2005B AMC 10 真题（P版）

1 答案 A

小队购买 \(1000 \div 5 = 200\) 组五根糖果棒，花费 \(200 \cdot 2 = 400\) 美元。

他们卖出 \(1000 \div 2 = 500\) 组两根糖果棒，收入 \(500 \cdot 1 = 500\) 美元。

利润为 \(\$500 - \$400 = \$100\)。

所以正确答案是 A。

English

The troop buys \(1000 \div 5 = 200\) groups of five bars, costing \(200 \cdot 2 = 400\) dollars.

They sell \(1000 \div 2 = 500\) pairs of bars, earning \(500 \cdot 1 = 500\) dollars.

The profit is \(\$500 - \$400 = \$100.\)

Thus, A is the correct answer.

2 答案 D

题意可写为 \[ \dfrac{x}{100}\cdot x = 4\text{，}\]所以 \(x^2 = 400\)。

因为 \(x\) 为正数，所以 \(x = 20\)。

所以正确答案是 D。

English

The statement translates to \[ \dfrac{x}{100}\cdot x = 4, \] so \(x^2 = 400.\)

Since \(x\) is positive, \(x = 20.\)

Thus, D is the correct answer.

3 答案 D

第一天之后，剩下 \(1 - \dfrac13 = \dfrac23\) 的油漆。

第二天用掉原来总量的 \(\dfrac13 \cdot \dfrac23 = \dfrac29\)。

第三天可用的部分为 \[ \dfrac23 - \dfrac29 = \dfrac49\text{。}\]

所以正确答案是 D。

English

After the first day, \(1 - \dfrac13 = \dfrac23\) of the paint remains.

On the second day, \(\dfrac13 \cdot \dfrac23 = \dfrac29\) of the original amount is used.

The fraction available on the third day is \[ \dfrac23 - \dfrac29 = \dfrac49. \]

Thus, D is the correct answer.

4 答案 D

第一个内层表达式为 \[ 5 \diamond 12 = \sqrt{5^2 + 12^2} = 13\text{，}\]而同理 \((-12) \diamond (-5)\) \(= \sqrt{144 + 25} = 13\)。

因此 \[ 13 \diamond 13 = \sqrt{13^2 + 13^2} = 13\sqrt{2}\text{。}\]

所以正确答案是 D。

English

Each inner expression evaluates to \[ 5 \diamond 12 = \sqrt{5^2 + 12^2} = 13 \] and similarly \((-12) \diamond (-5)\) \(= \sqrt{144 + 25} = 13.\)

Then \[ 13 \diamond 13 = \sqrt{13^2 + 13^2} = 13\sqrt{2}. \]

Thus, D is the correct answer.

5 答案 C

买完所有 CD 的花费是买三分之一 CD 的三倍，即她的钱的 \(3 \cdot \dfrac15 = \dfrac35\)。

剩下的部分为 \(1 - \dfrac35 = \dfrac25\)。

所以正确答案是 C。

English

Buying all the CDs costs three times as much as buying one third of them, namely \(3 \cdot \dfrac15 = \dfrac35\) of her money.

The fraction left over is \(1 - \dfrac35 = \dfrac25.\)

Thus, C is the correct answer.

6 答案 B

Lisa 至少需要 \(0.8 \cdot 50 = 40\) 次小测拿到 A。

她已经有 \(22\) 个 A，所以剩下 \(20\) 次中还需要 \(40 - 22 = 18\) 个 A。

因此最多有 \(20 - 18 = 2\) 次小测低于 A。

所以正确答案是 B。

English

Lisa needs an A on at least \(0.8 \cdot 50 = 40\) quizzes.

She already has \(22,\) so she needs \(40 - 22 = 18\) A’s among the remaining \(20\) quizzes.

That leaves at most \(20 - 18 = 2\) quizzes with a grade lower than an A.

Thus, B is the correct answer.

