(\tfrac{w}{2}\)、高度 \(h\) 和另一个 \(\tfrac{w}{2}\)，因此 \(\tfrac{s}{\sqrt2} = \tfrac{w}{2} + h + \tfrac{w}{2} = w + h\)。所以 \(s = \sqrt2\,(w + h)\)，包装纸面积为 \(s^2 = 2(w + h)^2\)。正确答案是 A。

English

Let the sheet have side \(s.\) The base sits as a square of side \(w\) turned \(45^\circ,\) so the center is \(\tfrac{w}{2}\) from each base edge. A corner of the sheet lies \(\tfrac{s}{\sqrt2}\) from the center. Folding that corner up to the top center traces a straight line: \(\tfrac{w}{2}\) out to the base edge, then \(h\) up the side, then \(\tfrac{w}{2}\) across the top. So \(\tfrac{s}{\sqrt2} = \tfrac{w}{2} + h + \tfrac{w}{2} = w + h.\) Then \(s = \sqrt2\,(w + h),\) and the area is \(s^2 = 2(w + h)^2.\) Thus, A is the correct answer.

16 答案 E

对任意整数 \(n\)，\(n^3 - n = (n-1)n(n+1)\) 是三个连续整数的乘积，所以能被 \(6\) 整除。因此 \(n^3 \equiv n \pmod 6\)。求和得到 \(\sum a_i^3 \equiv \sum a_i\) \(= 2018^{2018} \pmod 6\)。因为 \(2018 \equiv 2 \pmod 6\)，而 \(2\) 的幂模 \(6\) 的余数按 \(2, 4, 2, 4, \ldots\) 交替，指数 \(2018\) 为偶数，所以 \(2^{2018} \equiv 4 \pmod 6\)，余数为 \(4\)。正确答案是 E。

English

For any integer \(n,\) \(n^3 - n = (n-1)n(n+1)\) is a product of three consecutive integers, so it’s divisible by \(6.\) That means \(n^3 \equiv n \pmod 6.\) Summing, \(\sum a_i^3 \equiv \sum a_i\) \(= 2018^{2018} \pmod 6.\) Now \(2018 \equiv 2 \pmod 6,\) and powers of \(2\) mod \(6\) alternate \(2, 4, 2, 4, \ldots.\) The exponent \(2018\) is even, so \(2^{2018} \equiv 4 \pmod 6.\) The remainder is \(4.\) Therefore, the answer is E.

17 答案 B

设八边形的边长为 \(s\)，并设 \(x=AP=BQ=\frac{8-s}{2}\)。直角三角形 \(\triangle APH\) 和 \(\triangle BQC\) 的斜边同为 \(s\)，且都有一条长为 \(x\) 的直角边，所以它们全等；记 \(PH=QC=y\)。因为 \(CD=HG=s\)，且长方形两条竖边的长度都为 \(6\)，所以 \(DR=GS\)。于是 \(R\) 和 \(S\) 处的直角三角形全等，从而 \(RE=SF\)。因为 \(RS=8\)，且 \(EF=s\)，这两个相等长度都为 \(\frac{8-s}{2}=x\)。所以四个被切去的角的两条直角边都是 \(x\) 和 \(y\)。

八边形各边相等，给出 \[8-2x=6-2y=\sqrt{x^2+y^2}\text{。}\]第一个等式给出 \(y=x-1\)。代入并平方，得到 \(2x^2-30x+63=0\)，所以满足 \(x<4\) 的根为 \(x=\frac{15-3\sqrt{11}}{2}\)。

边长为 \(8-2x=-7+3\sqrt{11}\)，所以 \(k+m+n=-7+3+11=7\)。因此正确答案是 B。

English

Let \(s\) be the octagon’s side length and let \(x=AP=BQ=\frac{8-s}{2}.\) The right triangles \(\triangle APH\) and \(\triangle BQC\) have the same hypotenuse \(s\) and a leg of length \(x,\) so they are congruent; write \(PH=QC=y.\) Because \(CD=HG=s\) and the vertical sides of the rectangle both have length \(6,\) it follows that \(DR=GS.\) The right triangles at \(R\) and \(S\) are then congruent, so \(RE=SF.\) Since \(RS=8\) and \(EF=s,\) each of these equal lengths is \(\frac{8-s}{2}=x.\) Thus all four cut corners have legs \(x\) and \(y.\)

