2004 AMC 12B Solutions | LIVE by Po-Shen Loh

1.

At each basketball practice last week, Jenny made twice as many free throws as she made at the previous practice. At her fifth practice she made $48$ free throws. How many free throws did she make at the first practice?

$3$

$6$

$9$

$12$

$15$

Concepts:geometric sequence work backwards

Difficulty rating: 900

Small Hint:

Each practice she made half as many as the next practice

Big Hint:

Halve $48$ four times to reach the first practice

Solution:

Each practice she made twice the previous, so going backward we halve. From the fifth practice at $48,$ the earlier practices had $24,$ $12,$ $6,$ and $3$ free throws.

Thus, the correct answer is A.

2.

In the expression $c \cdot a^b - d,$ the values of $a,$ $b,$ $c,$ and $d$ are $0,$ $1,$ $2,$ and $3,$ although not necessarily in that order. What is the maximum possible value of the result?

$5$

$6$

$8$

$9$

$10$

Concepts:optimization casework

Difficulty rating: 1000

Small Hint:

To keep the result large, make $d$ as small as possible

Big Hint:

Set $d = 0$ and try each choice of which variable equals $1$

Solution:

To maximize, set $d = 0.$ With $a, b, c$ taking $1, 2, 3,$ the term $c \cdot a^b$ is largest when $c = 1$ and $a^b = 3^2 = 9.$ This gives $1 \cdot 9 - 0 = 9,$ which beats $2^3 = 8$ and the other assignments.

Thus, the correct answer is D.

3.

If $x$ and $y$ are positive integers for which $2^x 3^y = 1296,$ what is the value of $x + y?$

$8$

$9$

$10$

$11$

$12$

Concepts:prime factorization exponent

Difficulty rating: 980

Small Hint:

Factor $1296$ into powers of $2$ and $3$

Big Hint:

$1296 = 6^4 = 2^4 \cdot 3^4$

Solution:

Factoring, $1296 = 6^4 = 2^4 \cdot 3^4.$ Matching exponents gives $x = 4$ and $y = 4,$ so $x + y = 8.$

Thus, the correct answer is A.

4.

An integer $x,$ with $10 \le x \le 99,$ is to be chosen. If all choices are equally likely, what is the probability that at least one digit of $x$ is a $7?$

$\dfrac{1}{9}$

$\dfrac{1}{5}$

$\dfrac{19}{90}$

$\dfrac{2}{9}$

$\dfrac{1}{3}$

Concepts:inclusion-exclusion digits basic probability

Difficulty rating: 1100

Small Hint:

Count two-digit numbers with a $7$ in the units place and with a $7$ in the tens place

Big Hint:

$77$ is counted in both groups, so subtract it once

Solution:

There are $90$ integers from $10$ to $99.$ Ten have a units digit $7,$ and nine have a tens digit $7.$ Since $77$ is counted twice, there are $10 + 9 - 1 = 18$ with at least one $7.$ The probability is $\dfrac{18}{90} = \dfrac{1}{5}.$

Thus, the correct answer is B.

5.

On a trip from the United States to Canada, Isabella took $d$ U.S. dollars. At the border she exchanged them all, receiving $10$ Canadian dollars for every $7$ U.S. dollars. After spending $60$ Canadian dollars, she had $d$ Canadian dollars left. What is the sum of the digits of $d?$

$5$

$6$

$7$

$8$

$9$

Concepts:ratio and proportion linear equation

Difficulty rating: 1150

Small Hint:

Exchanging $d$ U.S. dollars gives $\dfrac{10d}{7}$ Canadian dollars

Big Hint:

Set $\dfrac{10d}{7} - 60 = d$ and solve for $d$

Solution:

Exchanging gives $\dfrac{10d}{7}$ Canadian dollars. After spending $60,$ she has $\dfrac{10d}{7} - 60 = d.$ Then $\dfrac{3d}{7} = 60,$ so $d = 140.$ The sum of its digits is $1 + 4 + 0 = 5.$

Thus, the correct answer is A.

6.

Minneapolis-St. Paul International Airport is $8$ miles southwest of downtown St. Paul and $10$ miles southeast of downtown Minneapolis. Which of the following is closest to the number of miles between downtown St. Paul and downtown Minneapolis?

