2014A AMC 12 真题（P版）

1 答案 C

括号内的和为 \(\dfrac12+\dfrac15+\dfrac{1}{10}=\dfrac{5+2+1}{10}=\dfrac{4}{5}\)。

它的倒数是 \(\dfrac54\)，所以原式等于 \(10\cdot\dfrac54=\dfrac{25}{2}\)。

所以正确答案是 C。

English

The sum inside the parentheses is \(\dfrac12+\dfrac15+\dfrac{1}{10}=\dfrac{5+2+1}{10}=\dfrac{4}{5}.\)

Its reciprocal is \(\dfrac54,\) so the expression equals \(10\cdot\dfrac54=\dfrac{25}{2}.\)

Thus, the correct answer is C.

2 答案 B

儿童票半价，所以 \(5\) 张成人票和 \(4\) 张儿童票相当于 \(5+\tfrac12\cdot4=7\) 张成人票，因此一张成人票价格为 \(\dfrac{24.50}{7}=\$3.50\)。

第二次购买相当于 \(8+\tfrac12\cdot6=11\) 张成人票，费用为 \(11\cdot 3.50=\$38.50\)。

所以正确答案是 B。

English

Since a child ticket is half an adult ticket, \(5\) adult and \(4\) child tickets equal \(5+\tfrac12\cdot4=7\) adult tickets, so one adult ticket costs \(\dfrac{24.50}{7}=\$3.50.\)

The second purchase equals \(8+\tfrac12\cdot6=11\) adult tickets, costing \(11\cdot 3.50=\$38.50.\)

Thus, the correct answer is B.

3 答案 B

若橙色先出现，则蓝色和黄色不能相邻，迫使顺序为橙、蓝、红、黄。

若蓝色先出现，则黄色可以在第三或第四个位置（不能在第二个位置以避免相邻），得到蓝、橙、黄、红和蓝、橙、红、黄。

这就是全部 \(3\) 种有效顺序。

所以正确答案是 B。

English

If orange comes first, then blue and yellow cannot be adjacent, forcing the order orange, blue, red, yellow.

If blue comes first, yellow can be in the third or fourth position (never second, to avoid adjacency), giving blue, orange, yellow, red and blue, orange, red, yellow.

These are the only \(3\) valid orderings.

Thus, the correct answer is B.

4 答案 A

产奶速率为每头奶牛每天 \(\dfrac{b}{ac}\) 加仑。

所以 \(d\) 头奶牛在 \(e\) 天内产 \(\dfrac{b}{ac}\cdot d\cdot e=\dfrac{bde}{ac}\) 加仑。

所以正确答案是 A。

English

The rate is \(\dfrac{b}{ac}\) gallons per cow per day.

So \(d\) cows over \(e\) days produce \(\dfrac{b}{ac}\cdot d\cdot e=\dfrac{bde}{ac}\) gallons.

Thus, the correct answer is A.

5 答案 C

其余 \(25\%\) 的学生得 \(100\) 分。由于 \(45\%\) 的学生得分不超过 \(80\)，而 \(75\%\) 的学生得分不超过 \(90\)，所以中位数是 \(90\)。

平均数为 \[ \begin{aligned} &0.10(70)+0.35(80)\\ &\quad{}+0.30(90)+0.25(100)\\ &\quad{}=7+28+27+25=87 \end{aligned}\text{。}\]

两者之差为 \(90-87=3\)。

所以正确答案是 C。

English

The remaining \(25\%\) scored \(100.\) Since \(45\%\) scored at most \(80\) and \(75\%\) scored at most \(90,\) the median is \(90.\)

The mean is \[ \begin{aligned} &0.10(70)+0.35(80)\\ &\quad{}+0.30(90)+0.25(100)\\ &\quad{}=7+28+27+25=87. \end{aligned} \]

The difference is \(90-87=3.\)

Thus, the correct answer is C.

