1985 AMC 12 真题（P版）

1 答案 A

由所给方程得到 \(2x=7\)，所以 \(x=\frac{7}{2}\)。因此 \(4x+1=4(\frac{7}{2})+1=15\)。

所以正确答案是 A。

English

The given equation gives \(2x=7,\) so \(x=\frac{7}{2}.\) Therefore \(4x+1=4(\frac{7}{2})+1=15.\)

Thus the correct answer is A.

2 答案 E

曲线部分是单位圆周长的 \(\frac{300^\circ}{360^\circ}=\frac{5}{6}\)，所以其长度为 \((\frac{5}{6})(2\pi)=\frac{5\pi}{3}\)。嘴巴的两条直边都是半径，总长为 \(2\)。因此周长为 \(\frac{5\pi}{3}+2\)。

所以正确答案是 E。

English

The curved part is \(\frac{300^\circ}{360^\circ}=\frac{5}{6}\) of a unit circle, so its length is \((\frac{5}{6})(2\pi)=\frac{5\pi}{3}.\) The two straight sides of the mouth are radii of total length \(2.\) Hence the perimeter is \(\frac{5\pi}{3}+2.\)

Thus the correct answer is E.

3 答案 D

斜边长为 \(13\)。由于 \(AM=12\)，点 \(M\) 距 \(A\) 为 \(12\)。又因为 \(BN=5\)，所以 \(AN=AB-BN=13-5=8\)。因此 \(MN=AM-AN\) \(=12-8=4\)。

所以正确答案是 D。

English

The hypotenuse has length \(13.\) Since \(AM=12,\) point \(M\) is \(12\) units from \(A.\) Also \(BN=5,\) so \(AN=AB-BN=13-5=8.\) Therefore \(MN=AM-AN\) \(=12-8=4.\)

Thus the correct answer is D.

4 答案 C

若有 \(p\) 枚一美分硬币，则有 \(2p\) 枚十美分硬币和 \(6p\) 枚二十五美分硬币。总价值为 \(p+10(2p)+25(6p)=171p\) 美分。所列美元数中，\(34200=171\cdot200\) 美分是可能的。

所以正确答案是 C。

English

If there are \(p\) pennies, there are \(2p\) dimes and \(6p\) quarters. Their total value is \(p+10(2p)+25(6p)=171p\) cents. Of the listed dollar amounts, \(34200=171\cdot200\) cents is possible.

Thus the correct answer is C.

5 答案 E

完整和式为 \[ \begin{aligned} \frac{\substack{60+30+20\\{}+15+12+10}}{120} &=\frac{147}{120}\text{。} \end{aligned} \] 因此，删去的两项之和必须为 \(\frac{27}{120}=\frac{9}{40}\)。由于 \(\frac{1}{8}+\frac{1}{10}\) \(=\frac{5}{40}+\frac{4}{40}\) \(=\frac{9}{40}\)，应删去这两项。

所以正确答案是 E。

English

The full sum is \[ \begin{aligned} \frac{\substack{60+30+20\\{}+15+12+10}}{120} &=\frac{147}{120}. \end{aligned} \] The removed terms must therefore total \(\frac{27}{120}=\frac{9}{40}.\) Since \(\frac{1}{8}+\frac{1}{10}\) \(=\frac{5}{40}+\frac{4}{40}\) \(=\frac{9}{40},\) those are the required terms.

Thus the correct answer is E.

6 答案 B

因为每名学生被选中的可能性相同，男生人数是女生人数的 \(\frac{2}{3}\)。所以男生与女生人数之比为 \(2:3\)，男生占全班的 \(\frac{2}{2+3}=\frac{2}{5}\)。

所以正确答案是 B。

English

Because every student is equally likely to be selected, the number of boys is \(\frac{2}{3}\) of the number of girls. Thus the boys-to-girls ratio is \(2:3,\) and boys make up \(\frac{2}{2+3}=\frac{2}{5}\) of the class.

Thus the correct answer is B.

