2007A AMC 12 真题（P版）

1 答案 C

Susan 支付 \((4)(0.75)(20)=60\) 美元。

Pam 支付 \((5)(0.70)(20)=70\) 美元。

所以 Pam 比 Susan 多付 \(70-60=10\) 美元。

因此，正确答案是 C。

English

Susan pays \((4)(0.75)(20)=60\) dollars.

Pam pays \((5)(0.70)(20)=70\) dollars.

So Pam pays \(70-60=10\) more dollars than Susan.

Thus, the correct answer is C.

2 答案 D

砖的体积是 \(40\cdot 20\cdot 10=8000\) 立方厘米。

如果水面上升 \(h\) 厘米，增加的水体积是 \(100\cdot 40\cdot h=4000h\) 立方厘米。

令它等于砖的体积，得 \(8000=4000h\)，所以 \(h=2\)。

因此，正确答案是 D。

English

The brick has a volume of \(40\cdot 20\cdot 10=8000\) cubic centimeters.

If the water rises by \(h\) centimeters, the added volume is \(100\cdot 40\cdot h=4000h\) cubic centimeters.

Setting this equal to the brick’s volume gives \(8000=4000h,\) so \(h=2.\)

Thus, the correct answer is D.

3 答案 A

设较小的整数为 \(x\)。那么较大的整数为 \(x+2\)。

于是 \(x+2=3x\)，得 \(x=1\)。

两个整数是 \(1\) 和 \(3\)，它们的和是 \(4\)。

因此，正确答案是 A。

English

Let the smaller integer be \(x.\) Then the larger is \(x+2.\)

So \(x+2=3x,\) which gives \(x=1.\)

The two integers are \(1\) and \(3,\) and their sum is \(4.\)

Thus, the correct answer is A.

4 答案 A

Kate 以每小时 \(16\) 英里的速度骑了 \(\tfrac12\) 小时，行进了 \(8\) 英里。

她以每小时 \(4\) 英里的速度走了 \(\tfrac32\) 小时，行进了 \(6\) 英里。

她在 \(2\) 小时内共行进 \(14\) 英里，所以平均速度是每小时 \(7\) 英里。

因此，正确答案是 A。

English

Kate rode for \(\tfrac12\) hour at \(16\) mph, covering \(8\) miles.

She walked for \(\tfrac32\) hours at \(4\) mph, covering \(6\) miles.

She covered \(14\) miles in \(2\) hours, so her average speed was \(7\) mph.

Thus, the correct answer is A.

5 答案 D

缴纳联邦税后，Public 先生保留遗产的 \(80\%\)。

他为这部分缴纳 \(10\%\) 的州税，也就是遗产的 \(8\%\)。

总税额是遗产的 \(20\%+8\%=28\%\)，所以遗产为 \(\frac{\$10{,}500}{0.28}=\$37{,}500\)。

因此，正确答案是 D。

English

After federal taxes, Mr. Public keeps \(80\%\) of his inheritance.

He pays \(10\%\) of that in state taxes, which is \(8\%\) of the inheritance.

His total tax is \(20\%+8\%=28\%\) of the inheritance, so the inheritance is \(\frac{\$10{,}500}{0.28}=\$37{,}500.\)

Thus, the correct answer is D.

6 答案 D

因为 \(\triangle ABC\) 是等腰三角形，\(\angle BAC=\tfrac12(180^\circ-\angle ABC)\) \(=70^\circ\)。

因为 \(\triangle ADC\) 是等腰三角形，\(\angle DAC=\tfrac12(180^\circ-\angle ADC)\) \(=20^\circ\)。

因此 \(\angle BAD=\angle BAC-\angle DAC\) \(=70^\circ-20^\circ\) \(=50^\circ\)。

因此，正确答案是 D。

English

Since \(\triangle ABC\) is isosceles, \(\angle BAC=\tfrac12(180^\circ-\angle ABC)\) \(=70^\circ.\)

Since \(\triangle ADC\) is isosceles, \(\angle DAC=\tfrac12(180^\circ-\angle ADC)\) \(=20^\circ.\)

Therefore \(\angle BAD=\angle BAC-\angle DAC\) \(=70^\circ-20^\circ\) \(=50^\circ.\)

Thus, the correct answer is D.