6 答案 D

设较大的数为 \(10a+b\)，则 \[ \begin{aligned} &(10a+b)-(10b+a)=9(a-b)\\ &\quad{}=5(a+b)\text{，} \end{aligned} \] 化简得 \(2a=7b\)。

满足条件的非零数字只有 \(a=7\)、\(b=2\)，所以这个数是 \(72\)，倒序数是 \(27\)。

两数之和为 \(72+27=99\)。

所以正确答案是 D。

English

Let the larger number be \(10a+b.\) Then \[ \begin{aligned} &(10a+b)-(10b+a)=9(a-b)\\ &\quad{}=5(a+b), \end{aligned} \] which simplifies to \(2a=7b.\)

The only nonzero digits satisfying this are \(a=7\) and \(b=2,\) so the number is \(72\) and its reverse is \(27.\)

Their sum is \(72+27=99.\)

Thus, the correct answer is D.

7 答案 A

把各项写成 \(3\) 的幂，前三项为 \(3^{\frac{1}{2}}\)、\(3^{\frac{1}{3}}\)、\(3^{\frac{1}{6}}\)。公比是 \(\dfrac{3^{\frac{1}{3}}}{3^{\frac{1}{2}}}=3^{-\frac{1}{6}}\)。

第四项为 \(3^{\frac{1}{6}}\cdot3^{-\frac{1}{6}}=3^{0}=1\)。

所以正确答案是 A。

English

Writing the terms as powers of \(3,\) they are \(3^{\frac{1}{2}},\) \(3^{\frac{1}{3}},\) \(3^{\frac{1}{6}}.\) The common ratio is \(\dfrac{3^{\frac{1}{3}}}{3^{\frac{1}{2}}}=3^{-\frac{1}{6}}.\)

The fourth term is \(3^{\frac{1}{6}}\cdot3^{-\frac{1}{6}}=3^{0}=1.\)

Thus, the correct answer is A.

8 答案 C

当价格 \(P\gt100\) 时，三张优惠券的降价金额分别为 \(\dfrac{P}{10}\)、\(20\) 和 \(\dfrac{18}{100}(P-100)\)。

优惠券 \(1\) 优于优惠券 \(2\) 需要 \(\dfrac{P}{10}\gt20\)，即 \(P\gt200\)。优惠券 \(1\) 优于优惠券 \(3\) 需要 \(\dfrac{P}{10}\gt\dfrac{18}{100}(P-100)\)，即 \(P\lt225\)。

所列价格中，只有 \(\$219.95\) 落在 \((200,225)\) 内。

所以正确答案是 C。

English

For a price \(P\gt100,\) the reductions are \(\dfrac{P}{10},\) \(20,\) and \(\dfrac{18}{100}(P-100).\)

Coupon \(1\) beats coupon \(2\) when \(\dfrac{P}{10}\gt20,\) that is \(P\gt200.\) Coupon \(1\) beats coupon \(3\) when \(\dfrac{P}{10}\gt\dfrac{18}{100}(P-100),\) that is \(P\lt225.\)

The only listed price in \((200,225)\) is \(\$219.95.\)

Thus, the correct answer is C.

9 答案 B

整数 \(a,a+1,a+2,a+3,a+4\) 的平均数为 \(a+2\)，所以 \(b=a+2\)。

从 \(b\) 开始的五个整数平均数为 \(b+2=(a+2)+2=a+4\)。

所以正确答案是 B。

English

The integers \(a,a+1,a+2,a+3,a+4\) have average \(a+2,\) so \(b=a+2.\)

The integers starting at \(b\) have average \(b+2=(a+2)+2=a+4.\)

Thus, the correct answer is B.

10 答案 B

等边三角形的面积为 \(\dfrac{\sqrt3}{4}\)。每个等腰三角形的底边为 \(1\)，设其高为 \(h\)，则 \(3\cdot\dfrac12 h=\dfrac{\sqrt3}{4}\)，所以 \(h=\dfrac{\sqrt3}{6}\)。

一条腰是从顶点到一个底边端点的斜边，因此其长度为 \[ \begin{gathered} \sqrt{\left(\dfrac12\right)^2+\left(\dfrac{\sqrt3}{6}\right)^2}\\ =\sqrt{\dfrac14+\dfrac{1}{12}}\\ =\sqrt{\dfrac13}=\dfrac{\sqrt3}{3} \end{gathered}\text{。}\]

