2006A AMC 12 真题（P版）

1 答案 A

五个三明治花费 \(5\cdot 3 = 15\) 美元，八杯汽水花费 \(8\cdot 2 = 16\) 美元。一共花费 \(15 + 16 = 31\) 美元。

因此，正确答案是 A。

English

Five sandwiches cost \(5\cdot 3 = 15\) dollars and eight sodas cost \(8\cdot 2 = 16\) dollars. Together they cost \(15 + 16 = 31\) dollars.

Thus, the correct answer is A.

2 答案 C

根据定义，\(h \otimes h = h^3 - h\)。因此 \[ \begin{gathered} h \otimes (h^3 - h) \\ = h^3 - (h^3 - h) \\ = h \end{gathered}\text{。}\]

因此，正确答案是 C。

English

By the definition, \(h \otimes h = h^3 - h.\) Then \[ \begin{gathered} h \otimes (h^3 - h) \\ = h^3 - (h^3 - h) \\ = h. \end{gathered} \]

Thus, the correct answer is C.

3 答案 B

Mary 的年龄是 Alice 年龄的 \(\tfrac{3}{5}\)，所以 Mary 是 \(\tfrac{3}{5}\cdot 30 = 18\) 岁。

因此，正确答案是 B。

English

Mary’s age is \(\tfrac{3}{5}\) of Alice’s, so Mary is \(\tfrac{3}{5}\cdot 30 = 18\) years old.

Thus, the correct answer is B.

4 答案 E

分钟的两个数字和最多为 \(5 + 9 = 14\)，出现在 \(59\) 分。小时方面，单个数字 \(9\) 的数字和为 \(9\)，比任何两位数小时都大；\((10, 11, 12\) 的数字和最多只有 \(1 + 2 = 3)\)。

最大总和为 \(14 + 9 = 23\)，出现在 \(9\!:\!59\)。

因此，正确答案是 E。

English

The two minutes digits sum to at most \(5 + 9 = 14,\) at \(59\) minutes past the hour. For the hour, a single digit \(9\) gives digit sum \(9,\) which beats any two-digit hour \((10, 11, 12\) give at most \(1 + 2 = 3).\)

The largest total is \(14 + 9 = 23,\) occurring at \(9\!:\!59.\)

Thus, the correct answer is E.

5 答案 D

每片原味披萨花费 \(\$1\)。\(\$2\) 的凤尾鱼费用分摊到 \(4\) 片凤尾鱼披萨上，每片多 \(\$0.50\)，所以每片凤尾鱼披萨花费 \(\$1.50\)。

Dave 吃了 \(4\) 片凤尾鱼披萨和 \(1\) 片原味披萨：\(4\cdot 1.5 + 1 = \$7\)。Doug 吃了剩下的 \(3\) 片原味披萨：\(\$3\)。Dave 多付 \(7 - 3 = \$4\)。

因此，正确答案是 D。

English

Each plain slice costs \(\$1.\) The \(\$2\) anchovy charge is spread over the \(4\) anchovy slices, adding \(\$0.50\) each, so an anchovy slice costs \(\$1.50.\)

Dave ate \(4\) anchovy slices and \(1\) plain slice: \(4\cdot 1.5 + 1 = \$7.\) Doug ate the \(3\) remaining plain slices: \(\$3.\) Dave paid \(7 - 3 = \$4\) more.

Thus, the correct answer is D.

6 答案 A

两个六边形组成一个面积为 \(8 \cdot 18 = 144\) 的正方形，所以正方形边长为 \(12\)。

阶梯形切线把宽分成三个长度为 \(y\) 的相等水平部分，它们合起来跨过整个宽度：\(y + y + y = 18\)，所以 \(y = 6\)。（两个竖直台阶各上升 \(12 - 8 = 4\)，形成正方形多出的高度。）

因此，正确答案是 A。

English

The two hexagons form a square of area \(8 \cdot 18 = 144,\) so the square has side \(12.\)

The staircase cut splits the width into three equal horizontal pieces of length \(y,\) which together span the full width: \(y + y + y = 18,\) so \(y = 6.\) (The two vertical steps each rise \(12 - 8 = 4,\) building the extra height of the square.)