7 答案 C

设 \(D\) 为公差。那么 \(a=c-2D\)，\(b=c-D\)，\(d=c+D\)，且 \(e=c+2D\)，所以 \[a+b+c+d+e=5c\text{。}\]

因此 \(5c=30\)，得 \(c=6\)。

其他项不能确定：数列 \(4,5,6,7,8\) 和 \(10,8,6,4,2\) 都满足条件，但除中间项外其余各项不同。

因此，正确答案是 C。

English

Let \(D\) be the common difference. Then \(a=c-2D,\) \(b=c-D,\) \(d=c+D,\) and \(e=c+2D,\) so \[a+b+c+d+e=5c.\]

Thus \(5c=30,\) giving \(c=6.\)

The other terms cannot be determined: the sequences \(4,5,6,7,8\) and \(10,8,6,4,2\) both satisfy the conditions but differ in every term except the middle one.

Thus, the correct answer is C.

8 答案 C

考虑在数字 \(5\) 处相交的两条弦。它们分别连到 \(12\) 和 \(10\)，因此它们所对的弧从 \(10\) 到 \(12\)。

这段弧跨过十二个小时刻度中的两个，所以其度数为 \(\tfrac{2}{12}\cdot 360^\circ=60^\circ\)。

根据圆周角定理，顶点角等于所对弧的一半，即 \(\tfrac12\cdot 60^\circ=30^\circ\)。由对称性，每个顶点角都是 \(30^\circ\)。

因此，正确答案是 C。

English

Consider the two chords meeting at the number \(5.\) They run to \(12\) and to \(10,\) so the arc they subtend extends from \(10\) to \(12.\)

That arc spans two of the twelve hour-marks, so its measure is \(\tfrac{2}{12}\cdot 360^\circ=60^\circ.\)

By the Inscribed Angle Theorem, the vertex angle is half the arc, or \(\tfrac12\cdot 60^\circ=30^\circ.\) By symmetry every vertex angle equals \(30^\circ.\)

Thus, the correct answer is C.

9 答案 B

设 \(w\) 为步行速度，\(x\) 和 \(y\) 分别为 Yan 到家和到体育场的距离。

步行到体育场需要 \(\dfrac{y}{w}\)。先走回家再骑车需要 \(\dfrac{x}{w}+\dfrac{x+y}{7w}=\dfrac{8x+y}{7w}\)。

令两者相等，得 \(7y=8x+y\)，所以 \(8x=6y\)，\(\dfrac{x}{y}=\dfrac{3}{4}\)。

因此，正确答案是 B。

English

Let \(w\) be the walking speed and let \(x\) and \(y\) be Yan’s distances from home and from the stadium.

Walking to the stadium takes \(\dfrac{y}{w}.\) Walking home then biking takes \(\dfrac{x}{w}+\dfrac{x+y}{7w}=\dfrac{8x+y}{7w}.\)

Setting these equal gives \(7y=8x+y,\) so \(8x=6y\) and \(\dfrac{x}{y}=\dfrac{3}{4}.\)

Thus, the correct answer is B.

10 答案 A

设三边为 \(3x\)、\(4x\) 和 \(5x\)。该三角形是直角三角形，所以斜边是直径。

因此 \(5x=2\cdot 3=6\)，得 \(x=\tfrac65\)。

面积为 \(\tfrac12\cdot 3x\cdot 4x=6x^2\) \(=6\cdot\tfrac{36}{25}\) \(=\tfrac{216}{25}=8.64\)。

因此，正确答案是 A。

English

Let the sides be \(3x,\) \(4x,\) and \(5x.\) The triangle is right, so its hypotenuse is a diameter.

Thus \(5x=2\cdot 3=6,\) giving \(x=\tfrac65.\)

The area is \(\tfrac12\cdot 3x\cdot 4x=6x^2\) \(=6\cdot\tfrac{36}{25}\) \(=\tfrac{216}{25}=8.64.\)

Thus, the correct answer is A.

11 答案 D

由于循环性质，每个出现的数字在百位、十位和个位上出现的次数相同。

设所有项个位数字之和为 \(k\)，则 \(S=100k+10k+k=111k\) \(=3\cdot 37\cdot k\)。

所以 \(S\) 总能被 \(37\) 整除。它不一定能被更大的数整除：数列 \(123,231,312\) 给出 \(S=666=2\cdot 3^2\cdot 37\)。

因此，正确答案是 D。

English

Because of the cycling property, each digit that appears is used the same number of times in the hundreds, tens, and units places.