所以正确答案是 B。

English

The equilateral triangle has area \(\dfrac{\sqrt3}{4}.\) Each isosceles triangle has base \(1\) and height \(h,\) so \(3\cdot\dfrac12 h=\dfrac{\sqrt3}{4},\) giving \(h=\dfrac{\sqrt3}{6}.\)

A congruent side is the hypotenuse from the apex to a base endpoint: \[ \begin{gathered} \sqrt{\left(\dfrac12\right)^2+\left(\dfrac{\sqrt3}{6}\right)^2}\\ =\sqrt{\dfrac14+\dfrac{1}{12}}\\ =\sqrt{\dfrac13}=\dfrac{\sqrt3}{3}. \end{gathered} \]

Thus, the correct answer is B.

11 答案 C

设第一小时后的剩余距离为 \(d\)，距离航班还有 \(t\) 小时。若继续以 \(35\) 英里每小时行驶，他会迟到一小时，所以 \(d=35(t+1)\)。改以 \(50\) 英里每小时行驶，他会提前半小时，所以 \(d=50\left(t-\tfrac12\right)\)。

令两式相等，得到 \(35t+35=50t-25\)，所以 \(t=4\)，且 \(d=175\)。

总路程为 \(175+35=210\) 英里。

所以正确答案是 C。

English

Let \(d\) be the remaining distance after one hour and \(t\) the remaining time until the flight. At \(35\) mph he would be an hour late, so \(d=35(t+1).\) At \(50\) mph he is half an hour early, so \(d=50\left(t-\tfrac12\right).\)

Setting these equal gives \(35t+35=50t-25,\) so \(t=4\) and \(d=175.\)

The total distance is \(175+35=210\) miles.

Thus, the correct answer is C.

12 答案 D

设对应 \(30^\circ\) 弧的圆半径为 \(R\)，对应 \(60^\circ\) 弧的圆半径为 \(r\)。公共弦的长度满足 \(2R\sin15^\circ=2r\sin30^\circ\)，所以 \(\dfrac{R}{r}=\dfrac{\sin30^\circ}{\sin15^\circ}\)。

较小的圆心角对应较大的半径，因此 \(R\gt r\)。所求面积比为 \[ \begin{gathered} \left(\dfrac{R}{r}\right)^2\\ =\dfrac{1}{4\sin^2 15^\circ}\\ =\dfrac{1}{2(1-\cos30^\circ)}\\ =\dfrac{1}{2-\sqrt3}=2+\sqrt3 \end{gathered}\text{。}\]

所以正确答案是 D。

English

Let the circles have radii \(R\) (with the \(30^\circ\) arc) and \(r\) (with the \(60^\circ\) arc). The common chord has length \(2R\sin15^\circ=2r\sin30^\circ,\) so \(\dfrac{R}{r}=\dfrac{\sin30^\circ}{\sin15^\circ}.\)

The smaller central angle gives the larger radius, so \(R\gt r.\) The area ratio is \[ \begin{gathered} \left(\dfrac{R}{r}\right)^2\\ =\dfrac{1}{4\sin^2 15^\circ}\\ =\dfrac{1}{2(1-\cos30^\circ)}\\ =\dfrac{1}{2-\sqrt3}=2+\sqrt3. \end{gathered} \]

Thus, the correct answer is D.

13 答案 B

全是单人：把 \(5\) 位朋友分到 \(5\) 个房间有 \(5!=120\) 种。

一对同住：选这对有 \(\binom52=10\) 种，再把 \(4\) 个组放进房间有 \(5\cdot4\cdot3\cdot2=120\) 种，共 \(10\cdot120=1200\) 种。

两对同住：选出单独住的人有 \(5\) 种，剩余四人分成两对有 \(3\) 种，因此共有 \(15\) 种分组方式。再把这 \(3\) 组安排进房间，有 \(5\cdot4\cdot3=60\) 种方式，共 \(15\cdot60=900\) 种。

总数为 \(120+1200+900=2220\)。

所以正确答案是 B。

English

All singles: assign \(5\) friends to \(5\) rooms in \(5!=120\) ways.