7 答案 E

从右边开始结合，先得到 \(c+d\)。用 \(b\) 减去这一结果，得到 \(b-(c+d)=b-c-d\)。最后用 \(a\) 除以该数，得到 \(\frac{a}{b-c-d}\)。

所以正确答案是 E。

English

Grouping from the right first gives \(c+d.\) Subtracting this result from \(b\) gives \(b-(c+d)=b-c-d.\) Finally, dividing \(a\) by that quantity gives \(\frac{a}{b-c-d}.\)

Thus the correct answer is E.

8 答案 E

两个解分别为 \(-\frac{b}{a}\) 和 \(-\frac{b'}{a'}\)。因此所给比较为 \[ -\frac ba\lt-\frac{b'}{a'}\text{。} \] 两边乘以 \(-1\)，不等号方向反转，得到 \(\frac{b}{a}\gt \frac{b'}{a'}\)，等价于 \(\frac{b'}{a'}\lt \frac{b}{a}\)。不能乘以符号未知的量 \(aa'\)。

所以正确答案是 E。

English

The two solutions are \(-\frac{b}{a}\) and \(-\frac{b'}{a'}.\) Thus the given comparison is \[ -\frac ba\lt-\frac{b'}{a'}. \] Multiplying both sides by \(-1\) reverses the inequality and gives \(\frac{b}{a}\gt \frac{b'}{a'},\) equivalently \(\frac{b'}{a'}\lt \frac{b}{a}.\) No multiplication by the unknown-sign quantity \(aa'\) is valid.

Thus the correct answer is E.

9 答案 B

每个从右向左排列的行的末项为 \(16k-1\)，下一行从第二列的 \(16k+1\) 开始。由于 \(1985=16\cdot124+1\)，它是这样一行的首项，位于第二列。

所以正确答案是 B。

English

The last number in each right-to-left row is \(16k-1,\) and the next row begins with \(16k+1\) in the second column. Since \(1985=16\cdot124+1,\) it begins such a row and lies in the second column.

Thus the correct answer is B.

10 答案 E

取一个在原点与 \(x\) 轴相切、圆心位于坐标轴上方很高处的圆。随着半径增大，在任意长的区间内，它的下弧都保持为正且可以任意接近 \(x\) 轴。在 \(y=\sin x\) 的每个正弧上，正弦曲线在两端为 \(0\)，而在中间远高于这段圆弧，从而产生两个交点。因此，选取足够大的半径便可得到超过 \(16\) 个交点。

所以正确答案是 E。

English

Take a circle tangent to the \(x\)-axis at the origin with its center high above the axis. As its radius grows, its lower arc stays positive but arbitrarily close to the \(x\)-axis over an arbitrarily long interval. On each positive arch of \(y=\sin x,\) the sine curve is \(0\) at the endpoints and rises well above this circular arc in between, producing two crossings. Choosing a sufficiently large radius therefore produces more than \(16\) intersections.

Thus the correct answer is E.

11 答案 B

元音 \(O,E\) 排在最前面共有 \(2!\) 种顺序。剩余字母 \(C,N,T,S,T\) 有 \(\frac{5!}{2!}=60\) 种可区分排列，因为两个 \(T\) 相同。总数为 \(2\cdot60=120\)。

所以正确答案是 B。

English

The vowels \(O,E\) can be ordered first in \(2!\) ways. The remaining letters \(C,N,T,S,T\) can be arranged in \(\frac{5!}{2!}=60\) distinguishable ways because the two \(T\)’s agree. The total is \(2\cdot60=120.\)

Thus the correct answer is B.

12 答案 D

完全立方数的素因子指数都能被 \(3\) 整除。不小于 \(1\)、\(2\)、\(4\) 的最小 \(3\) 的倍数依次为 \(3\)、\(3\)、\(6\)。因此最小的完全立方数为 \[ p^3q^3r^6=(pqr^2)^3\text{。} \] 所以正确答案是 D。

English

A perfect cube has prime exponents divisible by \(3.\) The least multiples of \(3\) that are at least \(1,\) \(2,\) and \(4\) are \(3,\) \(3,\) and \(6,\) respectively. Thus the smallest possible cube is \[ p^3q^3r^6=(pqr^2)^3. \] Therefore the correct answer is D.