Let \(k\) be the sum of the units digits over all terms. Then \(S=100k+10k+k=111k\) \(=3\cdot 37\cdot k.\)

So \(S\) is always divisible by \(37.\) It need not be divisible by anything larger: the sequence \(123,231,312\) gives \(S=666=2\cdot 3^2\cdot 37.\)

Thus, the correct answer is D.

12 答案 E

从 \(0\) 到 \(2007\) 的整数中恰好有一半是奇数。

乘积 \(ad\) 只有在两个因数都是奇数时才是奇数，概率为 \(\tfrac12\cdot\tfrac12=\tfrac14\)，因此为偶数的概率为 \(\tfrac34\)。\(bc\) 也同理。

因此 \(ad-bc\) 在两个乘积都为奇数或都为偶数时为偶数：\[\tfrac14\cdot\tfrac14+\tfrac34\cdot\tfrac34=\tfrac{10}{16}=\tfrac58\text{。}\]

因此，正确答案是 E。

English

Exactly half of the integers from \(0\) to \(2007\) are odd.

A product \(ad\) is odd only when both factors are odd, with probability \(\tfrac12\cdot\tfrac12=\tfrac14,\) and even with probability \(\tfrac34.\) The same holds for \(bc.\)

Then \(ad-bc\) is even when both products are odd or both are even: \[\tfrac14\cdot\tfrac14+\tfrac34\cdot\tfrac34=\tfrac{10}{16}=\tfrac58.\]

Thus, the correct answer is E.

13 答案 B

老鼠在从 \((12,10)\) 到该直线的垂足处最接近奶酪。

这条垂线斜率为 \(\tfrac15\)，所以方程为 \(y=10+\tfrac15(x-12)=\tfrac15 x+\tfrac{38}{5}\)。

令 \(\tfrac15 x+\tfrac{38}{5}=-5x+18\)，得 \(x=2\) 且 \(y=8\)。因此 \((a,b)=(2,8)\)，\(a+b=10\)。

因此，正确答案是 B。

English

The mouse is closest to the cheese at the foot of the perpendicular from \((12,10)\) to the line.

This perpendicular has slope \(\tfrac15,\) so its equation is \(y=10+\tfrac15(x-12)=\tfrac15 x+\tfrac{38}{5}.\)

Setting \(\tfrac15 x+\tfrac{38}{5}=-5x+18\) gives \(x=2\) and \(y=8.\) Thus \((a,b)=(2,8)\) and \(a+b=10.\)

Thus, the correct answer is B.

14 答案 C

五个因数是乘积为 \(45\) 的互不相同的非零整数，因此每个因数都是 \(45\) 的正因数或负因数。检验 \(\pm1,\pm3,\pm5,\pm9,\pm15,\pm45\)，只有五元集合 \(\{-3,-1,1,3,5\}\) 的乘积为 \(45\)。

那么 \(a,b,c,d,e\) 按某种顺序为 \(9,7,5,3,1\)，它们的和是 \(25\)。

因此，正确答案是 C。

English

The five factors are distinct nonzero integers with product \(45,\) so each is a positive or negative divisor of \(45.\) Checking \(\pm1,\pm3,\pm5,\pm9,\pm15,\pm45,\) only the five-element set \(\{-3,-1,1,3,5\}\) has product \(45.\)

Then \(a,b,c,d,e\) are \(9,7,5,3,1\) in some order, and their sum is \(25.\)

Thus, the correct answer is C.

15 答案 E

平均数是 \(\dfrac{28+n}{5}\)。

如果 \(n\lt 6\)，中位数是 \(6\)，所以 \(28+n=30\)，\(n=2\)。

如果 \(6\lt n\lt 9\)，中位数是 \(n\)，所以 \(28+n=5n\)，\(n=7\)。

如果 \(n\gt 9\)，中位数是 \(9\)，所以 \(28+n=45\)，\(n=17\)。

所有可能值之和为 \(2+7+17=26\)。

因此，正确答案是 E。

English

The mean is \(\dfrac{28+n}{5}.\)

If \(n\lt 6,\) the median is \(6,\) so \(28+n=30\) and \(n=2.\)

If \(6\lt n\lt 9,\) the median is \(n,\) so \(28+n=5n\) and \(n=7.\)

If \(n\gt 9,\) the median is \(9,\) so \(28+n=45\) and \(n=17.\)

The sum of all possible values is \(2+7+17=26.\)

Thus, the correct answer is E.