Thus, the correct answer is A.

7 答案 B

设 Danielle 为 \(x\) 岁。那么 Sally 为 \(0.6x\)，Mary 为 \(1.2(0.6x) = 0.72x\)。

由 \(x + 0.6x + 0.72x = 2.32x = 23.2\)，得 \(x = 10\)。所以 Mary 现在 \(0.72(10) = 7.2\) 岁，下一个生日时她将满 \(8\) 岁。

因此，正确答案是 B。

English

Let Danielle be \(x\) years old. Then Sally is \(0.6x\) and Mary is \(1.2(0.6x) = 0.72x.\)

The sum \(x + 0.6x + 0.72x = 2.32x = 23.2\) gives \(x = 10.\) So Mary is \(0.72(10) = 7.2\) years old, and on her next birthday she will be \(8.\)

Thus, the correct answer is B.

8 答案 C

\(n\) 个连续整数的和等于 \(n\) 乘以它们的中位数。和为 \(15\)：\(n = 2\) 给出 \(7 + 8\)，\(n = 3\) 给出 \(4 + 5 + 6\)，\(n = 5\) 给出 \(1 + 2 + 3 + 4 + 5\)。

四个连续整数的和为偶数，而超过五项时已经超过 \(1 + 2 + 3 + 4 + 5 = 15\)。所以共有 \(3\) 组。

因此，正确答案是 C。

English

The sum of \(n\) consecutive integers equals \(n\) times their median. For a sum of \(15:\) \(n = 2\) gives \(7 + 8,\) \(n = 3\) gives \(4 + 5 + 6,\) and \(n = 5\) gives \(1 + 2 + 3 + 4 + 5.\)

A run of four consecutive integers sums to an even number, and more than five terms already exceed \(1 + 2 + 3 + 4 + 5 = 15.\) So there are \(3\) sets.

Thus, the correct answer is C.

9 答案 A

设 \(p\) 为一支铅笔的价格，\(s\) 为一支铅笔加一块橡皮的总价，单位为美分。那么 \[ 13p + 3e = 3s + 10p = 100\text{，} \] 所以 \(3s\) 是 \(10\) 的倍数且小于 \(100\)。因此 \(s \in \{10, 20, 30\}\)，对应的 \(p = 7, 4, 1\)。

因为铅笔比橡皮贵，\(p \gt \tfrac{s}{2}\)，只有 \(s = 10\) 满足（铅笔 \(7\) 美分，橡皮 \(3\) 美分）。所以一支铅笔和一块橡皮共 \(10\) 美分。

因此，正确答案是 A。

English

Let \(p\) be a pencil’s cost and \(s\) the cost of one pencil plus one eraser, in cents. Then \[ 13p + 3e = 3s + 10p = 100, \] so \(3s\) is a multiple of \(10\) less than \(100.\) Hence \(s \in \{10, 20, 30\},\) with \(p = 7, 4, 1\) respectively.

Since a pencil costs more than an eraser, \(p \gt \tfrac{s}{2},\) which holds only for \(s = 10\) (pencil \(7,\) eraser \(3\)). So one pencil and one eraser cost \(10\) cents.

Thus, the correct answer is A.

10 答案 E

令 \(k = \sqrt{120 - \sqrt{x}}\) 为整数。那么 \(k \ge 0\)，且 \(k^2 = 120 - \sqrt{x} \le 120\)，所以 \(0 \le k \le 10\)。

每个这样的 \(k\) 都给出 \(\sqrt{x} = 120 - k^2 \ge 0\)，因而给出不同的 \(x = (120 - k^2)^2\)。共有 \(11\) 个值。

因此，正确答案是 E。

English

Let \(k = \sqrt{120 - \sqrt{x}}\) be an integer. Then \(k \ge 0\) and \(k^2 = 120 - \sqrt{x} \le 120,\) so \(0 \le k \le 10.\)

Each such \(k\) gives \(\sqrt{x} = 120 - k^2 \ge 0,\) hence a distinct value \(x = (120 - k^2)^2.\) That is \(11\) values.

Thus, the correct answer is E.