13 答案 E

以左下角的钉子为 \((0,0)\)，四边形按顺序的顶点为 \((1,3)\)、\((4,1)\)、\((3,0)\)、\((0,1)\)。鞋带公式给出 \[ \begin{aligned} &\frac12\left|(1+0+3+0)\right.\\ &\quad\left.{}-(12+3+0+1)\right|=6\text{。} \end{aligned} \] 所以正确答案是 E。

English

With the lower-left peg as \((0,0),\) the quadrilateral’s vertices in order are \((1,3),\) \((4,1),\) \((3,0),\) and \((0,1).\) The shoelace formula gives \[ \begin{aligned} &\frac12\left|(1+0+3+0)\right.\\ &\quad\left.{}-(12+3+0+1)\right|=6. \end{aligned} \] Therefore the correct answer is E.

14 答案 C

对于一个 \(n\) 边形，三个钝角之和小于 \(3\cdot180^\circ\)，其余 \(n-3\) 个角之和至多为 \((n-3)90^\circ\)。因此 \[ \begin{aligned} (n-2)180^\circ &\lt3(180^\circ)\\ &\quad{}+(n-3)90^\circ\text{，} \end{aligned} \] 化简得 \(n\lt7\)。六边形可以达到，例如其内角为 \(150^\circ\)、\(150^\circ\)、\(150^\circ\)、\(90^\circ\)、\(90^\circ\)、\(90^\circ\)。所以最大边数为 \(6\)。

因此正确答案是 C。

English

For an \(n\)-gon, the three obtuse angles have total less than \(3\cdot180^\circ,\) while the other \(n-3\) angles have total at most \((n-3)90^\circ.\) Hence \[ \begin{aligned} (n-2)180^\circ &\lt3(180^\circ)\\ &\quad{}+(n-3)90^\circ, \end{aligned} \] which simplifies to \(n\lt7.\) Six sides are attainable, for example with interior angles \(150^\circ,\) \(150^\circ,\) \(150^\circ,\) \(90^\circ,\) \(90^\circ,\) and \(90^\circ.\) Thus the maximum is \(6.\)

Therefore the correct answer is C.

15 答案 E

代入得到 \(a^{9a}=(9a)^a\)。两边取正的 \(a\) 次方根，得 \(a^9=9a\)，所以 \(a^8=9\)。因此 \(a=9^{\frac{1}{8}}=3^{\frac{1}{4}}=\sqrt[4]{3}\)。

所以正确答案是 E。

English

Substitution gives \(a^{9a}=(9a)^a.\) Taking the positive \(a\)-th root yields \(a^9=9a,\) so \(a^8=9.\) Therefore \(a=9^{\frac{1}{8}}=3^{\frac{1}{4}}=\sqrt[4]{3}.\)

Thus the correct answer is E.

16 答案 B

因为 \(\tan(A+B)=\tan45^\circ=1\)， \[ \frac{\tan A+\tan B}{1-\tan A\tan B}=1\text{，} \] 所以 \(\tan A+\tan B\) \(=1-\tan A\tan B\)。因此 \[ \begin{aligned} &(1+\tan A)(1+\tan B)\\ &\quad=1+(1-\tan A\tan B)\\ &\qquad+\tan A\tan B\\ &\quad=2\text{。} \end{aligned} \] 所以正确答案是 B。

English

Because \(\tan(A+B)=\tan45^\circ=1,\) \[ \frac{\tan A+\tan B}{1-\tan A\tan B}=1, \] so \(\tan A+\tan B\) \(=1-\tan A\tan B.\) Therefore \[ \begin{aligned} &(1+\tan A)(1+\tan B)\\ &\quad=1+(1-\tan A\tan B)\\ &\qquad+\tan A\tan B\\ &\quad=2. \end{aligned} \] Thus the correct answer is B.

