1997 AMC 12 真题（P版）

1 答案 C

由第一个部分积可得 \(3(20+a)=69\)，所以 \(a=3\)。由第二个部分积可得 \(23b=92\)，所以 \(b=4\)。因此 \(a+b=7\)，正确答案是 C。

English

The first partial product says \(3(20+a)=69,\) so \(a=3.\) The second says \(23b=92,\) so \(b=4.\) Thus \(a+b=7,\) and the correct answer is C.

2 答案 D

向右水平边的长度之和为 \(12\)，所以所有水平边的长度之和为 \(24\)。图形的总高度为 \(8+2=10\)，所以所有竖直边的长度之和为 \(20\)。因此周长为 \(24+20=44\)，正确答案是 D。

English

The total of the rightward horizontal sides is \(12,\) so all horizontal sides total \(24.\) The full height is \(8+2=10,\) so the vertical sides total \(20.\) Therefore the perimeter is \(24+20=44,\) and the correct answer is D.

3 答案 D

三个平方项都非负，因此每一项都必须为 \(0\)。于是 \(x=3\)、\(y=4\)、\(z=5\)，且 \(x+y+z=12\)。正确答案是 D。

English

All three squares are nonnegative, so each must be \(0.\) Hence \(x=3,\) \(y=4,\) \(z=5,\) and \(x+y+z=12.\) The correct answer is D.

4 答案 A

有 \(a=1.5c\) 且 \(b=1.25c\)。因此 \(\frac{a}{b}=\frac{1.5}{1.25}=1.2\)，所以 \(a\) 比 \(b\) 大 \(20\%\)。正确答案是 A。

English

We have \(a=1.5c\) and \(b=1.25c.\) Thus \(\frac{a}{b}=\frac{1.5}{1.25}=1.2,\) so \(a\) is \(20\%\) larger than \(b.\) The correct answer is A.

5 答案 C

设小长方形的两条边长为 \(x\) 和 \(y\)。由共同的宽度可得 \(3x=2y\)，而大长方形的长和宽分别为 \(3x\) 与 \(x+y\)。因此其周长满足 \(2(4x+y)=176\)。由于 \(y=\frac{3x}{2}\)，可得 \(11x=176\)，所以 \(x=16\) 且 \(y=24\)。一个小长方形的周长为 \(2(16+24)=80\)，故正确答案是 C。

English

Let a small rectangle have sides \(x\) and \(y.\) The common width gives \(3x=2y,\) while the large rectangle has dimensions \(3x\) by \(x+y.\) Its perimeter is \(2(4x+y)=176.\) Since \(y=\frac{3x}{2},\) this is \(11x=176,\) so \(x=16\) and \(y=24.\) One small perimeter is \(2(16+24)=80,\) making C correct.

6 答案 B

每一对 \((2k-1)-2k\) 的和都是 \(-1\)。因此 \(100\) 对的总和为 \(-100\)，这 \(200\) 项的平均数为 \(-\frac{100}{200}=-0.5\)。正确答案是 B。

English

Each pair \((2k-1)-2k\) has sum \(-1.\) The \(100\) pairs therefore total \(-100,\) and their \(200\)-term average is \(-\frac{100}{200}=-0.5.\) The correct answer is B.

7 答案 D

如果七个整数都大于 \(13\)，它们的和至少为 \(98\)。但可以有六个整数大于 \(13\)：取六个 \(14\)，再取一个 \(-85\)。因此最多有 \(6\) 个，正确答案是 D。

English

If all seven integers exceeded \(13,\) their sum would be at least \(98.\) Six can exceed \(13\): take six copies of \(14\) and a seventh integer of \(-85.\) Thus the maximum is \(6,\) and the correct answer is D.

8 答案 D

在第一个分界点，\(C(25)=275\) 小于 \(C(23)=276\) 和 \(C(24)=288\)，所以 \(n=23\) 和 \(n=24\) 符合条件。在第二个分界点，当 \(n=45\)、\(n=46\)、\(n=47\)、\(n=48\) 时，\(C(49)=490\) 小于 \(C(n)=11n\)。每一段的价格都随 \(n\) 增加，因此没有其他取值符合条件。共有 \(2+4=6\) 个，故正确答案是 D。

English

At the first break, \(C(25)=275\) is below \(C(23)=276\) and \(C(24)=288,\) giving \(n=23\) and \(n=24.\) At the second, \(C(49)=490\) is below \(C(n)=11n\) for \(n=45,\) \(n=46,\) \(n=47,\) \(n=48.\) No other \(n\) works because each piece increases. There are \(2+4=6,\) so D is correct.