11 答案 C

展开得 \(x^2 + 2xy + y^2 = x^2 + y^2\)，所以 \(2xy = 0\)，即 \(xy = 0\)。

这是两条坐标轴的并集，也就是一对直线。

因此，正确答案是 C。

English

Expanding, \(x^2 + 2xy + y^2 = x^2 + y^2,\) so \(2xy = 0,\) i.e. \(xy = 0.\)

This is the union of the two coordinate axes, a pair of lines.

Thus, the correct answer is C.

12 答案 B

最上面的环跨越 \(20\) cm。下面每个环与上方的环重叠 \(2\) cm（两倍的 \(1\)-cm 厚度），所以它增加的长度为外径减 \(2\)。

下面各环外径为 \(19, 18, \ldots, 3\)，贡献 \(17, 16, \ldots, 1\)。因此总距离为 \[ \begin{gathered} 20 + (17 + 16 + \cdots + 1) \\ = 20 + \frac{17 \cdot 18}{2} \\ = 20 + 153 \\ = 173 \text{ cm} \end{gathered}\text{。}\]

因此，正确答案是 B。

English

The top ring spans \(20\) cm. Each ring below overlaps the ring above by \(2\) cm (twice the \(1\)-cm thickness), so it adds its outside diameter minus \(2.\)

The lower rings have outside diameters \(19, 18, \ldots, 3,\) contributing \(17, 16, \ldots, 1.\) Thus the total distance is \[ \begin{gathered} 20 + (17 + 16 + \cdots + 1) \\ = 20 + \frac{17 \cdot 18}{2} \\ = 20 + 153 \\ = 173 \text{ cm}. \end{gathered} \]

Thus, the correct answer is B.

13 答案 E

若 \(r, s, t\) 是三个顶点处圆的半径，则 \(r + s = 3,\ r + t = 4,\ s + t = 5\)。三式相加得 \(r + s + t = 6\)，所以 \(r = 1,\ s = 2,\ t = 3\)。

面积之和为 \(\pi(1^2 + 2^2 + 3^2) = 14\pi\)。

因此，正确答案是 E。

English

If \(r, s, t\) are the radii at the vertices, then \(r + s = 3,\ r + t = 4,\ s + t = 5.\) Adding all three gives \(r + s + t = 6,\) so \(r = 1,\ s = 2,\ t = 3.\)

The sum of the areas is \(\pi(1^2 + 2^2 + 3^2) = 14\pi.\)

Thus, the correct answer is E.

14 答案 C

债务 \(D\) 能结清当且仅当 \(D = 300p + 210g\) \(= 30(10p + 7g)\)，其中 \(p, g\) 为整数。因此 \(D\) 是 \(\gcd(300, 210) = 30\) 的倍数，所以没有更小的正债务可行。

\(\$30\) 的债务可以做到，因为 \(30 = 300(-2) + 210(3)\)，即给出 \(3\) 只山羊并收到 \(2\) 头猪作为找零。

因此，正确答案是 C。

English

A debt \(D\) is resolvable if and only if \(D = 300p + 210g\) \(= 30(10p + 7g)\) for integers \(p, g.\) Thus \(D\) is a multiple of \(\gcd(300, 210) = 30,\) so no smaller positive debt works.

A debt of \(\$30\) is achievable since \(30 = 300(-2) + 210(3),\) i.e. give \(3\) goats and receive \(2\) pigs in change.

Thus, the correct answer is C.

15 答案 A

因为 \(\cos x = 0\)，所以 \(x = \tfrac{\pi}{2} + k\pi\)。又因为 \(\cos(x + z) = \tfrac{1}{2}\)，所以 \(x + z = 2n\pi \pm \tfrac{\pi}{3}\)。

在单位圆上，\(\tfrac{\pi}{2}\) 的奇数倍与同余于 \(\pm\tfrac{\pi}{3}\) 的角之间，最小正角距离为 \(\tfrac{\pi}{6}\)。取 \(x = -\tfrac{\pi}{2}\) 和 \(x + z = -\tfrac{\pi}{3}\)，可达到此值，此时 \(z = -\tfrac{\pi}{3} + \tfrac{\pi}{2} = \tfrac{\pi}{6}\)。