17 答案 B

设 \(D=(0,0)\)、\(B=(w,h)\)、\(A=(0,h)\)，且 \(C=(w,0)\)。因为 \(DB=3\)，有 \(w^2+h^2=9\)。\(DA\) 在 \(DB\) 上的投影长为 \(\frac{h^2}{3}\)，而第一段标记线段长为 \(1\)，所以 \(h^2=3\)。同理，\(DC\) 的投影到达第二个分点，所以 \(\frac{w^2}{3}=2\)，且 \(w^2=6\)。因此面积为 \[ wh=\sqrt{18}=3\sqrt2\approx4.2426\text{，} \] 四舍五入后为 \(4.2\)。

因此正确答案是 B。

English

Let \(D=(0,0),\) \(B=(w,h),\) \(A=(0,h)\) and \(C=(w,0).\) Since \(DB=3,\) we have \(w^2+h^2=9.\) The projection of \(DA\) onto \(DB\) has length \(\frac{h^2}{3},\) and the first marked segment has length \(1,\) so \(h^2=3.\) Similarly, the projection of \(DC\) reaches the second division point, so \(\frac{w^2}{3}=2\) and \(w^2=6.\) Hence the area is \[ wh=\sqrt{18}=3\sqrt2\approx4.2426, \] which rounds to \(4.2.\)

Therefore the correct answer is B.

18 答案 D

六袋共有 \(140\) 颗弹珠。若破损弹珠袋中有 \(c\) 颗，则其余袋的总数是乔治所得数量的三倍，所以 \(140-c\) 能被 \(3\) 整除。选项中只有 \(c=23\) 满足这一条件。它确实可以实现：乔治取 \(18+21=39\)，简取 \(19+25+34=78\)。

所以正确答案是 D。

English

The six bags total \(140.\) If the chipped bag contains \(c\) marbles, then the other bags total three times George’s amount, so \(140-c\) is divisible by \(3.\) Of the choices, only \(c=23\) has this property. It is attainable: George can take \(18+21=39,\) while Jane takes \(19+25+34=78.\)

Thus the correct answer is D.

19 答案 A

将 \(x^2=\frac{y}{A}\) 代入第二个方程，得到 \[ Ay^2-(4A+1)y+3A=0\text{。} \] 其判别式为 \(4A^2+8A+1\gt0\)。两根的积为正数 \(3\)，和为正数 \(\frac{4A+1}{A}\)，所以两根互不相同且均为正。对每个根 \(y\)，方程 \(x^2=\frac{y}{A}\) 给出两个不同的 \(x\) 值。因此恰有四个交点。

所以正确答案是 A。

English

Substituting \(x^2=\frac{y}{A}\) into the second equation gives \[ Ay^2-(4A+1)y+3A=0. \] Its discriminant is \(4A^2+8A+1\gt0.\) Its roots have positive product \(3\) and positive sum \(\frac{4A+1}{A},\) so both roots are positive and distinct. For each root \(y,\) the equation \(x^2=\frac{y}{A}\) gives two distinct values of \(x.\) Thus there are exactly four intersection points.

Therefore the correct answer is A.

20 答案 D

完全未涂漆的小正方体有 \((n-2)^3\) 个。六个面各贡献 \((n-2)^2\) 个恰有一个面涂黑的小正方体，所以相等条件给出 \[ 6(n-2)^2=(n-2)^3\text{。} \] 因为 \(n\gt2\)，两边除以 \((n-2)^2\)，得到 \(n-2=6\)，所以 \(n=8\)。

因此正确答案是 D。

English

There are \((n-2)^3\) completely unpainted cubes. Each of the six faces contributes \((n-2)^2\) cubes with exactly one painted face, so equality gives \[ 6(n-2)^2=(n-2)^3. \] Since \(n\gt2,\) division by \((n-2)^2\) gives \(n-2=6,\) hence \(n=8.\)

Therefore the correct answer is D.