7 答案 B

设较小圆半径为 \(r\)，则它的面积为 \(\pi r^2\)。

外切于这个小圆的较小正方形边长为 \(2r\)，其对角线 \(2\sqrt2\,r\) 是较大圆的直径。所以较大圆半径为 \(\sqrt2\,r\)。

外切于较大圆的较大正方形边长为 \(2\sqrt2\,r\)，面积为 \(8r^2\)。

所求比为 \[ \dfrac{\pi r^2}{8r^2} = \dfrac{\pi}{8}\text{。}\]

所以正确答案是 B。

English

Let the smaller circle have radius \(r,\) so its area is \(\pi r^2.\)

The smaller square, which circumscribes this circle, has side \(2r,\) and its diagonal \(2\sqrt2\,r\) is the diameter of the larger circle. So the larger circle has radius \(\sqrt2\,r.\)

The larger square circumscribes the larger circle, so it has side \(2\sqrt2\,r\) and area \(8r^2.\)

The desired ratio is \[ \dfrac{\pi r^2}{8r^2} = \dfrac{\pi}{8}. \]

Thus, B is the correct answer.

8 答案 A

一块瓷砖上的四个四分之一圆合起来是一个半径为 \(\dfrac12\) 的整圆，面积为 \(\pi\left(\dfrac12\right)^2 = \dfrac{\pi}{4}\)。

所以每块瓷砖的阴影面积为 \(1 - \dfrac{\pi}{4}\) 平方英尺。

地板共有 \(8 \cdot 10 = 80\) 块瓷砖，因此总阴影面积为 \[ 80\left(1 - \dfrac{\pi}{4}\right) = 80 - 20\pi\text{。}\]

所以正确答案是 A。

English

The four quarter circles on a tile together make one full circle of radius \(\dfrac12,\) with area \(\pi\left(\dfrac12\right)^2 = \dfrac{\pi}{4}.\)

So each tile has shaded area \(1 - \dfrac{\pi}{4}\) square feet.

There are \(8 \cdot 10 = 80\) tiles, so the total shaded area is \[ 80\left(1 - \dfrac{\pi}{4}\right) = 80 - 20\pi. \]

Thus, A is the correct answer.

9 答案 D

第一个骰子为奇数（\(1\) 或 \(3\)）的概率为 \(\dfrac23\)，为偶数的概率为 \(\dfrac13\)。第二个骰子为奇数（\(5\)）的概率为 \(\dfrac13\)，为偶数的概率为 \(\dfrac23\)。

两个数奇偶性不同时和为奇数，所以概率为 \[ \dfrac13 \cdot \dfrac13 + \dfrac23 \cdot \dfrac23 = \dfrac19 + \dfrac49 = \dfrac59\text{。}\]

所以正确答案是 D。

English

The first die is odd (a \(1\) or \(3\)) with probability \(\dfrac23\) and even with probability \(\dfrac13.\) The second die is odd (a \(5\)) with probability \(\dfrac13\) and even with probability \(\dfrac23.\)

The sum is odd when the two parities differ: \[ \dfrac13 \cdot \dfrac13 + \dfrac23 \cdot \dfrac23 = \dfrac19 + \dfrac49 = \dfrac59. \]

Thus, D is the correct answer.

10 答案 A

设 \(H\) 为从 \(C\) 到直线 \(AB\) 的垂足。因为 \(AC = BC\)，\(H\) 是 \(AB\) 的中点，所以 \(BH = 1\)，且 \(CH^2 = 7^2 - 1^2 = 48\)。

在 \(\triangle CHD\) 中应用勾股定理，其中 \(HD = BH + BD = 1 + BD\)，得到 \[ 8^2 = 48 + (1 + BD)^2\text{，}\]所以 \((1 + BD)^2 = 16\)。

因此 \(1 + BD = 4\)，所以 \(BD = 3\)。

所以正确答案是 A。

English

Let \(H\) be the foot of the altitude from \(C\) to line \(AB.\) Since \(AC = BC,\) \(H\) is the midpoint of \(AB,\) so \(BH = 1\) and \(CH^2 = 7^2 - 1^2 = 48.\)

Applying the Pythagorean theorem in \(\triangle CHD,\) where \(HD = BH + BD = 1 + BD,\) gives \[ 8^2 = 48 + (1 + BD)^2, \] so \((1 + BD)^2 = 16.\)

Then \(1 + BD = 4,\) so \(BD = 3.\)

Thus, A is the correct answer.