$13$

$14$

$15$

$16$

$17$

Concepts:Pythagorean Theorem estimation

Difficulty rating: 1190

Small Hint:

Southwest and southeast directions are perpendicular

Big Hint:

The distance is $\sqrt{10^2 + 8^2},$ then estimate

Solution:

Southwest and southeast are perpendicular, so the airport sits at the right angle of a right triangle with legs $8$ and $10.$ The distance between downtowns is $\sqrt{10^2 + 8^2} = \sqrt{164} \approx 12.8,$ closest to $13.$

Thus, the correct answer is A.

7.

A square has sides of length $10,$ and a circle centered at one of its vertices has radius $10.$ What is the area of the union of the regions enclosed by the square and the circle?

$200 + 25\pi$

$100 + 75\pi$

$75 + 100\pi$

$100 + 100\pi$

$100 + 125\pi$

Concepts:circle area area decomposition

Difficulty rating: 1220

Small Hint:

The overlap of the square and circle is a quarter of the circle

Big Hint:

Union $= [\text{square}]$ $+ [\text{circle}]$ $- [\text{quarter circle}]$

Solution:

The square has area $100$ and the circle has area $100\pi.$ Their overlap is the quarter of the circle lying inside the square, with area $25\pi.$ The union is $100 + 100\pi - 25\pi = 100 + 75\pi.$

Thus, the correct answer is B.

8.

A grocer makes a display of cans in which the top row has one can and each lower row has two more cans than the row above it. If the display contains $100$ cans, how many rows does it contain?

$5$

$8$

$9$

$10$

$11$

Concepts:sum of first n odd numbers

Difficulty rating: 1220

Small Hint:

The row counts are $1, 3, 5, \ldots,$ the odd numbers

Big Hint:

The sum of the first $n$ odd numbers is $n^2$

Solution:

The rows contain $1, 3, 5, \ldots, (2n - 1)$ cans, and the sum of the first $n$ odd numbers is $n^2.$ Setting $n^2 = 100$ gives $n = 10.$

Thus, the correct answer is D.

9.

The point $(-3, 2)$ is rotated $90^\circ$ clockwise around the origin to point $B.$ Point $B$ is then reflected in the line $y = x$ to point $C.$ What are the coordinates of $C?$

$(-3, -2)$

$(-2, -3)$

$(2, -3)$

$(2, 3)$

$(3, 2)$

Concepts:transformation coordinate geometry

Difficulty rating: 1350

Small Hint:

A $90^\circ$ clockwise rotation sends $(x, y)$ to $(y, -x)$

Big Hint:

Reflecting in $y = x$ swaps the coordinates

Solution:

Rotating $(-3, 2)$ by $90^\circ$ clockwise sends $(x, y) \to (y, -x),$ giving $B = (2, 3).$ Reflecting in $y = x$ swaps coordinates, giving $C = (3, 2).$

Thus, the correct answer is E.

10.

An annulus is the region between two concentric circles. The concentric circles in the figure have radii $b$ and $c,$ with $b \gt c.$ Let $\overline{OX}$ be a radius of the larger circle, let $\overline{XZ}$ be tangent to the smaller circle at $Z,$ and let $\overline{OY}$ be the radius of the larger circle that contains $Z.$ Let $a = XZ,$ $d = YZ,$ and $e = XY.$ What is the area of the annulus?

$\pi a^2$

$\pi b^2$

$\pi c^2$

$\pi d^2$

$\pi e^2$

Concepts:annulus tangent line Pythagorean Theorem

Difficulty rating: 1460

Small Hint:

The annulus area is $\pi b^2 - \pi c^2$

Big Hint:

Since $\overline{XZ}$ is tangent, $\triangle OZX$ is right-angled at $Z,$ so $b^2 - c^2 = a^2$

Solution:

The annulus area is $\pi b^2 - \pi c^2.$ Because $\overline{XZ}$ is tangent to the smaller circle at $Z,$ it is perpendicular to radius $\overline{OZ},$ so $\triangle OZX$ is right-angled at $Z.$ Then $b^2 = c^2 + a^2,$ giving $b^2 - c^2 = a^2.$ The area is $\pi a^2.$

Thus, the correct answer is A.

11.

All the students in an algebra class took a $100$ -point test. Five students scored $100,$ each student scored at least $60,$ and the mean score was $76.$ What is the smallest possible number of students in the class?