21 答案 C

若底数为 \(1\)，则 \(x^2-x-1=1\)，得到 \(x=-1\) 和 \(x=2\)。若底数为 \(-1\)，则 \(x(x-1)=0\)；其中只有 \(x=0\) 使指数 \(x+2\) 为偶数。若指数为 \(0\)，则 \(x=-2\)，此时底数为 \(5\)，所以也符合条件。因此四个解为 \(-2\)、\(-1\)、\(0\)、\(2\)。

所以正确答案是 C。

English

If the base is \(1,\) then \(x^2-x-1=1,\) giving \(x=-1\) and \(x=2.\) If the base is \(-1,\) then \(x(x-1)=0;\) only \(x=0\) makes the exponent \(x+2\) even. If the exponent is \(0,\) then \(x=-2,\) and its base is \(5,\) so it also works. Thus the four solutions are \(-2,\) \(-1,\) \(0,\) and \(2.\)

Therefore the correct answer is C.

22 答案 D

弧 \(CD\) 给出 \(\angle CAD=30^\circ\)。由于 \(A,B,C\) 共线，且 \(A,O,D\) 共线，\(\angle BAO=30^\circ\)。再结合 \(\angle ABO=60^\circ\)，三角形 \(ABO\) 是 \(30^\circ\)-\(60^\circ\)-\(90^\circ\) 三角形。由于 \(BO=5\)，得到 \(AB=10\)，且 \(AO=5\sqrt3\)。

又因为 \(AD\) 是直径，所以 \(\angle ACD=90^\circ\)。在 \(30^\circ\)-\(60^\circ\)-\(90^\circ\) 三角形 \(ACD\) 中，斜边 \(AD=10\sqrt3\)，所以 \(AC=15\)。因此 \(BC=AC-AB=5\)。

所以正确答案是 D。

English

Arc \(CD\) gives \(\angle CAD=30^\circ.\) Since \(A,B,C\) are collinear and \(A,O,D\) are collinear, \(\angle BAO=30^\circ.\) With \(\angle ABO=60^\circ,\) triangle \(ABO\) is a \(30^\circ\)-\(60^\circ\)-\(90^\circ\) triangle. Because \(BO=5,\) we get \(AB=10\) and \(AO=5\sqrt3.\)

Also \(\angle ACD=90^\circ\) because \(AD\) is a diameter. In the \(30^\circ\)-\(60^\circ\)-\(90^\circ\) triangle \(ACD,\) the hypotenuse is \(AD=10\sqrt3,\) so \(AC=15.\) Therefore \(BC=AC-AB=5.\)

Thus the correct answer is D.

23 答案 C

这两个数是两个非实的三次单位根，所以 \[ x^n+y^n=2\cos\left(\frac{2\pi n}{3}\right)\text{。} \] 该式等于 \(-1\)（当 \(n\) 不是 \(3\) 的倍数时），但等于 \(2\)（当 \(n\) 是 \(3\) 的倍数时）。所列指数中只有 \(9\) 能被 \(3\) 整除，所以对应的等式是不正确的一项。

因此正确答案是 C。

English

The numbers are the two nonreal cube roots of unity, so \[ x^n+y^n=2\cos\left(\frac{2\pi n}{3}\right). \] This equals \(-1\) when \(n\) is not divisible by \(3,\) but equals \(2\) when \(n\) is divisible by \(3.\) Of the listed exponents, only \(9\) is divisible by \(3,\) so its displayed equation is the one that is not correct.

Therefore the correct answer is C.

24 答案 C

选中 \(2\) 的概率为 \(\log_{10}(\frac{3}{2})\)，所以其两倍为 \(\log_{10}(\frac{9}{4})\)。对于从 \(4\) 到 \(8\) 的数字，概率和逐项消去： \[ \begin{aligned} &\sum_{d=4}^{8} \bigl(\log_{10}(d+1)\\ &\qquad{}-\log_{10}d\bigr) =\log_{10}\frac94\text{。} \end{aligned} \] 因此该集合的概率恰为选中数字 \(2\) 的概率的两倍。

所以正确答案是 C。

English

The probability of \(2\) is \(\log_{10}(\frac{3}{2}),\) so twice that probability is \(\log_{10}(\frac{9}{4}).\) For the digits \(4\) through \(8,\) the sum telescopes: \[ \begin{aligned} &\sum_{d=4}^{8} \bigl(\log_{10}(d+1)\\ &\qquad{}-\log_{10}d\bigr) =\log_{10}\frac94. \end{aligned} \] Therefore that set has exactly twice the probability of digit \(2.\)

Thus the correct answer is C.