11 答案 E

数列开头是 \(2005, 133, 55, 250, 133, \ldots\)，所以从第一项以后，它重复循环 \(133, 55, 250\)，其长度为 \(3\)。

第 \(2\)、\(3\) 和 \(4\) 项分别是这个循环的第一、第二和第三项。因为 \(2005 - 2 = 2003\) 除以 \(3\) 的余数是 \(2\)，所以第 \(2005\) 项对应循环的第三项，即 \(250\)。

所以正确答案是 E。

English

The sequence begins \(2005, 133, 55, 250, 133, \ldots,\) so after the first term it repeats the cycle \(133, 55, 250\) of length \(3.\)

Terms \(2,\) \(3,\) and \(4\) are the first, second, and third entries of this cycle. Because \(2005 - 2 = 2003\) leaves remainder \(2\) upon division by \(3,\) the \(2005\)th term matches the third entry, \(250.\)

Thus, E is the correct answer.

12 答案 E

乘积为质数当且仅当一个骰子显示质数（\(2\)、\(3\) 或 \(5\)），其余十一个骰子都显示 \(1\)。

指定某一个骰子为质数骰时，它显示质数的概率为 \(\dfrac36 = \dfrac12\)，其余每个骰子显示 \(1\) 的概率为 \(\dfrac16\)。考虑十二个骰子中哪一个显示质数，所求概率为 \[ \begin{aligned} 12 \cdot \dfrac12 \cdot \left(\dfrac16\right)^{11} &= 6 \cdot \left(\dfrac16\right)^{11} \\ &= \left(\dfrac16\right)^{10} \end{aligned}\text{。}\]

所以正确答案是 E。

English

The product is prime exactly when one die shows a prime (\(2,\) \(3,\) or \(5\)) and the other eleven all show \(1.\)

The probability that any single die is the prime one is \(\dfrac36 = \dfrac12,\) and each of the other eleven shows \(1\) with probability \(\dfrac16.\) Accounting for which of the twelve dice is prime, the probability is \[ \begin{aligned} 12 \cdot \dfrac12 \cdot \left(\dfrac16\right)^{11} &= 6 \cdot \left(\dfrac16\right)^{11} \\ &= \left(\dfrac16\right)^{10}. \end{aligned} \]

Thus, E is the correct answer.

13 答案 C

在 \(1\) 到 \(2005\) 之间，\(3\) 的倍数有 \(668\) 个，\(4\) 的倍数有 \(501\) 个，\(12\) 的倍数有 \(167\) 个。

每个 \(12\) 的倍数同时是 \(3\) 和 \(4\) 的倍数，所以分别从两个集合中去掉这些数，得到 \[ \begin{aligned} &(668 - 167) \\ &\quad {}+ (501 - 167) = 835 \end{aligned} \] 个是 \(3\) 或 \(4\) 的倍数但不是 \(12\) 的倍数的数。

所以正确答案是 C。

English

Between \(1\) and \(2005\) there are \(668\) multiples of \(3,\) \(501\) multiples of \(4,\) and \(167\) multiples of \(12.\)

Every multiple of \(12\) is both a multiple of \(3\) and of \(4,\) so removing them from each group gives \[ \begin{aligned} &(668 - 167) \\ &\quad {}+ (501 - 167) = 835 \end{aligned} \] numbers that are multiples of \(3\) or \(4\) but not \(12.\)

Thus, C is the correct answer.