$10$

$11$

$12$

$13$

$14$

Concepts:mean bounding to limit cases

Difficulty rating: 1440

Small Hint:

Measure each score as its deviation from the mean $76$

Big Hint:

The five $100$ s are $120$ points above the mean; each other student is at most $16$ below it

Solution:

Each score of $100$ is $24$ above the mean, so the five contribute $120$ points above $76.$ These must be balanced by points below the mean, and each remaining student is at most $76 - 60 = 16$ below. So at least $\dfrac{120}{16} = 7.5,$ hence $8$ more students are needed, for a total of $13.$ Five $100$ s and eight $61$ s achieve this.

Thus, the correct answer is D.

12.

In the sequence $2001,$ $2002,$ $2003,$ $\ldots,$ each term after the third is found by subtracting the previous term from the sum of the two terms that precede that term. For example, the fourth term is $2001 + 2002 - 2003 = 2000.$ What is the $2004$ th term in this sequence?

$-2004$

$-2$

$0$

$4003$

$6007$

Concepts:recursion arithmetic sequence pattern recognition

Difficulty rating: 1500

Small Hint:

Write out several terms to spot a pattern

Big Hint:

The even-indexed terms $2002, 2000, 1998, \ldots$ form an arithmetic progression

Solution:

The rule gives $2001,$ $2002,$ $2003,$ $2000,$ $2005,$ $1998, \ldots$ The even-indexed terms are $2002, 2000, 1998, \ldots,$ decreasing by $2.$

More precisely, the recurrence verifies inductively that $a_{2k} = 2004 - 2k$ and $a_{2k+1} = 2001 + 2k.$ Therefore $\begin{aligned} a_{2004} &= a_{2\cdot1002} \\ &= 2004 - 2(1002) = 0. \end{aligned}$

Thus, the correct answer is C.

13.

If $f(x) = ax + b$ and $f^{-1}(x) = bx + a$ with $a$ and $b$ real, what is the value of $a + b?$

$-2$

$-1$

$0$

$1$

$2$

Concepts:function system of equations

Difficulty rating: 1580

Small Hint:

Use $f(f^{-1}(x)) = x$

Big Hint:

Expanding $a(bx + a) + b = x$ gives $ab = 1$ and $a^2 + b = 0$

Solution:

Since $f(f^{-1}(x)) = x,$ we have $a(bx + a) + b = x.$ Matching terms gives $ab = 1$ and $a^2 + b = 0.$ Then $b = \frac{1}{a}$ and $a^2 + \frac{1}{a} = 0,$ so $a^3 = -1,$ giving $a = -1$ and $b = -1.$ Thus $a + b = -2.$

Thus, the correct answer is A.

14.

In $\triangle ABC,$ $AB = 13,$ $AC = 5$ and $BC = 12.$ Points $M$ and $N$ lie on $\overline{AC}$ and $\overline{BC},$ respectively, with $CM = CN = 4.$ Points $J$ and $K$ are on $\overline{AB}$ so that $\overline{MJ}$ and $\overline{NK}$ are perpendicular to $\overline{AB}.$ What is the area of pentagon $CMJKN?$

$15$

$\dfrac{81}{5}$

$\dfrac{205}{12}$

$\dfrac{240}{13}$

$20$

Difficulty rating: 1680

Small Hint:

$\triangle ABC$ is right-angled at $C$ since $5^2 + 12^2 = 13^2$

Big Hint:

$\triangle AMJ$ and $\triangle NBK$ are similar to $\triangle ABC$ with hypotenuses $AM = 1$ and $BN = 8$

Solution:

Since $5^2 + 12^2 = 13^2,$ $\triangle ABC$ is right-angled at $C$ with area $\tfrac12 (5)(12) = 30.$ The small right triangles $\triangle AMJ$ and $\triangle NBK$ are each similar to $\triangle ABC,$ with hypotenuses $AM = 5 - 4 = 1$ and $BN = 12 - 4 = 8.$ Their areas are $\left(\dfrac{1}{13}\right)^2 (30)$ and $\left(\dfrac{8}{13}\right)^2 (30).$

The pentagon is what remains: $\begin{gathered} \left(1 - \dfrac{1}{169} - \dfrac{64}{169}\right)(30) \\ {}= \dfrac{104}{169}(30) = \dfrac{240}{13}. \end{gathered}$

Thus, the correct answer is D.