所以正确答案是 A。

English

Since \(\cos x = 0,\) we have \(x = \tfrac{\pi}{2} + k\pi.\) Since \(\cos(x + z) = \tfrac{1}{2},\) we have \(x + z = 2n\pi \pm \tfrac{\pi}{3}.\)

On the unit circle, the smallest positive angular separation between an odd multiple of \(\tfrac{\pi}{2}\) and an angle congruent to \(\pm\tfrac{\pi}{3}\) is \(\tfrac{\pi}{6}.\) It is attained by taking \(x = -\tfrac{\pi}{2}\) and \(x + z = -\tfrac{\pi}{3},\) which gives \(z = -\tfrac{\pi}{3} + \tfrac{\pi}{2} = \tfrac{\pi}{6}.\)

Thus, the correct answer is A.

16 答案 B

半径满足 \(AC \perp CD\) 和 \(BD \perp CD\)。由勾股定理，\(CE = \sqrt{5^2 - 3^2} = 4\)。

因为 \(\triangle ACE \sim \triangle BDE\)，得 \(\tfrac{DE}{CE} = \tfrac{BD}{AC} = \tfrac{8}{3}\)，所以 \(DE = 4 \cdot \tfrac{8}{3} = \tfrac{32}{3}\)。因此 \[ \begin{gathered} CD = CE + DE \\ = 4 + \frac{32}{3} \\ = \frac{44}{3} \end{gathered}\text{。}\]

因此，正确答案是 B。

English

The radii satisfy \(AC \perp CD\) and \(BD \perp CD.\) By the Pythagorean theorem, \(CE = \sqrt{5^2 - 3^2} = 4.\)

Since \(\triangle ACE \sim \triangle BDE,\) we get \(\tfrac{DE}{CE} = \tfrac{BD}{AC} = \tfrac{8}{3},\) so \(DE = 4 \cdot \tfrac{8}{3} = \tfrac{32}{3}.\) Then \[ \begin{gathered} CD = CE + DE \\ = 4 + \frac{32}{3} \\ = \frac{44}{3}. \end{gathered} \]

Thus, the correct answer is B.

17 答案 B

设 \(B = (0, 0)\)，\(C = (s, 0)\)，\(A = (0, s)\)，\(D = (s, s)\)，因此 \(E = \left(s + \tfrac{r}{\sqrt{2}},\ s + \tfrac{r}{\sqrt{2}}\right)\) 在射线 \(BD\) 上。

因为 \(AF\) 是圆的切线，\(AF^2 = AE^2 - r^2\)。计算 \(AE^2\) 并化简，得到 \(9 + 5\sqrt{2} = s^2 + rs\sqrt{2}\)。

因为 \(r\) 和 \(s\) 都是有理数，有理部分和无理部分分别相等：\(s^2 = 9\)，\(rs = 5\)。因此 \(s = 3,\ r = \tfrac{5}{3}\)，所以 \(\frac{r}{s} = \tfrac{5}{9}\)。

因此，正确答案是 B。

English

Set \(B = (0, 0),\) \(C = (s, 0),\) \(A = (0, s),\) \(D = (s, s),\) so that \(E = \left(s + \tfrac{r}{\sqrt{2}},\ s + \tfrac{r}{\sqrt{2}}\right)\) lies on ray \(BD.\)

Since \(AF\) is tangent to the circle, \(AF^2 = AE^2 - r^2.\) Computing \(AE^2\) and simplifying gives \(9 + 5\sqrt{2} = s^2 + rs\sqrt{2}.\)

Because \(r\) and \(s\) are rational, the rational and irrational parts match: \(s^2 = 9\) and \(rs = 5.\) Thus \(s = 3,\ r = \tfrac{5}{3},\) and \(\frac{r}{s} = \tfrac{5}{9}.\)

Thus, the correct answer is B.