14 答案 C

以 \(\overline{CD}\) 为底。因为 \(C\) 是 \(\overline{BD}\) 的中点且 \(BC = 2\)，所以 \(CD = 2\)。

\(\triangle CDM\) 的高是从 \(M\) 到直线 \(BD\) 的距离。由于 \(M\) 是 \(\overline{AC}\) 的中点，这个距离是 \(\triangle ABC\) 高的一半，即 \(\dfrac12 \cdot \sqrt3 = \dfrac{\sqrt3}{2}\)。

面积为 \[ \dfrac12 \cdot 2 \cdot \dfrac{\sqrt3}{2} = \dfrac{\sqrt3}{2}\text{。}\]

所以正确答案是 C。

English

Take \(\overline{CD}\) as the base. Since \(C\) is the midpoint of \(\overline{BD}\) and \(BC = 2,\) we have \(CD = 2.\)

The height of \(\triangle CDM\) is the distance from \(M\) to line \(BD.\) Because \(M\) is the midpoint of \(\overline{AC},\) this distance is half the height of \(\triangle ABC,\) which is \(\dfrac12 \cdot \sqrt3 = \dfrac{\sqrt3}{2}.\)

The area is \[ \dfrac12 \cdot 2 \cdot \dfrac{\sqrt3}{2} = \dfrac{\sqrt3}{2}. \]

Thus, C is the correct answer.

15 答案 D

共有 \(\binom82 = 28\) 对等可能的纸币组合。

金额至少为 \(\$20\) 的情况包括两张二十美元（\(1\) 种）、一张二十美元配六张较小纸币中的任意一张（\(2 \cdot 6 = 12\) 种），或两张十美元（\(1\) 种）。

概率为 \[ \dfrac{1 + 12 + 1}{28} = \dfrac{14}{28} = \dfrac12\text{。}\]

所以正确答案是 D。

English

There are \(\binom82 = 28\) equally likely pairs.

A sum of at least \(\$20\) comes from both twenties (\(1\) way), a twenty paired with any of the six smaller bills (\(2 \cdot 6 = 12\) ways), or both tens (\(1\) way).

The probability is \[ \dfrac{1 + 12 + 1}{28} = \dfrac{14}{28} = \dfrac12. \]

Thus, D is the correct answer.

16 答案 D

设 \(r_1\) 和 \(r_2\) 为 \(x^2 + px + m = 0\) 的根，则 \(m = r_1 r_2\)，且 \(p = -(r_1 + r_2)\)。

\(x^2 + mx + n = 0\) 的根为 \(2r_1\) 和 \(2r_2\)，所以 \(n = 4r_1 r_2\)，且 \(-m = 2(r_1 + r_2)\)。

因此 \(n = 4m\)，并且 \(m = -2(r_1 + r_2) = 2p\)，所以 \(p = \dfrac{m}{2}\)。于是 \[ \dfrac{n}{p} = \dfrac{4m}{\frac{m}{2}} = 8\text{。}\]

所以正确答案是 D。

English

Let \(r_1\) and \(r_2\) be the roots of \(x^2 + px + m = 0,\) so \(m = r_1 r_2\) and \(p = -(r_1 + r_2).\)

The roots of \(x^2 + mx + n = 0\) are \(2r_1\) and \(2r_2,\) so \(n = 4r_1 r_2\) and \(-m = 2(r_1 + r_2).\)

Then \(n = 4m\) and \(m = -2(r_1 + r_2) = 2p,\) so \(p = \dfrac{m}{2}.\) Therefore \[ \dfrac{n}{p} = \dfrac{4m}{\frac{m}{2}} = 8. \]

Thus, D is the correct answer.