16 答案 C

三个互不相同的数字构成一个递增等差数列。按公差计数：公差为 \(1\) 有 \(8\) 组，公差为 \(2\) 有 \(6\) 组，公差为 \(3\) 有 \(4\) 组，公差为 \(4\) 有 \(2\) 组，共 \(20\) 组。

其中 \(4\) 组含有 \(0\)（即 \(\{0,1,2\}\)，\(\{0,2,4\}\)，\(\{0,3,6\}\)，\(\{0,4,8\}\)）；每组产生 \(2\cdot 2!=4\) 个有效三位数，因为 \(0\) 不能放在首位。

另外 \(16\) 组各产生 \(3!=6\) 个数。总数为 \(4\cdot 4+16\cdot 6=112\)。

因此，正确答案是 C。

English

The three distinct digits form an increasing arithmetic progression. Counting by common difference: \(8\) with difference \(1,\) \(6\) with difference \(2,\) \(4\) with difference \(3,\) and \(2\) with difference \(4,\) for \(20\) sets.

Of these, \(4\) sets contain \(0\) (namely \(\{0,1,2\},\) \(\{0,2,4\},\) \(\{0,3,6\},\) \(\{0,4,8\}\)); each yields \(2\cdot 2!=4\) valid numbers since \(0\) cannot lead.

The other \(16\) sets each yield \(3!=6\) numbers. The total is \(4\cdot 4+16\cdot 6=112.\)

Thus, the correct answer is C.

17 答案 B

将两个等式平方，得 \(\sin^2 a+2\sin a\sin b+\sin^2 b=\tfrac53\) 和 \(\cos^2 a+2\cos a\cos b\) \(+\cos^2 b=1\)。

相加并两次使用 \(\sin^2\theta+\cos^2\theta=1\)，得 \[ \begin{aligned} &2+2(\sin a\sin b+\cos a\cos b) \\ &=\tfrac83 \end{aligned}\text{。}\]

所以 \(\cos(a-b)=\sin a\sin b\) \(+\cos a\cos b\) \(=\tfrac13\)。

因此，正确答案是 B。

English

Squaring both equations gives \(\sin^2 a+2\sin a\sin b+\sin^2 b=\tfrac53\) and \(\cos^2 a+2\cos a\cos b\) \(+\cos^2 b=1.\)

Adding and using \(\sin^2\theta+\cos^2\theta=1\) twice, \[ \begin{aligned} &2+2(\sin a\sin b+\cos a\cos b) \\ &=\tfrac83. \end{aligned} \]

So \(\cos(a-b)=\sin a\sin b\) \(+\cos a\cos b\) \(=\tfrac13.\)

Thus, the correct answer is B.

18 答案 D

因为 \(f\) 的系数为实数，所以共轭数 \(-2i\) 和 \(2-i\) 也是根。因此 \[ \begin{gathered} f(x) = (x^2+4)(x^2-4x+5) \\ = x^4-4x^3+9x^2 \\ {}-16x+20 \end{gathered}\text{。}\]

所以 \(a+b+c+d=-4+9\) \(-16+20\) \(=9\)。等价地，\(a+b+c+d=f(1)-1\) \(=(1+4)(1+1)-1\) \(=9\)。

因此，正确答案是 D。

English

Since \(f\) has real coefficients, the conjugates \(-2i\) and \(2-i\) are also roots. Thus \[ \begin{gathered} f(x) = (x^2+4)(x^2-4x+5) \\ = x^4-4x^3+9x^2 \\ {}-16x+20. \end{gathered} \]

Then \(a+b+c+d=-4+9\) \(-16+20\) \(=9.\) Equivalently, \(a+b+c+d=f(1)-1\) \(=(1+4)(1+1)-1\) \(=9.\)

Thus, the correct answer is D.

19 答案 E

\(\triangle ABC\) 中从 \(A\) 出发的高 \(h\) 满足 \(2007=\tfrac12\cdot 223\cdot h\)，所以 \(h=18\)。因此 \(A\) 在 \(y=18\) 或 \(y=-18\) 上。

直线 \(DE\) 的方程为 \(x-y-300=0\)。\(\triangle ADE\) 的面积条件同样把 \(A\) 限制在两条与 \(DE\) 平行的直线之一上。

\(A\) 的四个可能位置是一个平行四边形的顶点，其中心是 \(y=0\) 与直线 \(DE\) 的交点，即 \((300,0)\)。因此四个 \(x\) 坐标之和为 \(4\cdot 300=1200\)。

因此，正确答案是 E。

English

The altitude \(h\) from \(A\) in \(\triangle ABC\) satisfies \(2007=\tfrac12\cdot 223\cdot h,\) so \(h=18.\) Thus \(A\) lies on \(y=18\) or \(y=-18.\)

Line \(DE\) has equation \(x-y-300=0.\) The condition on \(\triangle ADE\) similarly places \(A\) on one of two lines parallel to \(DE.\)

The four possible positions of \(A\) are the vertices of a parallelogram whose center is the intersection of \(y=0\) with line \(DE,\) namely \((300,0).\) Hence the sum of the four \(x\)-coordinates is \(4\cdot 300=1200.\)

Thus, the correct answer is E.