9 答案 C

按第一个提示建立坐标系后，直线 \(BE\) 的方程为 \(y=2x\)，过 \(C\) 的垂线方程为 \(y=-\frac{(x-2)}{2}\)。两直线交于 \(F=(\frac{2}{5},\frac{4}{5})\)。正方形的面积为 \(4\)，而三角形 \(AEF\) 和 \(BCF\) 的面积分别为 \(\frac{3}{5}\) 和 \(\frac{4}{5}\)。因此 \([CDEF]=4-\frac{3}{5}\) \({}-\frac{4}{5}=\frac{11}{5}\)，故正确答案是 C。

English

With the coordinates in the first hint, \(BE\) has equation \(y=2x,\) and the perpendicular through \(C\) is \(y=-\frac{(x-2)}{2}.\) Their intersection is \(F=(\frac{2}{5},\frac{4}{5}).\) The square has area \(4,\) while triangles \(AEF\) and \(BCF\) have areas \(\frac{3}{5}\) and \(\frac{4}{5},\) respectively. Hence \([CDEF]=4-\frac{3}{5}\) \({}-\frac{4}{5}=\frac{11}{5},\) so C is correct.

10 答案 D

第一枚改过的骰子有 \(4\) 个奇数面和 \(2\) 个偶数面；第二枚有 \(2\) 个奇数面和 \(4\) 个偶数面。因此所求概率为 \((\frac{4}{6})(\frac{4}{6})+(\frac{2}{6})(\frac{2}{6})\) \(=\frac{20}{36}=\frac{5}{9}\)。正确答案是 D。

English

The first modified die has \(4\) odd faces and \(2\) even faces; the second has \(2\) odd and \(4\) even. Thus the probability is \((\frac{4}{6})(\frac{4}{6})+(\frac{2}{6})(\frac{2}{6})\) \(=\frac{20}{36}=\frac{5}{9}.\) The correct answer is D.

11 答案 D

第 \(6\) 至第 \(9\) 场的总得分为 \(68\)。由条件 \(\frac{S+68}{9}>\frac{S}{5}\) 可得 \(S<85\)，所以整数 \(S\) 的最大值为 \(84\)。十场后的场均得分大于 \(18\)，意味着总得分至少为 \(181\)，因此第十场至少得到 \(181-(84+68)=29\) 分。这个分数可以达到，所以正确答案是 D。

English

Games \(6\)–\(9\) total \(68.\) The condition \(\frac{S+68}{9}>\frac{S}{5}\) gives \(S<85,\) so the greatest integral \(S\) is \(84.\) A ten-game average above \(18\) requires a total at least \(181,\) hence the tenth score is at least \(181-(84+68)=29.\) This is attainable, so D is correct.

12 答案 E

若 \((1997,0)\) 在直线上，则 \(0=1997m+b\)，所以 \(b=-1997m\)，从而 \(mb=-1997m^2\lt0\)，产生矛盾。对于其余每个点，都可以选取适当的同号 \(m,b\) 使直线经过该点。因此正确答案是 E。

English

If \((1997,0)\) were on the line, then \(0=1997m+b,\) so \(b=-1997m\) and \(mb=-1997m^2\lt0,\) a contradiction. Each other point can occur for suitable same-sign \(m,b.\) Thus the correct answer is E.

13 答案 E

这个和为 \(11(x+y)\)。由于 \(1\le x+y\le18\)，它只有在 \(x+y=11\) 时才是完全平方数，此时等于 \(121\)。十位数字 \(x\) 可以是 \(2,3,\ldots,9\) 中的任意一个，而 \(y=11-x\)。共有 \(8\) 个这样的整数，所以正确答案是 E。

English

The sum is \(11(x+y).\) Since \(1\le x+y\le18,\) this is square only when \(x+y=11,\) giving \(121.\) The tens digit \(x\) can be any of \(2,3,\ldots,9,\) with \(y=11-x.\) There are \(8\) integers, so E is correct.