17 答案 B

将方程串联起来：\[ \begin{aligned} 4^{abcd} &= \left(\left(\left(4^a\right)^b\right)^c\right)^d \\ &= \left(\left(5^b\right)^c\right)^d \\ &= \left(6^c\right)^d = 7^d = 8 \end{aligned}\text{。}\]

因为 \(8 = 4^{\frac{3}{2}}\)，所以 \(a \cdot b \cdot c \cdot d = \dfrac32\)。

所以正确答案是 B。

English

Chaining the equations, \[ \begin{aligned} 4^{abcd} &= \left(\left(\left(4^a\right)^b\right)^c\right)^d \\ &= \left(\left(5^b\right)^c\right)^d \\ &= \left(6^c\right)^d = 7^d = 8. \end{aligned} \]

Since \(8 = 4^{\frac{3}{2}},\) we conclude \(a \cdot b \cdot c \cdot d = \dfrac32.\)

Thus, B is the correct answer.

18 答案 D

这七个数字从集合 \(\{2, 3, 4, 5, 6, 7, 8, 9\}\) 中选择，且一旦选定就必须按递增顺序写出，所以只需要考虑选哪些数字。

从这八个数字中选择七个，等价于选择留下不用的那一个数字，共有 \(8\) 种。

所以正确答案是 D。

English

The seven digits are chosen from \(\{2, 3, 4, 5, 6, 7, 8, 9\},\) and once chosen they must be written in increasing order, so only the choice of digits matters.

Choosing seven of these eight digits is the same as choosing the one digit to leave out, which can be done in \(8\) ways.

Thus, D is the correct answer.

19 答案 B

得 \(95\) 分的百分比为 \(100 - 10 - 25 - 20 - 15 = 30\)。

平均分为 \[ \begin{aligned} &0.10(70) + 0.25(80) \\ &\quad {}+ 0.20(85) + 0.15(90) \\ &\quad {}+ 0.30(95) = 86 \end{aligned}\text{。}\]

因为 \(35\%\) 的学生低于 \(85\) 分，且 \(35\%\) 的学生高于 \(85\) 分，中间的学生得 \(85\) 分，所以中位数为 \(85\)。

差为 \(86 - 85 = 1\)。

所以正确答案是 B。

English

The percentage scoring \(95\) is \(100 - 10 - 25 - 20 - 15 = 30.\)

The mean is \[ \begin{aligned} &0.10(70) + 0.25(80) \\ &\quad {}+ 0.20(85) + 0.15(90) \\ &\quad {}+ 0.30(95) = 86. \end{aligned} \]

Since \(35\%\) scored below \(85\) and \(35\%\) scored above \(85,\) the middle student scored \(85,\) so the median is \(85.\)

The difference is \(86 - 85 = 1.\)

Thus, B is the correct answer.

20 答案 C

由对称性，五个数字在每个数位上出现的次数相同，所以每个数位上的平均数字为 \[ \dfrac{1 + 3 + 5 + 7 + 8}{5} = 4.8\text{。}\]

因此平均数为 \[ \begin{gathered} \small 4.8(1 + 10 + 100 + 1000 + 10000) \\ = 4.8 \cdot 11111 = 53332.8 \end{gathered}\text{。}\]

所以正确答案是 C。

English

By symmetry, each of the five digits appears equally often in each place, so the average digit in every place is \[ \dfrac{1 + 3 + 5 + 7 + 8}{5} = 4.8. \]

The average number is therefore \[ \begin{gathered} \small 4.8(1 + 10 + 100 + 1000 + 10000) \\ = 4.8 \cdot 11111 = 53332.8. \end{gathered} \]

Thus, C is the correct answer.

21 答案 A

两个事件都从 \(\binom{40}{4}\) 个等可能选择中抽取，所以 \(\dfrac{q}{p}\) 等于它们有利情况数量之比。

四张纸条都写同一个数字的抽法有 \(10\) 种，每个数字一种。

若是两个 \(a\) 和两个 \(b\)，先选两个数字，有 \(\binom{10}{2}\) 种，再从四张 \(a\) 纸条中选两张、从四张 \(b\) 纸条中选两张：\[ \begin{aligned} \binom{10}{2}\binom{4}{2}\binom{4}{2} &= 45 \cdot 6 \cdot 6 \\ &= 1620 \end{aligned}\text{。}\]

因此 \(\dfrac{q}{p} = \dfrac{1620}{10} = 162\)。

所以正确答案是 A。

English

Both events draw from \(\binom{40}{4}\) equally likely selections, so \(\dfrac{q}{p}\) is the ratio of their favorable counts.