18 答案 E

用 \(\frac{1}{x}\) 替换 \(x\) 得 \(f\!\left(\tfrac{1}{x}\right) + f(x) = \tfrac{1}{x}\)。与 \(f(x) + f\!\left(\tfrac{1}{x}\right) = x\) 合在一起，就要求 \(x = \tfrac{1}{x}\)，所以 \(x = \pm 1\)。

这两个值都一致，其中 \(f(1) = \tfrac{1}{2}\)，\(f(-1) = -\tfrac{1}{2}\)。因此最大可能定义域为 \(\{-1, 1\}\)。

因此，正确答案是 E。

English

Replacing \(x\) by \(\frac{1}{x}\) gives \(f\!\left(\tfrac{1}{x}\right) + f(x) = \tfrac{1}{x}.\) Together with \(f(x) + f\!\left(\tfrac{1}{x}\right) = x,\) this requires \(x = \tfrac{1}{x},\) so \(x = \pm 1.\)

Both values are consistent, with \(f(1) = \tfrac{1}{2}\) and \(f(-1) = -\tfrac{1}{2}.\) So the largest possible domain is \(\{-1, 1\}.\)

Thus, the correct answer is E.

19 答案 E

每个圆的半径等于其圆心的 \(y\)-坐标，所以两圆都与 \(x\)-轴相切，这条直线就是一条公共外切线。两条外切线交于圆心连线与 \(x\)-轴的交点。

该直线斜率为 \(\tfrac{9 - 4}{14 - 2} = \tfrac{5}{12} = \tan\theta\)，且过 \((2, 4)\)， 与 \(x\)-轴交于 \(\left(-\tfrac{38}{5}, 0\right)\)。

另一条切线与 \(x\)-轴所成的角为 \(2\theta\)，所以其斜率为 \[ \tan 2\theta = \frac{2 \cdot \tfrac{5}{12}}{1 - \left(\tfrac{5}{12}\right)^2} = \frac{120}{119}\text{。} \] 因此 \(b = \tfrac{120}{119} \cdot \tfrac{38}{5} = \tfrac{912}{119}\)。

因此，正确答案是 E。

English

Each circle’s radius equals its center’s \(y\)-coordinate, so both are tangent to the \(x\)-axis, which is a common external tangent. The two external tangents meet at the \(x\)-intercept of the line through the centers.

That line has slope \(\tfrac{9 - 4}{14 - 2} = \tfrac{5}{12} = \tan\theta\) and passes through \((2, 4),\) meeting the \(x\)-axis at \(\left(-\tfrac{38}{5}, 0\right).\)

The other tangent makes angle \(2\theta\) with the \(x\)-axis, so its slope is \[ \tan 2\theta = \frac{2 \cdot \tfrac{5}{12}}{1 - \left(\tfrac{5}{12}\right)^2} = \frac{120}{119}. \] Then \(b = \tfrac{120}{119} \cdot \tfrac{38}{5} = \tfrac{912}{119}.\)

Thus, the correct answer is E.

20 答案 C

从起点出发共有 \(3^7\) 条等可能的 \(7\)-步路径。考虑一条访问全部 \(8\) 个顶点的路径：第一步有 \(3\) 种选择，第二步有 \(2\) 种选择。

用三位二进制串给立方体的顶点编号。由对称性，固定前两步之后，可设前三个顶点依次为 \(000,001,011\)。逐个分支检验可知，恰好只有以下三种走完全程的方式：\[ \begin{aligned} &010,110,111,101,100,\\ &010,110,100,101,111,\\ &111,101,100,110,010 \end{aligned}\text{。}\] 因此这样的路径共有 \(3 \cdot 2 \cdot 3 = 18\) 条。

概率为 \(\dfrac{18}{3^7} = \dfrac{18}{2187} = \dfrac{2}{243}\)。

因此，正确答案是 C。

English

From the start there are \(3^7\) equally likely \(7\)-move walks. For a walk visiting all \(8\) vertices, there are \(3\) choices for the first move and \(2\) for the second, since it cannot return to the starting vertex.

Label cube vertices by three-bit strings. By symmetry, after fixing those first two moves we may take the first three vertices to be \(000,001,011.\) A branch check gives exactly these three completions: \[ \begin{aligned} &010,110,111,101,100,\\ &010,110,100,101,111,\\ &111,101,100,110,010. \end{aligned} \] Thus there are \(3 \cdot 2 \cdot 3 = 18\) such walks.