25 答案 B

将三个尺寸写成 \(\frac{2}{r},2,2r\)，因为它们的乘积为 \(8\)。表面积的一半等于三个两两乘积之和，所以 \[ \frac4r+4+4r=16\text{，} \] 得到 \(r+\frac{1}{r}=3\)。三个尺寸之和为 \(2(r+1+\frac{1}{r})=8\)。每个尺寸在四条棱上出现，所以所有棱长之和为 \(4\cdot8=32\)。

所以正确答案是 B。

English

Write the dimensions as \(\frac{2}{r},2,2r,\) since their product is \(8.\) Half the surface area is the sum of the three pairwise products, so \[ \frac4r+4+4r=16, \] giving \(r+\frac{1}{r}=3.\) The sum of the dimensions is \(2(r+1+\frac{1}{r})=8.\) Since each dimension occurs on four edges, the sum of all edge lengths is \(4\cdot8=32.\)

Thus the correct answer is B.

26 答案 E

欧几里得算法给出 \[ \begin{aligned} &\gcd(n-13,5n+6)\\ &\quad=\gcd(n-13,71)\text{。} \end{aligned} \] 由于 \(71\) 是素数，该非零分数可约当且仅当 \(n-13\) 是 \(71\) 的非零倍数。最小的正数情形为 \(n=13+71=84\)，它不在前四个数值选项中。

所以正确答案是 E。

English

Euclid’s algorithm gives \[ \begin{aligned} &\gcd(n-13,5n+6)\\ &\quad=\gcd(n-13,71). \end{aligned} \] Since \(71\) is prime, the nonzero fraction is reducible exactly when \(n-13\) is a nonzero multiple of \(71.\) The least positive possibility is \(n=13+71=84,\) which is not among the four numerical choices.

Thus the correct answer is E.

27 答案 C

令 \(c=\sqrt[3]{3}\)。反复应用递推式得到 \[ x_n=3^{\frac{c^{\,n-1}}{3}}\text{，} \] 所以 \(x_4=3^{\frac{c^3}{3}}=3\)。还需排除前三项。该数列严格递增。又因为 \(c\lt\frac{3}{2}\)， \[ x_2=c^c \lt\left(\frac32\right)^{\frac{3}{2}} \lt2\text{。} \] 最后，\(x_3=3^{\frac{1}{c}}\)。由于 \(2^c\lt2^{\frac{3}{2}}\lt3\)，有 \(x_3\gt2\)；而 \(\frac{1}{c}\lt1\) 给出 \(x_3\lt3\)。所以 \(x_1,x_2,x_3\) 都不是整数，第一项整数是 \(x_4\)。

因此正确答案是 C。

English

Let \(c=\sqrt[3]{3}.\) Repeated application of the recurrence gives \[ x_n=3^{\frac{c^{\,n-1}}{3}}, \] so \(x_4=3^{\frac{c^3}{3}}=3.\) It remains to rule out the first three terms. The sequence is strictly increasing. Also \(c\lt\frac{3}{2},\) so \[ x_2=c^c \lt\left(\frac32\right)^{\frac{3}{2}} \lt2. \] Finally \(x_3=3^{\frac{1}{c}}.\) Since \(2^c\lt2^{\frac{3}{2}}\lt3,\) we have \(x_3\gt2,\) while \(\frac{1}{c}\lt1\) gives \(x_3\lt3.\) Thus \(x_1,x_2,x_3\) are not integers, and the first integral term is \(x_4.\)

Therefore the correct answer is C.