14 答案 B

根据规律可得 \(x-39=60k\) 和 \(123-60=kx\)。消去 \(k\)，得到 \(x(x-39)=3780\)，即 \((x-84)(x+45)=0\)。数量为正，因此 \(x=84\)，正确答案是 B。

English

The rule gives \(x-39=60k\) and \(123-60=kx.\) Eliminating \(k\) yields \(x(x-39)=3780,\) or \((x-84)(x+45)=0.\) The population is positive, so \(x=84,\) and B is correct.

15 答案 D

四边形 \(BCDE\) 的两条对角线 \(BD=8\) 和 \(CE=12\) 互相垂直，所以其面积为 \(\frac12(8)(12)=48\)。因为 \(D,E\) 都是中点，三角形 \(ADE\) 的面积是三角形 \(ABC\) 面积的四分之一，因此四边形 \(BCDE\) 的面积是它的四分之三。于是 \([ABC]=\frac{48\cdot4}{3}=64\)，所以正确答案是 D。

English

Quadrilateral \(BCDE\) has perpendicular diagonals \(BD=8\) and \(CE=12,\) so its area is \(\frac12(8)(12)=48.\) Since \(D,E\) are midpoints, triangle \(ADE\) has one fourth the area of \(ABC,\) making \(BCDE\) three fourths of it. Thus \([ABC]=\frac{48\cdot4}{3}=64,\) so D is correct.

16 答案 D

如果至多改变三个元素，那么要么六条行列中至少有两条没有改变，要么某个被改变的元素是它所在行与所在列中唯一被改变的元素。前一种情况下，两条未改变的行或列的和仍然相等；后一种情况下，该元素使其所在行与列的和发生同样的变化，所以这两个和仍然相等。改变四个元素即可：分别将 \(4\)、\(1\)、\(2\)、\(6\) 改为 \(5\)、\(3\)、\(7\)、\(9\)。新的三个行和为 \(21\)、\(20\)、\(15\)，三个列和为 \(16\)、\(17\)、\(23\)。因此最小值为 \(4\)，正确答案是 D。

English

With at most three alterations, either two of the six lines are unchanged, or some altered entry is the only alteration in both its row and its column. In the first case those two sums remain equal; in the second, that entry changes its row sum and column sum by the same amount, so those two sums remain equal. Four suffice: replace \(4,\) \(1,\) \(2,\) \(6\) by \(5,\) \(3,\) \(7,\) \(9,\) respectively. The resulting row sums are \(21,\) \(20,\) \(15\) and column sums are \(16,\) \(17,\) \(23.\) Hence the minimum is \(4,\) and D is correct.

17 答案 A

两个交点的竖直距离满足 \[ \begin{aligned} \log_5(k+4)-\log_5k &=\frac12,\\ \frac{k+4}{k}&=\sqrt5 \end{aligned}\text{。}\]因此 \(\frac{k+4}{k}=\sqrt5\)，所以 \(k=\frac{4}{\sqrt5-1}=1+\sqrt5\)。于是 \(a+b=1+5=6\)，正确答案是 A。

English

The vertical distance is \[ \begin{aligned} \log_5(k+4)-\log_5k &=\frac12,\\ \frac{k+4}{k}&=\sqrt5. \end{aligned} \] Thus \(\frac{k+4}{k}=\sqrt5,\) so \(k=\frac{4}{\sqrt5-1}=1+\sqrt5.\) Therefore \(a+b=1+5=6,\) and A is correct.

18 答案 E

这列数共有 \(\frac{10}{2}=5\) 项。将它们写成 \(10\le a\le m\le b\le c\)。把 \(m\) 替换为 \(m+10\) 后，这个新值成为中位数，所以 \(b\ge m+10\) 且 \(c\ge m+10\)；又因为众数是 \(32\)，必有 \(b=c=32\)。原数列的总和为 \(110\)，所以 \(a+m=36\)。把 \(m\) 替换为 \(m-8\) 后，中位数为 \(a=m-4\)，所以 \(2m-4=36\)，解得 \(m=20\)。因此正确答案是 E。

English

The list has \(\frac{10}{2}=5\) entries. Write them \(10\le a\le m\le b\le c.\) Replacing \(m\) by \(m+10\) makes that value the median, so \(b\ge m+10\) and \(c\ge m+10;\) because the mode is \(32,\) we must have \(b=c=32.\) The original total is \(110,\) giving \(a+m=36.\) Replacing \(m\) by \(m-8\) makes the median \(a=m-4,\) so \(2m-4=36\) and \(m=20.\) Thus E is correct.