One pair: choose the pair in \(\binom52=10\) ways, then place the \(4\) groups into rooms in \(5\cdot4\cdot3\cdot2=120\) ways, giving \(10\cdot120=1200.\)

Two pairs: choose the solo friend in \(5\) ways and split the rest into two pairs in \(3\) ways (\(15\) groupings), then place the \(3\) groups into rooms in \(5\cdot4\cdot3=60\) ways, giving \(15\cdot60=900.\)

The total is \(120+1200+900=2220.\)

Thus, the correct answer is B.

14 答案 C

设 \(d=b-a\gt0\)，则 \(b=a+d\)，\(c=a+2d\)。因为 \(a,c,b\) 成等比数列， \[ \begin{gathered} \dfrac{c}{a}=\dfrac{b}{c}\\ \implies(a+2d)^2=a(a+d)\text{，} \end{gathered} \] 化简得 \(3ad+4d^2=0\)，所以 \(3a+4d=0\)。

于是 \(a=-4k\)、\(d=3k\)，其中 \(k\) 为正整数，因此 \(c=a+2d=2k\)。最小值为 \(c=2\)，此时 \(a=-4\)、\(b=-1\)、\(c=2\)。

所以正确答案是 C。

English

Let \(d=b-a\gt0,\) so \(b=a+d\) and \(c=a+2d.\) Since \(a,c,b\) is geometric, \[ \begin{gathered} \dfrac{c}{a}=\dfrac{b}{c}\\ \implies(a+2d)^2=a(a+d), \end{gathered} \] which simplifies to \(3ad+4d^2=0,\) so \(3a+4d=0.\)

Then \(a=-4k\) and \(d=3k\) for a positive integer \(k,\) giving \(c=a+2d=2k.\) The smallest value is \(c=2\) (with \(a=-4,\) \(b=-1,\) \(c=2\)).

Thus, the correct answer is C.

15 答案 B

写 \(\overline{abcba}=10001a+1010b+100c\)。对所有回文数求和时，每个 \(a\in\{1,\dots,9\}\) 与 \(10\cdot10=100\) 种 \(b,c\) 搭配；每个 \(b\) 或 \(c\) 的值与另外两位有 \(9\cdot10=90\) 种搭配。

利用 \(\sum a=\sum b=\sum c=45\)，\[ \begin{gathered} \scriptsize S=45\big(10001\cdot100+1010\cdot90+100\cdot90\big)\\ =45\cdot1{,}100{,}000\\ =49{,}500{,}000 \end{gathered}\text{。}\]

\(S\) 的各位数字之和为 \(4+9+5=18\)。

所以正确答案是 B。

English

Write \(\overline{abcba}=10001a+1010b+100c.\) Summing over all palindromes, each value of \(a\in\{1,\dots,9\}\) occurs with \(10\cdot10=100\) choices of \(b,c,\) and each value of \(b\) or \(c\) occurs with \(9\cdot10=90\) choices of the other two digits.

Using \(\sum a=\sum b=\sum c=45,\) \[ \begin{gathered} \scriptsize S=45\big(10001\cdot100+1010\cdot90+100\cdot90\big)\\ =45\cdot1{,}100{,}000\\ =49{,}500{,}000. \end{gathered} \]

The sum of the digits of \(S\) is \(4+9+5=18.\)

Thus, the correct answer is B.

16 答案 D

按位乘法并处理进位可得 \(8\cdot\underbrace{88\ldots8}_{k}=7\underbrace{1\ldots1}_{k-2}04\)，其中有 \(k-2\) 个数字一。

各位数字之和为 \(7+(k-2)+0+4=k+9\)。令 \(k+9=1000\)，得到 \(k=991\)。

所以正确答案是 D。

English

By carrying out the multiplication, \(8\cdot\underbrace{88\ldots8}_{k}=7\underbrace{1\ldots1}_{k-2}04,\) which has \(k-2\) ones.

The digit sum is \(7+(k-2)+0+4=k+9.\) Setting \(k+9=1000\) gives \(k=991.\)

Thus, the correct answer is D.