20 答案 B

切角会从每条棱的两端各去掉长度 \(x\)。每个八边形的边长为 \(x\sqrt2\)，且棱长满足 \(1=2x+x\sqrt2\)，所以 \[x=\frac{1}{2+\sqrt2}=\frac{2-\sqrt2}{2}\text{。}\]

每个被切去的角都是三条互相垂直的棱长为 \(x\) 的四面体，体积为 \(\tfrac16 x^3\)。共有 \(8\) 个角，所以总体积为 \[ \begin{aligned} &8\cdot\tfrac16 x^3 \\ &=\tfrac43\left(\tfrac{2-\sqrt2}{2}\right)^3 \\ &=\frac{10-7\sqrt2}{3} \end{aligned}\text{。}\]

因此，正确答案是 B。

English

Slicing removes two equal segments of length \(x\) from each edge. Each octagon then has side length \(x\sqrt2,\) and the edge satisfies \(1=2x+x\sqrt2,\) so \[x=\frac{1}{2+\sqrt2}=\frac{2-\sqrt2}{2}.\]

Each removed corner is a tetrahedron with three mutually perpendicular legs of length \(x,\) so its volume is \(\tfrac16 x^3.\) There are \(8\) corners, giving total volume \[ \begin{aligned} &8\cdot\tfrac16 x^3 \\ &=\tfrac43\left(\tfrac{2-\sqrt2}{2}\right)^3 \\ &=\frac{10-7\sqrt2}{3}. \end{aligned} \]

Thus, the correct answer is B.

21 答案 A

零点之积为 \(\tfrac ca\)，零点之和为 \(-\tfrac ba\)。令它们相等，得 \(c=-b\)。

于是系数之和为 \(a+b+c=a\)，这就是 \(x^2\) 的系数。

其他选项一般不成立：例如 \(f(x)=-2x^2-4x+4\) 的公共值为 \(-2\)，但 \(x\) 的系数是 \(-4\)，\(y\) 截距是 \(4\)，\(x\) 截距是 \(-1\pm\sqrt3\)，它们的平均数是 \(-1\)。

因此，正确答案是 A。

English

The product of the zeros is \(\tfrac ca\) and the sum of the zeros is \(-\tfrac ba.\) Equating them gives \(c=-b.\)

Then the sum of the coefficients is \(a+b+c=a,\) which is the coefficient of \(x^2.\)

The other choices fail in general: for \(f(x)=-2x^2-4x+4\) the common value is \(-2,\) but the coefficient of \(x\) is \(-4,\) the \(y\)-intercept is \(4,\) the \(x\)-intercepts are \(-1\pm\sqrt3,\) and their mean is \(-1.\)

Thus, the correct answer is A.

22 答案 D

对 \(n\le 2007\) 有 \(S(n)\le S(1999)=28\)，进而 \(S(S(n))\le S(28)=10\)。所以任何解都满足 \(n\ge 2007-28-10=1969\)。

又因为 \(n\)、\(S(n)\) 和 \(S(S(n))\) 模 \(9\) 同余，而 \(2007\) 是 \(9\) 的倍数，所以三者都必须是 \(3\) 的倍数。

检查 \(1969\) 到 \(2007\) 之间的 \(3\) 的倍数（许多数会因 \(n+S(n)\) 已超过 \(2007\) 而被排除），剩下 \(1977,1980,1983\) 和 \(2001\)。共有 \(4\) 个值。

因此，正确答案是 D。

English

For \(n\le 2007,\) \(S(n)\le S(1999)=28,\) and then \(S(S(n))\le S(28)=10.\) So any solution has \(n\ge 2007-28-10=1969.\)

Also \(n,\) \(S(n),\) and \(S(S(n))\) are congruent modulo \(9,\) and \(2007\) is a multiple of \(9,\) so all three must be multiples of \(3.\)

Checking the multiples of \(3\) between \(1969\) and \(2007\) (many are eliminated because \(n+S(n)\) already exceeds \(2007\)) leaves \(1977,1980,1983,\) and \(2001.\) That is \(4\) values.