The equal octagon sides give \[8-2x=6-2y=\sqrt{x^2+y^2}.\] The first equality gives \(y=x-1.\) Substituting and squaring gives \(2x^2-30x+63=0,\) so the root with \(x<4\) is \(x=\frac{15-3\sqrt{11}}{2}.\)

The side length is \(8-2x=-7+3\sqrt{11},\) so \(k+m+n=-7+3+11=7.\) Thus, B is the correct answer.

18 答案 D

若某个家庭的两个孩子坐在同一排，他们只能坐在该排的第 \(1\) 和第 \(3\) 个座位，这会迫使另一个家庭的两个孩子坐在同一列，违反条件。因此每排必须恰好有每个家庭的一个孩子。第二排安排三个家庭有 \(3! = 6\) 种。第三排必须在每一列都与第二排家庭不同，是三个元素的错排，共 \(2\) 种。最后每对兄妹可互换座位，共 \(2^3 = 8\) 种。因此总数为 \(6 \cdot 2 \cdot 8 = 96\)。正确答案是 D。

English

Suppose some family put both children in one row. They’d have to take the non-adjacent seats \(1\) and \(3,\) which forces the middle family’s two children into the same column. Not allowed. So each row holds exactly one child from each family. The second row is a permutation of the three families, \(3! = 6\) ways. The third row needs a different family in every column, a derangement of the second row’s order, and there are \(2\) of those. Finally, each pair can swap its two children between their seats, \(2^3 = 8\) ways. The total is \(6 \cdot 2 \cdot 8 = 96.\) Therefore, the answer is D.

19 答案 E

设 Chloe 今天 \(n\) 岁，Zoe 今天 \(1\) 岁。\(t\) 年后，她们的年龄之比为 \[\dfrac{n+t}{1+t}=1+\dfrac{n-1}{1+t}\text{，}\] 它是整数当且仅当 \(1+t\) 整除 \(n-1\)。因此 \(n-1\) 恰好有 \(9\) 个正因子。

有 \(9\) 个正因子的数形如 \(p^8\) 或 \(p^2q^2\)。其中唯一的两位数是 \(2^2\cdot3^2=36\)，所以 Chloe 今天 \(37\) 岁，Joey 今天 \(38\) 岁。

现在 Joey 的年龄 \(38+t\) 是 \(1+t\)（即 Zoe 年龄）的倍数，当且仅当 \(1+t\) 整除 \(37\)。所以下一次发生在 \(t=36\)，此时 Joey \(74\) 岁，数字和为 \(7+4=11\)。正确答案是 E。

English

Let Chloe be \(n\) today; Zoe is \(1.\) In \(t\) years their age ratio is \[\dfrac{n+t}{1+t}=1+\dfrac{n-1}{1+t},\] which is an integer exactly when \(1+t\) divides \(n-1.\) Thus \(n-1\) has exactly \(9\) positive divisors.

A number with \(9\) divisors has the form \(p^8\) or \(p^2q^2.\) The only two-digit possibility is \(2^2\cdot3^2=36,\) so Chloe is \(37\) and Joey is \(38.\)

Joey’s age \(38+t\) is a multiple of Zoe’s age \(1+t\) exactly when \(1+t\) divides \(37.\) The next time is \(t=36,\) when Joey is \(74.\) Its digit sum is \(7+4=11.\) Thus, E is the correct answer.

20 答案 B

注意 \(f(n) = n + 1\) 本身就满足这个递推式，因此令 \(f(n) = (n + 1) + g(n)\)，则 \(g\) 满足对应的齐次递推 \(g(n) = g(n-1) - g(n-2)\)。由初值得 \(g(1) = -1\) 且 \(g(2) = -2\)，于是序列以 \(6\) 为周期循环：\(-1, -2, -1, 1, 2, 1, \ldots\)。因为 \(2018 \equiv 2 \pmod 6\)，所以 \(g(2018) = -2\)，于是 \(f(2018) = 2019 - 2 = 2017\)。正确答案是 B。

English

Notice \(f(n) = n + 1\) solves the recurrence on its own, so write \(f(n) = (n + 1) + g(n).\) Then \(g\) satisfies the homogeneous version \(g(n) = g(n-1) - g(n-2).\) With \(g(1) = -1\) and \(g(2) = -2,\) it cycles with period \(6\): \(-1, -2, -1, 1, 2, 1, \ldots.\) Since \(2018 \equiv 2 \pmod 6,\) we get \(g(2018) = -2,\) so \(f(2018) = 2019 - 2 = 2017.\) Therefore, the answer is B.