Exactly \(10\) draws give four slips of the same number, one for each value.

For two \(a\)’s and two \(b\)’s, choose the two values in \(\binom{10}{2}\) ways, then two of the four \(a\)-slips and two of the four \(b\)-slips: \[ \begin{aligned} \binom{10}{2}\binom{4}{2}\binom{4}{2} &= 45 \cdot 6 \cdot 6 \\ &= 1620. \end{aligned} \]

Therefore \(\dfrac{q}{p} = \dfrac{1620}{10} = 162.\)

Thus, A is the correct answer.

22 答案 C

由 \(1 + 2 + \cdots + n = \dfrac{n(n+1)}{2}\) 可知，整除条件等价于 \[ \dfrac{n!}{\frac{n(n+1)}{2}} = \dfrac{2(n-1)!}{n+1} \] 是整数。

令 \(N=n+1\)。若 \(N\) 是合数但不是完全平方数，它有两个不同的真因数，其乘积为 \(N\)；而这两个因数都出现在 \((N-2)!=(n-1)!\) 中。若 \(N=k^2\) 且 \(k\ge3\)，因数 \(k\) 和 \(2k\) 都出现在该阶乘中，所以阶乘含有 \(2N\) 的倍数。剩下的合数情形 \(N=4\) 也能整除 \(2(N-2)!=4\)。因此当 \(N\) 是合数时，分式为整数。若 \(N=n+1\) 是奇质数，它既不整除 \((n-1)!\)，也不整除 \(2\)，所以分式不是整数。偶质数 \(N=2\) 给出 \(n=1\)，符合条件。

不超过 \(25\) 的奇质数是 \(3\)、\(5\)、\(7\)、\(11\)、\(13\)、\(17\)、\(19\)、\(23\)，对应 \(8\) 个不符合条件的 \(n\)。所以有 \(24 - 8 = 16\) 个值符合条件。

所以正确答案是 C。

English

Since \(1 + 2 + \cdots + n = \dfrac{n(n+1)}{2},\) divisibility is equivalent to \[ \dfrac{n!}{\frac{n(n+1)}{2}} = \dfrac{2(n-1)!}{n+1} \] being an integer.

Put \(N=n+1.\) If \(N\) is composite and not a square, it has two distinct proper factors whose product is \(N;\) both occur in \((N-2)!=(n-1)!.\) If \(N=k^2\) with \(k\ge3,\) the factors \(k\) and \(2k\) occur in that factorial, so it contains a multiple of \(2N.\) The remaining composite case, \(N=4,\) also divides \(2(N-2)!=4.\) Thus the fraction is an integer whenever \(N\) is composite. If \(N=n+1\) is an odd prime, it divides neither \((n-1)!\) nor \(2,\) so the fraction is not an integer. The even prime \(N=2\) gives \(n=1,\) which works.

The odd primes at most \(25\) are \(3,\) \(5,\) \(7,\) \(11,\) \(13,\) \(17,\) \(19,\) \(23,\) giving \(8\) failing values of \(n.\) Hence \(24 - 8 = 16\) values work.

Thus, C is the correct answer.