19 答案 D

取 \(A=(0,0)\)、\(B=(1,0)\)、\(C=(0,\sqrt3)\)。斜边所在直线的方程为 \(\sqrt3x+y=\sqrt3\)，圆心为 \(O=(r,r)\)。由相切可得 \[ \frac{\left|(\sqrt3+1)r-\sqrt3\right|}{2}=r\text{。}\]对于图中三角形外部的圆，有 \((\sqrt3+1)r-\sqrt3=2r\)，所以 \(r=\frac{\sqrt3}{\sqrt3-1}\) \(=\frac{3+\sqrt3}{2}\approx2.37\)。因此正确答案是 D。

English

Take \(A=(0,0),\) \(B=(1,0),\) and \(C=(0,\sqrt3).\) The hypotenuse is \(\sqrt3x+y=\sqrt3,\) and the circle center is \(O=(r,r).\) Tangency gives \[ \frac{\left|(\sqrt3+1)r-\sqrt3\right|}{2}=r. \] The pictured external circle has \((\sqrt3+1)r-\sqrt3=2r,\) so \(r=\frac{\sqrt3}{\sqrt3-1}\) \(=\frac{3+\sqrt3}{2}\approx2.37.\) Thus D is correct.

20 答案 A

它们的和为 \(100a+(1+\cdots+100)\) \(=100a+5050\)，所以末两位是 \(50\)。只有 \(1,627,384,950\) 具有这一性质，并且对应的 \(a=\frac{1,627,384,950-5050}{100}\) 是正整数。因此正确答案是 A。

English

The sum is \(100a+(1+\cdots+100)\) \(=100a+5050,\) so it ends in \(50.\) Only \(1,627,384,950\) has that property, and it gives the positive integer \(a=\frac{1,627,384,950-5050}{100}.\) Thus A is correct.

21 答案 C

对整数 \(n\)，当且仅当 \(n\) 是 \(2\) 的幂时，\(\log_8n\) 才是有理数。符合条件的数为 \(n=2^k\)，其中 \(0\le k\le10\)，并且 \(f(2^k)=\frac{k}{3}\)。因此总和为 \(\frac{0+1+\cdots+10}{3}=\frac{55}{3}\)，所以正确答案是 C。

English

For integer \(n,\) \(\log_8n\) is rational exactly when \(n\) is a power of \(2.\) The relevant values are \(n=2^k\) for \(0\le k\le10,\) and \(f(2^k)=\frac{k}{3}.\) Therefore the sum is \(\frac{0+1+\cdots+10}{3}=\frac{55}{3},\) so C is correct.

22 答案 E

环上各差的绝对值为 \(19\)、\(7\)、\(5\)、\(4\)、\(11\)，而带符号的差之和为 \(0\)。因此正负两组中的一组之和必须是 \(46\) 的一半，即 \(23\)。唯一的分组是 \(19+4=7+5+11\)。按一种方向取符号时，各人的金额满足 \(A=E+11\)、\(B=E-8\)、\(C=E-1\)、\(D=E+4\)。总金额为 \(5E+6\)，所以 \(5E+6=56\)，从而 \(E=10\)。若把所有符号反向，则有 \(5E-6=56\)，得不到整数 \(E\)。因此埃尔金有 \(\$10\)，正确答案是 E。

English

The signed differences around the cycle have magnitudes \(19,\) \(7,\) \(5,\) \(4,\) \(11\) and sum \(0.\) Thus one side of the sign split must total half of \(46,\) namely \(23.\) The only split is \(19+4=7+5+11.\) In one orientation, the amounts are \(A=E+11,\) \(B=E-8,\) \(C=E-1,\) and \(D=E+4.\) Their total is \(5E+6,\) so \(5E+6=56\) and \(E=10.\) Reversing every sign would give \(5E-6=56,\) not an integral \(E.\) Thus Elgin had \(\$10\) and E is correct.