21 答案 C

设 \(d\) 是 \(323\) 之后的下一个因数。若 \(\gcd(d,323)=1\)，则 \(n\) 是 \(323d>323^2>9999\) 的倍数，这对四位数来说不可能。因此 \(\gcd(d,323)>1\)。

因为 \(323=17\cdot19\)，这个最大公因数至少为 \(17\)。它也整除 \(d-323\)，所以 \(d-323\ge17\)，从而 \(d\ge340\)。

这个界可以达到：当 \(n=6460=2^2\cdot5\cdot17\cdot19\) 时，\(323\) 和 \(340\) 都是因数。下界说明二者之间不存在因数。因此最小的下一个因数是 \(340\)，正确答案是 C。

English

Let \(d\) be the next divisor after \(323.\) If \(\gcd(d,323)=1,\) then \(n\) is a multiple of \(323d>323^2>9999,\) impossible for a four-digit number. Thus \(\gcd(d,323)>1.\)

Since \(323=17\cdot19,\) this gcd is at least \(17.\) It also divides \(d-323,\) so \(d-323\ge17\) and hence \(d\ge340.\)

This bound is attained: for \(n=6460=2^2\cdot5\cdot17\cdot19,\) both \(323\) and \(340\) are divisors. The lower bound shows there is no divisor between them. Thus the smallest possible next divisor is \(340,\) and C is the correct answer.

22 答案 C

三边 \(x, y, 1\) 构成三角形，当且仅当 \(x + y > 1\)。又因为 \(1\) 是最长边，该三角形为钝角，当且仅当 \(x^2 + y^2 < 1\)。因此在单位正方形中，所求区域位于四分之一圆 \(x^2 + y^2 = 1\) 内、直线 \(x + y = 1\) 上方，也就是四分之一圆盘去掉弦下方的直角三角形，面积为 \(\tfrac{\pi}{4} - \tfrac12 \approx 0.285\)。最接近的选项是 \(0.29\)。正确答案是 C。

English

The three lengths \(x, y, 1\) make a triangle iff \(x + y > 1.\) Since \(1\) is the longest side, that triangle is obtuse iff \(x^2 + y^2 < 1.\) So in the unit square we want the region inside the quarter circle \(x^2 + y^2 = 1\) but above the line \(x + y = 1.\) That’s the quarter disk with the right triangle under the chord removed: \(\tfrac{\pi}{4} - \tfrac12 \approx 0.285.\) The closest choice is \(0.29.\) Therefore, the answer is C.

23 答案 B

回忆 \(ab=\gcd(a,b)\operatorname{lcm}(a,b)\)。令 \(x=\operatorname{lcm}(a,b)\)，\(y=\gcd(a,b)\)，则原方程变为 \[(x-12)(y-20)=177=3\cdot59\text{。}\]

正因数分解给出 \((x,y)=(13,197)\)、\((189,21)\)、\((15,79)\) 和 \((71,23)\)。负因数分解会使 \(x\) 或 \(y\) 为负数，因此不可能。此外 \(y\) 必须整除 \(x\)，只有 \((x,y)=(189,21)\) 满足这一点。

写成 \(a=21u\) 和 \(b=21v\)。于是 \(\gcd(u,v)=1\) 且 \(uv=\frac{189}{21}=9\)，所以 \((u,v)=(1,9)\) 或 \((9,1)\)。因此符合条件的有序对是 \((21,189)\) 和 \((189,21)\)，正确答案是 B。

English

Recall \(ab=\gcd(a,b)\operatorname{lcm}(a,b).\) Let \(x=\operatorname{lcm}(a,b)\) and \(y=\gcd(a,b).\) The equation becomes \[(x-12)(y-20)=177=3\cdot59.\]

The positive factor pairs give \((x,y)=(13,197),\) \((189,21),\) \((15,79),\) and \((71,23).\) The negative factor pairs make either \(x\) or \(y\) negative, so they are impossible. Also \(y\) must divide \(x,\) and only \((x,y)=(189,21)\) passes.