23 答案 C

设 \(AB = a\)，\(DC = c\)。中位线 \(\overline{FE}\) 的长度为 \(\dfrac{a + c}{2}\)，且 \(ABEF\) 和 \(FECD\) 高度相同。

两个小梯形的面积与其平行边平均长度成正比，所以 \[ \dfrac{a + \frac{a+c}{2}}{\frac{a+c}{2} + c} = \dfrac{3a + c}{a + 3c} = 2\text{。}\]

由此 \(3a + c = 2a + 6c\)，所以 \(a = 5c\)，从而 \(\dfrac{AB}{DC} = 5\)。

所以正确答案是 C。

English

Let \(AB = a\) and \(DC = c.\) The midsegment \(\overline{FE}\) has length \(\dfrac{a + c}{2},\) and \(ABEF\) and \(FECD\) have the same height.

Their areas are proportional to the averages of their parallel sides, so \[ \dfrac{a + \frac{a+c}{2}}{\frac{a+c}{2} + c} = \dfrac{3a + c}{a + 3c} = 2. \]

Then \(3a + c = 2a + 6c,\) so \(a = 5c\) and \(\dfrac{AB}{DC} = 5.\)

Thus, C is the correct answer.

24 答案 E

写成 \(x = 10a + b\)、\(y = 10b + a\)，其中 \(a \gt b\)。于是 \[ \begin{aligned} m^2 &= x^2 - y^2 \\ &= 99(a^2 - b^2) \\ &= 99(a+b)(a-b) \end{aligned}\text{。}\]

由于 \(99 = 9 \cdot 11\)，要使 \(m^2\) 为完全平方数，需要 \((a+b)(a-b)\) 能被 \(11\) 整除。又因 \(a + b \le 17\)，这迫使 \(a + b = 11\)，然后 \(a - b\) 本身必须是完全平方数。

由于 \(a - b \le 8\)，唯一可行的是 \(a - b = 1\)，得到 \((a, b) = (6, 5)\)。于是 \(x = 65\)，\(y = 56\)，且 \(m^2 = 99 \cdot 11 = 33^2\)，所以 \(m = 33\)。

因此 \(x + y + m\) \(= 65 + 56 + 33\) \(= 154\)。

所以正确答案是 E。

English

Write \(x = 10a + b\) and \(y = 10b + a\) with \(a \gt b.\) Then \[ \begin{aligned} m^2 &= x^2 - y^2 \\ &= 99(a^2 - b^2) \\ &= 99(a+b)(a-b). \end{aligned} \]

Since \(99 = 9 \cdot 11,\) for \(m^2\) to be a perfect square we need \((a+b)(a-b)\) to be divisible by \(11.\) As \(a + b \le 17,\) this forces \(a + b = 11,\) and then \(a - b\) must itself be a perfect square.

With \(a - b \le 8,\) the only workable case is \(a - b = 1,\) giving \((a, b) = (6, 5).\) Then \(x = 65,\) \(y = 56,\) and \(m^2 = 99 \cdot 11 = 33^2,\) so \(m = 33.\)

Therefore \(x + y + m\) \(= 65 + 56 + 33\) \(= 154.\)

Thus, E is the correct answer.

25 答案 C

和为 \(125\) 的配对为 \((25, 100), (26, 99), \ldots, (62, 63)\)，共有 \(62 - 25 + 1 = 38\) 对。每一对中，\(B\) 至多包含一个元素。

数字 \(1\) 到 \(24\) 无法与范围内任何数字配成 \(125\)，所以这 \(24\) 个数字都可以加入。

因此 \(B\) 至多有 \(38 + 24 = 62\) 个元素，而集合 \(\{1, 2, \ldots, 62\}\) 可以达到这个数量。

所以正确答案是 C。

English

The pairs summing to \(125\) are \((25, 100), (26, 99), \ldots, (62, 63),\) which is \(62 - 25 + 1 = 38\) pairs. From each pair, \(B\) may contain at most one element.

The numbers \(1\) through \(24\) cannot pair with anything in range to sum to \(125,\) so all \(24\) of them may be included.

Thus \(B\) has at most \(38 + 24 = 62\) elements, and the set \(\{1, 2, \ldots, 62\}\) achieves this.

Thus, C is the correct answer.