17 答案 A

把盒子的一个角放在原点。每个小球位于一个角落，其球心到相邻三个面各相距 \(1\)。上方四个小球的球心构成边长为 \(2\) 的正方形，正方形中心在盒子的中轴线上，一个顶点到中心的距离为 \(\sqrt2\)。

大球球心在中轴线上，到每个上方小球球心的距离为 \(2+1=3\)，因此两者的竖直距离为 \(\sqrt{3^2-\sqrt2^2}=\sqrt7\)。

大球球心的高度是 \(\dfrac h2\)，上方小球球心的高度是 \(h-1\)，所以 \((h-1)-\dfrac h2=\sqrt7\)。由此 \(\dfrac h2=1+\sqrt7\)，并得到 \(h=2+2\sqrt7\)。

所以正确答案是 A。

English

Place the box with a corner at the origin. Each small sphere sits in a corner with center \(1\) unit from three faces. The four top small-sphere centers form a square of side \(2,\) whose center lies on the box axis; a corner of that square is \(\sqrt2\) from the center.

The big sphere’s center is on the axis, at distance \(2+1=3\) from each top small center. The vertical gap between them is \(\sqrt{3^2-\sqrt2^2}=\sqrt7.\)

The big center is at height \(\dfrac h2\) and the top small centers at height \(h-1,\) so \((h-1)-\dfrac h2=\sqrt7,\) giving \(\dfrac h2=1+\sqrt7\) and \(h=2+2\sqrt7.\)

Thus, the correct answer is A.

18 答案 C

从外向内看，\(f\) 有定义当且仅当 \(\log_4\!\left(\log_{\frac{1}{4}}\!\left(\log_{16}\!\left(\log_{\frac{1}{16}}x\right)\right)\right)\) \(\gt0\)，这等价于 \(\log_{\frac{1}{4}}\!\left(\log_{16}\!\left(\log_{\frac{1}{16}}x\right)\right)\gt1\)。

因为底数 \(\tfrac14\lt1\)，这意味着 \(0\lt\log_{16}\!\left(\log_{\frac{1}{16}}x\right)\lt\tfrac14\)，于是 \(1\lt\log_{\frac{1}{16}}x\lt16^{\frac{1}{4}}=2\)。

又因为 \(\tfrac{1}{16}\lt1\)，不等号反向，得到 \(\left(\tfrac{1}{16}\right)^2\lt x\lt\left(\tfrac{1}{16}\right)^1\)，即 \(\tfrac{1}{256}\lt x\lt\tfrac{1}{16}\)。区间长度为 \(\tfrac{1}{16}-\tfrac{1}{256}=\tfrac{15}{256}\)，所以 \(m+n=15+256=271\)。

所以正确答案是 C。

English

Working from the outside, \(f\) is defined exactly when \(\log_4\!\left(\log_{\frac{1}{4}}\!\left(\log_{16}\!\left(\log_{\frac{1}{16}}x\right)\right)\right)\) \(\gt0,\) which is equivalent to \(\log_{\frac{1}{4}}\!\left(\log_{16}\!\left(\log_{\frac{1}{16}}x\right)\right)\gt1.\)

Since the base \(\tfrac14\lt1,\) this means \(0\lt\log_{16}\!\left(\log_{\frac{1}{16}}x\right)\lt\tfrac14,\) hence \(1\lt\log_{\frac{1}{16}}x\lt16^{\frac{1}{4}}=2.\)

As \(\tfrac{1}{16}\lt1,\) this reverses to \(\left(\tfrac{1}{16}\right)^2\lt x\lt\left(\tfrac{1}{16}\right)^1,\) i.e. \(\tfrac{1}{256}\lt x\lt\tfrac{1}{16}.\) The length is \(\tfrac{1}{16}-\tfrac{1}{256}=\tfrac{15}{256},\) so \(m+n=15+256=271.\)

Thus, the correct answer is C.