28 答案 B

由正弦定理和三倍角公式， \[ \frac{48}{27} =\frac{\sin3A}{\sin A} =3-4\sin^2A\text{。} \] 因此 \(\sin^2A=\frac{11}{36}\)。由于 \(A\lt60^\circ\)，\(\cos A=\frac{5}{6}\)，从而 \(\cos2A=\frac{7}{18}\)。又有 \(B=180^\circ-4A\)，再次使用正弦定理，得到 \[ \begin{aligned} b &=27\frac{\sin B}{\sin A}\\ &=27\frac{\sin4A}{\sin A}\\ &=27\cdot4\cos A\cos2A\\ &=35\text{。} \end{aligned} \] 所以正确答案是 B。

English

By the sine law and the triple-angle identity, \[ \frac{48}{27} =\frac{\sin3A}{\sin A} =3-4\sin^2A. \] Hence \(\sin^2A=\frac{11}{36}.\) Since \(A\lt60^\circ,\) \(\cos A=\frac{5}{6},\) and therefore \(\cos2A=\frac{7}{18}.\) Also \(B=180^\circ-4A,\) so another application of the sine law gives \[ \begin{aligned} b &=27\frac{\sin B}{\sin A}\\ &=27\frac{\sin4A}{\sin A}\\ &=27\cdot4\cos A\cos2A\\ &=35. \end{aligned} \] Thus the correct answer is B.

29 答案 C

令 \(k=1985\)，且 \(R=\frac{10^k-1}{9}\)。则 \(a=8R\)、\(b=5R\)，而 \(9ab\) 以 \(0\) 结尾。去掉这个零不改变数字和，剩下 \[ \begin{aligned} N=\frac{9ab}{10} &=36R^2\\ &=\frac49(10^{2k}-1)\\ &\quad{}-\frac89(10^k-1)\text{。} \end{aligned} \] 因此，\(N\) 是由 \(2k\) 个四组成的数减去由 \(k\) 个八组成的数。相减后得到 \(k-1\) 个四，再接一个 \(3\)、\(k-1\) 个五，最后接一个 \(6\)。其数字和为 \[ \begin{aligned} &4(k-1)+3+5(k-1)+6\\ &\quad=9k=17865\text{。} \end{aligned} \] 所以正确答案是 C。

English

Let \(k=1985\) and \(R=\frac{10^k-1}{9}.\) Then \(a=8R,\) \(b=5R,\) and \(9ab\) ends in \(0.\) Removing that zero does not change the digit sum and leaves \[ \begin{aligned} N=\frac{9ab}{10} &=36R^2\\ &=\frac49(10^{2k}-1)\\ &\quad{}-\frac89(10^k-1). \end{aligned} \] Thus \(N\) is a string of \(2k\) fours minus a string of \(k\) eights. The subtraction produces \(k-1\) fours, then \(3,\) then \(k-1\) fives, then \(6.\) Its digit sum is \[ \begin{aligned} &4(k-1)+3+5(k-1)+6\\ &\quad=9k=17865. \end{aligned} \] Therefore the correct answer is C.

30 答案 E

令 \(n=\lfloor x\rfloor\)。解必须为正，并满足 \[ \begin{aligned} x&=\sqrt{10n-\frac{51}{4}},\\ n&\le x\lt n+1\text{。} \end{aligned} \] 下界不等式等价于 \[ 4n^2-40n+51\le0\text{，} \] 所以 \(n=2,3,\ldots,8\)。上界不等式等价于 \[ 4n^2-32n+55\gt0\text{。} \] 在 \(2,\ldots,8\) 中，它恰对 \(n=2,6,7,8\) 成立。每个区间贡献一个根，所以共有四个实数解。

因此正确答案是 E。

English

Let \(n=\lfloor x\rfloor.\) A solution must be positive and satisfy \[ \begin{aligned} x&=\sqrt{10n-\frac{51}{4}},\\ n&\le x\lt n+1. \end{aligned} \] The lower inequality is equivalent to \[ 4n^2-40n+51\le0, \] so \(n=2,3,\ldots,8.\) The upper inequality is equivalent to \[ 4n^2-32n+55\gt0. \] Among \(2,\ldots,8,\) this holds exactly for \(n=2,6,7,8.\) Each interval contributes one root, so there are four real solutions.

Therefore the correct answer is E.