15.

The two digits in Jack’s age are the same as the digits in Bill’s age, but in reverse order. In five years Jack will be twice as old as Bill will be then. What is the difference in their current ages?

$9$

$18$

$27$

$36$

$45$

Concepts:ages place value linear equation

Difficulty rating: 1510

Small Hint:

Let Jack be $10x + y$ and Bill be $10y + x$

Big Hint:

$10x + y + 5 = 2(10y + x + 5)$ simplifies to $8x = 19y + 5$

Solution:

Let Jack be $10x + y$ and Bill be $10y + x.$ Then $10x + y + 5 = 2(10y + x + 5),$ so $8x = 19y + 5.$ Testing digits, only $y = 1, x = 3$ works, so Jack is $31$ and Bill is $13.$ The difference is $31 - 13 = 18.$

Thus, the correct answer is B.

16.

A function $f$ is defined by $f(z) = i\overline{z},$ where $i = \sqrt{-1}$ and $\overline{z}$ is the complex conjugate of $z.$ How many values of $z$ satisfy both $|z| = 5$ and $f(z) = z?$

$0$

$1$

$2$

$4$

$8$

Concepts:complex number coordinate geometry

Difficulty rating: 1610

Small Hint:

Write $z = x + iy$ and compute $i\overline{z}$

Big Hint:

$f(z) = z$ forces $y = x,$ a line; intersect it with the circle $|z| = 5$

Solution:

Writing $z = x + iy,$ we get $f(z) = i(x - iy) = y + ix.$ Setting $f(z) = z$ gives $y = x,$ which is a line through the origin. The condition $|z| = 5$ is a circle, and a line through the center meets the circle in $2$ points.

Thus, the correct answer is C.

17.

For some real numbers $a$ and $b,$ the equation $8x^3 + 4ax^2 + 2bx + a = 0$ has three distinct positive roots. If the sum of the base- $2$ logarithms of the roots is $5,$ what is the value of $a?$

$-256$

$-64$

$-8$

$64$

$256$

Concepts:Vieta’s Formulas logarithm

Difficulty rating: 1770

Small Hint:

The sum of $\log_2$ of the roots equals $\log_2$ of their product

Big Hint:

For $8x^3 + \cdots + a,$ the product of the roots is $-\dfrac{a}{8}$

Solution:

The sum of the base- $2$ logarithms is $\log_2(r_1 r_2 r_3) = 5,$ so $r_1 r_2 r_3 = 2^5 = 32.$ By Vieta’s formulas on $8x^3 + 4ax^2 + 2bx + a,$ the product of the roots is $-\dfrac{a}{8}.$ Thus $-\dfrac{a}{8} = 32,$ giving $a = -256.$

Thus, the correct answer is A.

18.

Points $A$ and $B$ are on the parabola $y = 4x^2 + 7x - 1,$ and the origin is the midpoint of $\overline{AB}.$ What is the length of $AB?$

$2\sqrt{5}$

$5 + \dfrac{\sqrt{2}}{2}$

$5 + \sqrt{2}$

$7$

$5\sqrt{2}$

Concepts:parabola midpoint distance formula

Difficulty rating: 1740

Small Hint:

If $B = (a, b),$ then $A = (-a, -b)$

Big Hint:

Substitute both points into the parabola and subtract the equations

Solution:

Let $B = (a, b)$ and $A = (-a, -b).$ Then $4a^2 + 7a - 1 = b$ and $4a^2 - 7a - 1 = -b.$ Subtracting gives $14a = 2b,$ so $b = 7a.$ Then $4a^2 + 7a - 1 = 7a$ gives $a^2 = \dfrac14,$ and $b^2 = 49a^2 = \dfrac{49}{4}.$ So $AB = 2\sqrt{a^2 + b^2}$ $= 2\sqrt{\dfrac{50}{4}} = 5\sqrt{2}.$

Thus, the correct answer is E.

19.

A truncated cone has horizontal bases with radii $18$ and $2.$ A sphere is tangent to the top, bottom, and lateral surface of the truncated cone. What is the radius of the sphere?