The probability is \(\dfrac{18}{3^7} = \dfrac{18}{2187} = \dfrac{2}{243}.\)

Thus, the correct answer is C.

21 答案 E

对 \(j = 1, 2\)，条件变为 \(j + x^2 + y^2 \le 10^j(x + y)\)，即 \[ \begin{gathered} \left(x - \frac{10^j}{2}\right)^2 \\ {}+ \left(y - \frac{10^j}{2}\right)^2 \\ \le \frac{10^{2j}}{2} - j \end{gathered}\text{。}\]

这是两个圆盘，\(S_1\) 的半径平方为 \(\tfrac{100}{2} - 1 = 49\)，\(S_2\) 的半径平方为 \(\tfrac{10000}{2} - 2 = 4998\)。

对数要求 \(x+y>0\)。对每个圆盘，其圆心到直线 \(x+y=0\) 的距离平方为 \(\tfrac{10^{2j}}{2}\)，比半径平方恰好多 \(j\)。所以两个圆盘都完全位于 \(x+y>0\) 内，改写不等式时没有加入额外点。面积之比为 \(\dfrac{4998}{49} = 102\)。

所以正确答案是 E。

English

For \(j = 1, 2,\) the condition becomes \(j + x^2 + y^2 \le 10^j(x + y),\) i.e. \[ \begin{gathered} \left(x - \frac{10^j}{2}\right)^2 \\ {}+ \left(y - \frac{10^j}{2}\right)^2 \\ \le \frac{10^{2j}}{2} - j. \end{gathered} \]

These are disks with squared radii \(\tfrac{100}{2} - 1 = 49\) for \(S_1\) and \(\tfrac{10000}{2} - 2 = 4998\) for \(S_2.\)

The logarithms require \(x+y>0.\) For each disk, the squared distance from its center to the line \(x+y=0\) is \(\tfrac{10^{2j}}{2},\) which is \(j\) more than the squared radius. Hence both disks lie entirely in \(x+y>0,\) so no points were added when the inequalities were rewritten. The area ratio is \(\dfrac{4998}{49} = 102.\)

Thus, the correct answer is E.

22 答案 D

把正六边形放在圆的中心。每个顶点都对应一段圆弧，从该弧上只能完整看到在此顶点相交的两条边。这六段全等圆弧组成补事件，其概率为 \(\tfrac{1}{2}\)，所以每段弧的度数为 \(30^\circ\)。

取以圆心 \(O\) 到顶点 \(A\) 的射线为中心的那段弧，并设 \(P\) 为它的上端点。则 \(\angle POA = 15^\circ\)，而在 \(P\) 处第三条边刚好开始可见，所以 \(P\) 位于该边所在的直线上。这条直线到 \(O\) 的距离为边心距 \(\sqrt{3}\)。

因此 \(\sqrt{3} = r\sin 15^\circ = r \cdot \dfrac{\sqrt{6} - \sqrt{2}}{4}\)，所以 \[ r = \frac{4\sqrt{3}}{\sqrt{6} - \sqrt{2}} = 3\sqrt{2} + \sqrt{6}\text{。}\]

所以正确答案是 D。

English

Place the hexagon at the center of the circle. Corresponding to each vertex is an arc from which only the two sides meeting there are entirely visible. These six congruent arcs make up the complementary probability \(\tfrac{1}{2},\) so each arc measures \(30^\circ.\)

Take the arc centered on the ray from the center \(O\) through a vertex \(A,\) and let \(P\) be its upper endpoint. Then \(\angle POA = 15^\circ,\) and at \(P\) a third side is just becoming visible, so \(P\) lies on that side’s supporting line. Its distance from \(O\) is the apothem \(\sqrt{3}.\)

Hence \(\sqrt{3} = r\sin 15^\circ = r \cdot \dfrac{\sqrt{6} - \sqrt{2}}{4},\) giving \[ r = \frac{4\sqrt{3}}{\sqrt{6} - \sqrt{2}} = 3\sqrt{2} + \sqrt{6}. \]

Thus, the correct answer is D.