23 答案 D

这个展开图形成一个切去一角的单位立方体。被切去的部分是一个三条单位棱两两垂直的三棱锥，因此其体积为 \(\frac13\cdot\frac12\cdot1=\frac{1}{6}\)。剩余多面体的体积为 \(1-\frac{1}{6}=\frac{5}{6}\)，正确答案是 D。

English

The net forms a unit cube with one corner cut off. The removed corner is a triangular pyramid with three mutually perpendicular unit edges, so its volume is \(\frac13\cdot\frac12\cdot1=\frac{1}{6}.\) The remaining polyhedron has volume \(1-\frac{1}{6}=\frac{5}{6},\) and the correct answer is D.

24 答案 B

以 \(1\) 开头的共有 \(\binom84=70\) 个。在以 \(2\) 开头的数中，最前面的 \(\binom63=20\) 个以 \(23\) 开头，占据第 \(71\) 至第 \(90\) 位。因此总列表中的第 \(97\) 个数，是以 \(24\) 开头的第七个数：\[ \begin{gathered} 24567,24568,24569,24578,\\ 24579,24589,24678 \end{gathered}\text{。}\]数 \(24678\) 不含 \(5\)，所以正确答案是 B。

English

There are \(\binom84=70\) beginning with \(1.\) Among those beginning with \(2,\) the first \(\binom63=20\) begin with \(23,\) occupying positions \(71\)–\(90.\) Thus the \(97\)th overall is the seventh beginning with \(24:\) \[ \begin{gathered} 24567,24568,24569,24578,\\ 24579,24589,24678. \end{gathered} \] The number \(24678\) omits \(5,\) so B is correct.

25 答案 B

取各射线的共同方向为单位向量 \(u\)。于是 \[ \begin{aligned} M&=\frac{A+C}{2}+14u,\\ N&=\frac{B+D}{2}+15u \end{aligned}\text{。}\]对于平行四边形有 \(A+C=B+D\)，所以 \(N-M=u\)，从而 \(MN=1\)。正确答案是 B。

English

Choose the common ray direction as a unit vector \(u.\) Then \[ \begin{aligned} M&=\frac{A+C}{2}+14u,\\ N&=\frac{B+D}{2}+15u. \end{aligned} \] Since \(A+C=B+D\) for a parallelogram, \(N-M=u,\) so \(MN=1.\) The correct answer is B.

26 答案 A

因为 \(PA=PB\)，作以 \(P\) 为圆心并经过这两点的圆。条件 \(\angle APB=2\angle ACB\) 正是圆心角与圆周角的关系，所以 \(C\) 也在同一个圆上。直线 \(PB\) 与圆的另一个交点记为 \(E\)，则 \(PE=3\)。由点 \(D\) 的幂可得 \[ \begin{aligned} AD\cdot CD&=DE\cdot DB\\ &=(3-2)(3+2)=5 \end{aligned}\text{。}\]因此正确答案是 A。

English

Because \(PA=PB,\) draw their circle with center \(P.\) The condition \(\angle APB=2\angle ACB\) is the central-inscribed angle relation, so \(C\) lies on the same circle. Along line \(PB,\) the other circle intersection is \(E\) with \(PE=3.\) Power of \(D\) gives \[ \begin{aligned} AD\cdot CD&=DE\cdot DB\\ &=(3-2)(3+2)=5. \end{aligned} \] Thus A is correct.

27 答案 D

对 \(u_n=f(x+4n)\)，原方程化为 \(u_{n+1}=u_n-u_{n-1}\)。从 \(u_0,u_1\) 出发，数列依次为 \[ \begin{gathered} u_0,u_1,u_1-u_0,-u_0,\\ -u_1,u_0-u_1,u_0,u_1,\ldots \end{gathered}\text{，}\]因此每个这样的函数都有周期 \(6\cdot4=24\)。又因为 \(f(x)=\sin(\frac{\pi x}{12})\) 满足原方程且最小正周期为 \(24\)，所以这个公周期不能再小。故正确答案是 D。

English

For \(u_n=f(x+4n),\) the equation is \(u_{n+1}=u_n-u_{n-1}.\) Starting from \(u_0,u_1,\) the sequence is \[ \begin{gathered} u_0,u_1,u_1-u_0,-u_0,\\ -u_1,u_0-u_1,u_0,u_1,\ldots, \end{gathered} \] so every such function has period \(6\cdot4=24.\) This is least because \(f(x)=\sin(\frac{\pi x}{12})\) satisfies the equation and has fundamental period \(24.\) Hence D is correct.