Thus, the correct answer is D.

23 答案 A

设 \(A=(p,\log_a p)\) 且 \(B=(q,2\log_a q)\)。因为 \(AB\) 水平，\(\log_a p=2\log_a q=\log_a q^2\)，所以 \(p=q^2\)。

边长为 \(6=|p-q|=|q^2-q|\)，其唯一正解是 \(q=3\)。

因为 \(C=(q,3\log_a q)\)，竖直边给出 \(BC=6=\log_a q=\log_a 3\)。因此 \(a^6=3\)，所以 \(a=\sqrt[6]{3}\)。

因此，正确答案是 A。

English

Let \(A=(p,\log_a p)\) and \(B=(q,2\log_a q).\) Since \(AB\) is horizontal, \(\log_a p=2\log_a q=\log_a q^2,\) so \(p=q^2.\)

The side length is \(6=|p-q|=|q^2-q|,\) whose only positive solution is \(q=3.\)

Since \(C=(q,3\log_a q),\) the vertical side gives \(BC=6=\log_a q=\log_a 3.\) Thus \(a^6=3,\) so \(a=\sqrt[6]{3}.\)

Thus, the correct answer is A.

24 答案 D

在每个 \(\sin nx\ge 0\) 的区间内，\(\sin x\) 和 \(\sin nx\) 的图像相交两次，除非它们在那里共享值 \(1\)，此时只相交一次。数出这些波峰并加上端点 \((\pi,0)\)，可得到以下结论。

当 \(n\) 为偶数或 \(n\equiv 3\pmod 4\) 时，\(F(n)=n+1\)；当 \(n\equiv 1\pmod 4\) 时， \(F(n)=n\)。

因此 \[ \begin{aligned} &\sum_{n=2}^{2007}F(n) \\ &=\sum_{n=2}^{2007}(n+1) \\ &\quad {}-\#\{n\equiv 1\!\!\pmod 4\}\text{。} \end{aligned} \] 第一个和为 \(2{,}017{,}033\)，该范围内有 \(501\) 个 \(n\equiv 1\pmod 4\)，所以结果为 \(2{,}017{,}033-501=2{,}016{,}532\)。

因此，正确答案是 D。

English

On each interval where \(\sin nx\ge 0,\) the graphs of \(\sin x\) and \(\sin nx\) meet twice, unless they share the value \(1\) there, in which case they meet once. Counting the humps and the endpoint at \((\pi,0)\) gives

\(F(n)=n+1\) when \(n\) is even or \(n\equiv 3\pmod 4,\) and \(F(n)=n\) when \(n\equiv 1\pmod 4.\)

Thus \[ \begin{aligned} &\sum_{n=2}^{2007}F(n) \\ &=\sum_{n=2}^{2007}(n+1) \\ &\quad {}-\#\{n\equiv 1\!\!\pmod 4\}. \end{aligned} \] The first sum is \(2{,}017{,}033,\) and there are \(501\) values \(n\equiv 1\pmod 4\) in the range, giving \(2{,}017{,}033-501=2{,}016{,}532.\)

Thus, the correct answer is D.

25 答案 E

设 \(c_n\) 为 \(\{1,\ldots,n\}\) 的稀疏子集个数。一个稀疏子集要么不包含 \(n\)（有 \(c_{n-1}\) 种），要么包含 \(n\)，此时它不能包含 \(n-1\) 和 \(n-2\)（有 \(c_{n-3}\) 种）。

因此 \(c_n=c_{n-1}+c_{n-3}\)，且 \(c_1=2\)，\(c_2=3\)，\(c_3=4\)。

这个数列继续为 \(6,9,13,19,28,41,60,88,129\)，所以 \(c_{12}=129\)。

因此，正确答案是 E。

English

Let \(c_n\) be the number of spacy subsets of \(\{1,\ldots,n\}.\) A spacy subset either omits \(n\) (there are \(c_{n-1}\) of these) or contains \(n,\) in which case it omits \(n-1\) and \(n-2\) (there are \(c_{n-3}\) of these).

Hence \(c_n=c_{n-1}+c_{n-3},\) with \(c_1=2,\) \(c_2=3,\) \(c_3=4.\)

The sequence continues \(6,9,13,19,28,41,60,88,129,\) so \(c_{12}=129.\)

Thus, the correct answer is E.