19 答案 E

若整数 \(x\) 是根，则 \(k=-\left(5x+\dfrac{12}{x}\right)\)，所以 \(x\ne0\)。对 \(x\ge2\)，\(|k|=5|x|+\dfrac{12}{|x|}\) 递增；\(|x|=39\) 时 \(|k|\approx195.3\lt200\) 而 \(|x|=40\) 时 \(|k|\gt200\)。

因此 \(x\) 可取 \(\pm1,\pm2,\dots,\pm39\)，共 \(78\) 个值。若两个不同整数 \(a\ne b\) 给出相同的 \(k\) 则会使 \(5a+\tfrac{12}{a}=5b+\tfrac{12}{b}\)，从而推出 \(5ab=12\)。这没有整数解，所以所有 \(78\) 个 \(k\) 互不相同。

所以正确答案是 E。

English

If an integer \(x\) is a root, then \(k=-\left(5x+\dfrac{12}{x}\right),\) so \(x\ne0.\) For \(x\ge2,\) \(|k|=5|x|+\dfrac{12}{|x|}\) increases, and \(|x|=39\) gives \(|k|\approx195.3\lt200,\) while \(|x|=40\) gives \(|k|\gt200.\)

Thus \(x\) ranges over \(\pm1,\pm2,\dots,\pm39,\) which is \(78\) values. If two integers \(a\ne b\) gave the same \(k,\) then \(5a+\tfrac{12}{a}=5b+\tfrac{12}{b}\) forces \(5ab=12,\) which has no integer solutions, so all \(78\) values of \(k\) are distinct.

Thus, the correct answer is E.

20 答案 D

将 \(B\) 关于直线 \(AC\) 反射到 \(B'\)，将 \(C\) 关于直线 \(AB\) 反射到 \(C'\)。于是 \(BE=B'E\)、\(CD=C'D\)，从而 \(BE+DE+CD\) \(=B'E+ED+DC'\)，这是从 \(B'\) 到 \(C'\) 的一条折线路径。

当这条路径成为直线段 \(B'C'\) 时长度最小。此时 \(AB'=AB=10\)、\(AC'=AC=6\)，并且 \(\angle B'AC'=3\cdot40^\circ=120^\circ\)。

由余弦定理， \[ \begin{gathered} B'C'^2=10^2+6^2\\ {}-2\cdot10\cdot6\cos120^\circ\\ =136+60\\ =196\text{，} \end{gathered} \] 所以 \(B'C'=14\)。

所以正确答案是 D。

English

Reflect \(B\) across line \(AC\) to get \(B',\) and reflect \(C\) across line \(AB\) to get \(C'.\) Then \(BE=B'E\) and \(CD=C'D,\) so \(BE+DE+CD\) \(=B'E+ED+DC',\) a broken path from \(B'\) to \(C'.\)

This is minimized when the path is the straight segment \(B'C'.\) We have \(AB'=AB=10,\) \(AC'=AC=6,\) and \(\angle B'AC'=3\cdot40^\circ=120^\circ.\)

By the Law of Cosines, \[ \begin{gathered} B'C'^2=10^2+6^2\\ {}-2\cdot10\cdot6\cos120^\circ\\ =136+60\\ =196, \end{gathered} \] so \(B'C'=14.\)

Thus, the correct answer is D.

21 答案 A

写 \(x=n+r\)，其中 \(n\) 为整数且 \(1\le n\le2013\)，\(0\le r\lt1\)。于是 \(f(x)=n\left(2014^{\,r}-1\right)\)，而 \(f(x)\le1\) 等价于 \(2014^{\,r}\le1+\dfrac1n\)，也就是 \(0\le r\le\log_{2014}\dfrac{n+1}{n}\)。

因此每个 \(n\) 对应的区间长度为 \(\log_{2014}\dfrac{n+1}{n}\)，所以总长度为 \[ \begin{gathered} \sum_{n=1}^{2013}\log_{2014}\dfrac{n+1}{n}\\ =\log_{2014}\!\left(\dfrac21\cdot\dfrac32\cdots\dfrac{2014}{2013}\right)\\ =\log_{2014}2014=1 \end{gathered}\text{。}\]

所以正确答案是 A。

English

Write \(x=n+r\) with integer \(n\) (\(1\le n\le2013\)) and \(0\le r\lt1.\) Then \(f(x)=n\left(2014^{\,r}-1\right),\) and \(f(x)\le1\) becomes \(2014^{\,r}\le1+\dfrac1n,\) i.e. \(0\le r\le\log_{2014}\dfrac{n+1}{n}.\)