28 答案 E

若 \(c\ge0\)，代入并因式分解后，没有任何数对符合 \(a+b\) 所需的符号。因此 \(c\lt0\)，所以 \(s\gt19\) 且 \(ab+s=116\)。若 \(a+b=s\)，则 \((a+1)(b+1)=117\)，得到无序数对 \(\{0,116\}\)、\(\{2,38\}\)、\(\{8,12\}\)。若 \(a+b=-s\)，则 \((a-1)(b-1)=117\)，得到 \(\{-116,0\}\)、\(\{-38,-2\}\)、\(\{-12,-8\}\)。每个无序数对都有两种次序，因此共有 \(6\cdot2=12\) 个三元组。正确答案是 E。

English

If \(c\ge0,\) substitution and factoring lead to no pair consistent with the required sign of \(a+b.\) Thus \(c\lt0,\) so \(s\gt19\) and \(ab+s=116.\) If \(a+b=s,\) then \((a+1)(b+1)=117,\) producing the unordered pairs \(\{0,116\},\) \(\{2,38\},\) \(\{8,12\}.\) If \(a+b=-s,\) then \((a-1)(b-1)=117,\) producing \(\{-116,0\},\) \(\{-38,-2\},\) \(\{-12,-8\}.\) Each unordered pair has two orders, giving \(6\cdot2=12\) triples. The correct answer is E.

29 答案 B

假设 \(1\) 是 \(n\) 个特殊数之和，并令 \(a_k\) 表示在第 \(k\) 个小数位上取数字 \(7\) 的加数个数。将等式两边除以 \(7\)，可得 \[ \begin{aligned} \frac17&=\frac{a_1}{10}+\frac{a_2}{10^2}+\cdots\\ &=0.\overline{142857} \end{aligned}\text{。}\]当 \(n\le9\) 时，每个 \(a_k\) 都是一位数字，所以 \(a_1,a_2,\ldots=1,4,2,8,5,7,\ldots\)；因此 \(n\ge8\)。八个就足够，因为以下六位循环块所表示的特殊循环小数满足 \[ \begin{aligned} 700700+2(070707)\\ {}+2(077777)\\ {}+3(000777)&=999999 \end{aligned}\text{。}\]它们的和为 \(1\)。因此最小值为 \(8\)，正确答案是 B。

English

Suppose \(1\) is a sum of \(n\) special numbers, and let \(a_k\) count summands having a \(7\) in the \(k\)th decimal place. Dividing by \(7\) gives \[ \begin{aligned} \frac17&=\frac{a_1}{10}+\frac{a_2}{10^2}+\cdots\\ &=0.\overline{142857}. \end{aligned} \] For \(n\le9,\) each \(a_k\) is a digit, so \(a_1,a_2,\ldots=1,4,2,8,5,7,\ldots;\) hence \(n\ge8.\) Eight suffice because the repeating special decimals represented by \[ \begin{aligned} 700700+2(070707)\\ {}+2(077777)\\ {}+3(000777)&=999999. \end{aligned} \] Their sum is \(1.\) Therefore the minimum is \(8,\) and B is correct.

30 答案 C

一个恰好发生两次数字变化的 \(d\) 位二进制数必为 \(1^a0^b1^c\) 的形式，其中 \(a\)、\(b\)、\(c\) 均为正整数，因此共有 \(\binom{d-1}{2}\) 种。对 \(d=3\)、\(d=4\)、\(d=5\)、\(d=6\)，总数为 \(1+3+6+10=20\)。在不超过 \(1100001_2=97\) 的七位数中，以单个 \(1\) 开头的五种形式都符合条件；以至少两个 \(1\) 开头的形式中，只有 \(1100001_2\) 本身符合上界。因此共有 \(20+6=26\) 个，正确答案是 C。

English

A \(d\)-bit numeral with exactly two changes has form \(1^a0^b1^c\) with positive \(a,\) \(b,\) \(c,\) giving \(\binom{d-1}{2}\) choices. For \(d=3,\) \(d=4,\) \(d=5,\) \(d=6,\) the total is \(1+3+6+10=20.\) Among seven-bit numbers at most \(1100001_2=97,\) the five forms beginning with one \(1\) all work, and the only form beginning with at least two \(1\)s is \(1100001_2\) itself. Thus there are \(20+6=26,\) and C is correct.