23 答案 B

每次应用 \(A\) 都是在平均相邻项，所以 \(100\) 步后剩下的唯一一项为 \[ \begin{gathered} \frac{1}{2^{100}} \sum_{m=0}^{100} \binom{100}{m} x^m \\ = \frac{(1 + x)^{100}}{2^{100}} \end{gathered}\text{。}\]

将其设为 \(\dfrac{1}{2^{50}}\)，得 \((1 + x)^{100} = 2^{50}\)，所以 \(1 + x = 2^{\frac{1}{2}} = \sqrt{2}\)。因为 \(x \gt 0\)，得 \(x = \sqrt{2} - 1\)。

因此，正确答案是 B。

English

Each application of \(A\) averages adjacent terms, so after \(100\) steps the single remaining term is \[ \begin{gathered} \frac{1}{2^{100}} \sum_{m=0}^{100} \binom{100}{m} x^m \\ = \frac{(1 + x)^{100}}{2^{100}}. \end{gathered} \]

Setting this equal to \(\dfrac{1}{2^{50}}\) gives \((1 + x)^{100} = 2^{50},\) so \(1 + x = 2^{\frac{1}{2}} = \sqrt{2}.\) Since \(x \gt 0,\) we get \(x = \sqrt{2} - 1.\)

Thus, the correct answer is B.

24 答案 D

单项式 \(x^a y^b z^c\) 只有在 \(a\) 为偶数时保留，因为 \(a\) 为奇数的项会在两个展开式之间抵消。

对每个满足 \(0 \le a \le 2006\) 的偶数 \(a\)，指数 \(b\) 有 \(2007 - a\) 个取值，而 \(c = 2006 - a - b\) 随后确定。对所有偶数 \(a\) 求和： \[ \begin{gathered} (2007 - 0) + (2007 - 2) \\ {}+ \cdots + (2007 - 2006) \\ = 2007 + 2005 + \cdots + 1\text{，} \end{gathered} \] 这是前 \(1004\) 个正奇数之和，等于 \(1004^2 = 1{,}008{,}016\)。

因此，正确答案是 D。

English

A term \(x^a y^b z^c\) survives only when \(a\) is even, since terms with odd \(a\) cancel between the two expansions.

For each even \(a\) with \(0 \le a \le 2006,\) the exponent \(b\) ranges over \(2007 - a\) values and \(c = 2006 - a - b\) is then determined. Summing over even \(a:\) \[ \begin{gathered} (2007 - 0) + (2007 - 2) \\ {}+ \cdots + (2007 - 2006) \\ = 2007 + 2005 + \cdots + 1, \end{gathered} \] the sum of the first \(1004\) odd positive integers, which is \(1004^2 = 1{,}008{,}016.\)

Thus, the correct answer is D.

25 答案 E

由性质 \((2)\)，一个有效的 \(k\)-元素集合是 \(\{k, k+1, \ldots, 15\}\) 的一个 \(k\)-子集，且没有两个连续元素。

将被选元素之间的间隔压缩后，这些集合与一个 \((17 - 2k)\)-元素集合的 \(k\)-子集一一对应，数量为 \(\binom{17 - 2k}{k}\)。只有 \(k \le 5\) 时该数非零，所以总数为 \[ \begin{gathered} \binom{15}{1} + \binom{13}{2} \\ {}+ \binom{11}{3} + \binom{9}{4} \\ {}+ \binom{7}{5} \\ = 15 + 78 + 165 + 126 + 21 \\ = 405 \end{gathered}\text{。}\]

因此，正确答案是 E。

English

By property \((2),\) a valid \(k\)-element set is a \(k\)-subset of \(\{k, k+1, \ldots, 15\}\) with no two consecutive elements.

Collapsing the gaps between chosen elements, these correspond bijectively to \(k\)-subsets of a \((17 - 2k)\)-element set, counted by \(\binom{17 - 2k}{k}.\) This is nonzero only for \(k \le 5,\) so the total is \[ \begin{gathered} \binom{15}{1} + \binom{13}{2} \\ {}+ \binom{11}{3} + \binom{9}{4} \\ {}+ \binom{7}{5} \\ = 15 + 78 + 165 + 126 + 21 \\ = 405. \end{gathered} \]

Thus, the correct answer is E.