Each \(n\) contributes an interval of length \(\log_{2014}\dfrac{n+1}{n},\) so the total is \[ \begin{gathered} \sum_{n=1}^{2013}\log_{2014}\dfrac{n+1}{n}\\ =\log_{2014}\!\left(\dfrac21\cdot\dfrac32\cdots\dfrac{2014}{2013}\right)\\ =\log_{2014}2014=1. \end{gathered} \]

Thus, the correct answer is A.

22 答案 B

因为 \(2^2\lt5\lt2^3\)，每个区间 \((5^n,5^{n+1})\) 内含有两个或三个 \(2\) 的幂。不等式链 \(5^n\lt2^m\lt2^{m+2}\lt5^{n+1}\) 恰好在该区间含有三个连续的 \(2\) 的幂时成立，而且此时 \(m\) 唯一。

设 \(d\) 和 \(t\) 分别为 \(0\le n\le866\) 时区间 \((5^n,5^{n+1})\) 内含两个和三个 \(2\) 的幂的区间数。由于 \(2^{2013}\lt5^{867}\lt2^{2014}\)，这些区间内共有 \(2013\) 个 \(2\) 的幂，因此 \(d+t=867\)，且 \(2d+3t=2013\)。

解得 \(t=2013-2\cdot867=279\)。

所以正确答案是 B。

English

Because \(2^2\lt5\lt2^3,\) each interval \((5^n,5^{n+1})\) contains either two or three powers of \(2.\) The chain \(5^n\lt2^m\lt2^{m+2}\lt5^{n+1}\) holds exactly when the interval contains three consecutive powers of \(2,\) and then there is a unique such \(m.\)

Let \(d\) and \(t\) be the numbers of intervals \((5^n,5^{n+1})\) for \(0\le n\le866\) containing two and three powers of \(2,\) respectively. Since \(2^{2013}\lt5^{867}\lt2^{2014}\) there are \(2013\) powers of \(2\) in total, giving \(d+t=867\) and \(2d+3t=2013.\)

Solving, \(t=2013-2\cdot867=279.\)

Thus, the correct answer is B.

23 答案 B

把循环节按每两位一组读取（即用 \(100\) 进制），\(\dfrac{1}{9801}=\dfrac{1}{99^2}\) 的展开是 \(00,01,02,\ldots\)，因为 \(\dfrac{1}{(100-1)^2}=\sum_{k\ge1}k\cdot100^{-k}\)。令 \(a_j\) 为循环节的第 \(j\) 个 \(100\) 进制数字。将循环节乘以 \(99^2\)，首先得到 \(a_0=99\)。随后的进位给出 \(a_1=97\)，此后不再进位，并依次得到 \(a_j=98-j\)，其中 \(1\le j\le98\)。下一个数字又是 \(99\)，所以循环节是 \(00,01,02,\ldots,96,97,99\)，其中缺少 \(98\)。

如果从 \(00\) 到 \(99\) 的所有数块都出现，数字和将是 \((0+1+\cdots+9)\cdot20=900\)。去掉缺少的 \(98\) 要减去 \(9+8\)，得到 \(900-9-8=883\)。

因此，正确答案是 B。

English

Reading the block in pairs of digits (base \(100\)), \(\dfrac{1}{9801}=\dfrac{1}{99^2}\) expands as \(00,01,02,\ldots,\) since \(\dfrac{1}{(100-1)^2}=\sum_{k\ge1}k\cdot100^{-k}.\) Let \(a_j\) be the \(j\)th base-\(100\) digit of the repeating block. Multiplying the block by \(99^2\) shows first that \(a_0=99.\) The resulting carry gives \(a_1=97,\) after which there is no carry and successively \(a_j=98-j\) for \(1\le j\le98.\) The next digit is again \(99,\) so the period is \(00,01,02,\ldots,96,97,99,\) with \(98\) omitted.