Write \(a=21u\) and \(b=21v.\) Then \(\gcd(u,v)=1\) and \(uv=\frac{189}{21}=9,\) so \((u,v)=(1,9)\) or \((9,1).\) Hence the two ordered pairs are \((21,189)\) and \((189,21),\) and B is the correct answer.

24 答案 C

三角形 \(XYZ\) 是边长为 \(\frac{3}{2}\) 的等边三角形，所以面积为 \[\dfrac{\sqrt3}{4}\left(\dfrac32\right)^2=\dfrac{9\sqrt3}{16}\text{。}\]

三角形 \(ACE\) 和 \(XYZ\) 同心，且彼此旋转 \(30^\circ\)。在 \(XYZ\) 的每个顶点处，\(ACE\) 的两边截出一个 \(30\)-\(60\)-\(90\) 三角形，其斜边是半边长线段 \(AX=\frac{1}{2}\)。两条直角边分别为 \(\frac{1}{4}\) 和 \(\frac{\sqrt3}{4}\)，所以每个角的面积为 \(\frac{\sqrt3}{32}\)。

去掉三个角后，面积为 \[\dfrac{9\sqrt3}{16}-3\cdot\dfrac{\sqrt3}{32}=\dfrac{15\sqrt3}{32}\text{。}\]因此答案是 C。

English

The triangle \(XYZ\) is equilateral with side \(\frac{3}{2},\) so its area is \[\dfrac{\sqrt3}{4}\left(\dfrac32\right)^2=\dfrac{9\sqrt3}{16}.\]

The triangles \(ACE\) and \(XYZ\) are concentric and rotated \(30^\circ\) from each other. At each vertex of \(XYZ,\) the sides of \(ACE\) cut off a \(30\)-\(60\)-\(90\) triangle whose hypotenuse is the half-side segment \(AX=\frac{1}{2}.\) Its legs are \(\frac{1}{4}\) and \(\frac{\sqrt3}{4},\) so each corner has area \(\frac{\sqrt3}{32}.\)

Removing the three corners gives \[\dfrac{9\sqrt3}{16}-3\cdot\dfrac{\sqrt3}{32}=\dfrac{15\sqrt3}{32}.\] Therefore, the answer is C.

25 答案 C

设 \(a = \lfloor x \rfloor\)。方程可写成 \(x^2 = 10{,}000(x - a)\) \(= 10{,}000\{x\}\)。因为 \(0 \le \{x\} < 1\)，所以 \(0 \le x^2 < 10{,}000\)，即 \(-100 < x < 100\)。在每个区间 \([a, a + 1)\) 上，函数 \(10{,}000x - x^2\) 从 \(10{,}000a - a^2\) 单调增加并趋近于但不取到 \(10{,}000(a+1) - (a+1)^2\)。它恰好在 \((a + 1)^2 < 10{,}000\) 时取到 \(10{,}000a\)。满足条件的整数为 \(-100 \le a \le 98\)，共 \(199\) 个。正确答案是 C。

English

Let \(a = \lfloor x \rfloor.\) The equation reads \(x^2 = 10{,}000(x - a)\) \(= 10{,}000\{x\},\) and since \(0 \le \{x\} < 1,\) this forces \(0 \le x^2 < 10{,}000,\) so \(-100 < x < 100.\) On each interval \([a, a + 1)\) the quantity \(10{,}000x - x^2\) increases from \(10{,}000a - a^2\) and approaches, but does not reach, \(10{,}000(a+1) - (a+1)^2.\) It hits \(10{,}000a\) exactly once precisely when \((a + 1)^2 < 10{,}000.\) That holds for the integers \(-100 \le a \le 98,\) which is \(199\) solutions. Thus, C is the correct answer.