If the blocks \(00\) through \(99\) all appeared, the digit sum would be \((0+1+\cdots+9)\cdot20=900.\) Removing the missing \(98\) subtracts \(9+8,\) giving \(900-9-8=883.\)

Thus, the correct answer is B.

24 答案 C

若 \(f_{n-1}(x)=\pm k\)，则 \(f_n(x)=k-1\)。因此，如果 \(f_0(x)=\pm k\)，其中 \(k\) 是非负整数，那么 \(f_k(x)=0\)，此后的值按 \(0,-1,0,\ldots\) 交替出现。所以，\(f_{100}(x)=0\) 当且仅当对某个整数 \(-50\le k\le50\)，有 \(f_0(x)=2k\)。

函数 \(f_0(x)=x+|x-100|\) \(-|x+100|\) 在 \(x\lt-100\) 时等于 \(x+200\)，在 \(-100\le x\lt100\) 时等于 \(-x\)，在 \(x\ge100\) 时等于 \(x-200\)。它的图像是分段直线，转折点为 \((-100,100)\) 和 \((100,-100)\)。

当 \(-49\le k\le49\) 时，直线 \(y=2k\) 与图像有三个交点；当 \(k=\pm50\) 时，各有两个交点。交点总数为 \(99\cdot3+2\cdot2=301\)。

所以正确答案是 C。

English

If \(f_{n-1}(x)=\pm k,\) then \(f_n(x)=k-1.\) So if \(f_0(x)=\pm k\) for a nonnegative integer \(k,\) then \(f_k(x)=0,\) after which the sequence alternates \(0,-1,0,\ldots\) Thus \(f_{100}(x)=0\) exactly when \(f_0(x)=2k\) for some integer \(-50\le k\le50.\)

Now \(f_0(x)=x+|x-100|\) \(-|x+100|\) equals \(x+200\) for \(x\lt-100,\) \(-x\) for \(-100\le x\lt100,\) and \(x-200\) for \(x\ge100.\) Its graph is piecewise linear with turning points \((-100,100)\) and \((100,-100).\)

A line \(y=2k\) meets this graph three times for \(-49\le k\le49\) and twice for \(k=\pm50.\) The total is \(99\cdot3+2\cdot2=301.\)

Thus, the correct answer is C.

25 答案 B

因为 \((0,0)\) 是 \(A=(4,3)\) 与 \(B=(-4,-3)\) 的中点，所以线段 \(AB\) 是通径。准线与 \(AB\) 平行，并位于焦点另一侧、相距 \(5\)，其方程为 \(4y-3x+25=0\)。

令点到焦点和准线的距离相等，得到 \((4x+3y)^2\) \(=25\big(25+2(4y-3x)\big)\)。令 \(4x+3y=5s\)，可推出 \(s\) 是 \(5\) 的倍数；再令 \(s=5t\)，可推出 \(t\) 为奇数。写成 \(t=2u+1\) 后，所有整数点可表示为 \[ \begin{aligned} x&=-6u^2+2u+4,\\ y&=8u^2+14u+3 \end{aligned}\text{。}\]

于是 \(|4x+3y|=|50u+25|\le1000\) 等价于 \(|2u+1|\le39\)，即 \(-20\le u\le19\)。因此共有 \(40\) 个整数坐标点。

所以正确答案是 B。

English

Since \((0,0)\) is the midpoint of \(A=(4,3)\) and \(B=(-4,-3),\) the segment \(AB\) is the latus rectum, so the directrix is parallel to \(AB\) at distance \(5\) on the far side, namely \(4y-3x+25=0.\)

Equating distances to focus and directrix gives \((4x+3y)^2\) \(=25\big(25+2(4y-3x)\big).\) Writing \(4x+3y=5s\) forces \(s\) to be a multiple of \(5,\) and \(s=5t\) forces \(t\) odd; with \(t=2u+1\) the integer points are \[ \begin{aligned} x&=-6u^2+2u+4,\\ y&=8u^2+14u+3. \end{aligned} \]

Then \(|4x+3y|=|50u+25|\le1000\) iff \(|2u+1|\le39,\) i.e. \(-20\le u\le19.\) That gives \(40\) lattice points.

Thus, the correct answer is B.