$6$

$4\sqrt{5}$

$9$

$10$

$6\sqrt{3}$

Concepts:cone sphere tangent line

Difficulty rating: 1870

Small Hint:

Take the cross-section through the axis: an isosceles trapezoid with an inscribed circle

Big Hint:

Tangent lengths give a slant side of $18 + 2 = 20;$ drop a height to form a right triangle

Solution:

The axial cross-section is a trapezoid $ABCD$ with parallel sides $2$ and $18$ and an inscribed circle (a great circle of the sphere). By equal tangent lengths from $B$ and $C,$ the slant side $BC = 18 + 2 = 20.$ Dropping a perpendicular from $C$ to the bottom base gives a right triangle with horizontal leg $18 - 2 = 16,$ so the height is $\sqrt{20^2 - 16^2} = 12.$ The sphere’s radius is half the height, $6.$

Thus, the correct answer is A.

20.

Each face of a cube is painted either red or blue, each with probability $\tfrac12.$ The color of each face is determined independently. What is the probability that the painted cube can be placed on a horizontal surface so that the four vertical faces are all the same color?

$\dfrac{1}{4}$

$\dfrac{5}{16}$

$\dfrac{3}{8}$

$\dfrac{7}{16}$

$\dfrac{1}{2}$

Concepts:cube geometry casework basic probability

Difficulty rating: 1890

Small Hint:

There are $2^6 = 64$ equally likely colorings

Big Hint:

Count arrangements with all $6,$ exactly $5,$ or a $4\text{-}2$ split on opposite faces

Solution:

There are $2^6 = 64$ colorings. A suitable orientation exists when all six faces are one color ( $2$ ways), exactly five faces are one color ( $2 \cdot 6 = 12$ ways), or four faces are one color with the other color on a pair of opposite faces ( $2 \cdot 3 = 6$ ways). That is $2 + 12 + 6 = 20$ favorable colorings, so the probability is $\dfrac{20}{64} = \dfrac{5}{16}.$

Thus, the correct answer is B.

21.

The graph of $2x^2 + xy + 3y^2$ $- 11x - 20y + 40 = 0$ is an ellipse in the first quadrant of the $xy$ -plane. Let $a$ and $b$ be the maximum and minimum values of $\dfrac{y}{x}$ over all points $(x, y)$ on the ellipse. What is the value of $a + b?$

$3$

$\sqrt{10}$

$\dfrac{7}{2}$

$\dfrac{9}{2}$

$2\sqrt{14}$

Concepts:ellipse tangent line Vieta’s Formulas

Difficulty rating: 2080

Small Hint:

The extreme values of $\dfrac{y}{x}$ come from lines $y = mx$ tangent to the ellipse

Big Hint:

Substitute $y = mx$ and set the discriminant of the resulting quadratic in $x$ to zero

Solution:

The slopes $a$ and $b$ are the values of $m$ for which $y = mx$ meets the ellipse in exactly one point. Substituting gives $\begin{aligned} &(3m^2 + m + 2)x^2 \\ &\quad {}- (20m + 11)x + 40 = 0. \end{aligned}$ Setting its discriminant to zero yields $-80m^2 + 280m - 199 = 0.$ By Vieta’s formulas, $a + b = \dfrac{280}{80} = \dfrac{7}{2}.$

Thus, the correct answer is C.

22.

The square $\begin{array}{|c|c|c|} \hline 50 & b & c \\ \hline d & e & f \\ \hline g & h & 2 \\ \hline \end{array}$ is a multiplicative magic square. That is, the product of the numbers in each row, column, and diagonal is the same. If all the entries are positive integers, what is the sum of the possible values of $g?$

$10$

$25$

$35$

$62$

$136$

Concepts:magic square system of equations divisibility

Difficulty rating: 1940

Small Hint:

Express every entry in terms of $b$ using the equal products

Big Hint:

The corner relations force $c = \dfrac{20}{b}$ and $d = \dfrac{4}{b},$ so $b \in \{1, 2, 4\}$

Solution:

From the equal row, column, and diagonal products, every entry can be written in terms of $b:$ $h = \dfrac{100}{b},$ $g = \dfrac{100}{c},$ $f = \dfrac{100}{d}.$ Comparing rows and columns gives $c = \dfrac{20}{b}$ and $d = \dfrac{4}{b},$ hence $g = 5b$ and $e = 10.$

All entries are positive integers exactly when $b = 1, 2,$ or $4,$ giving $g = 5, 10, 20.$ Their sum is $35.$

Thus, the correct answer is C.

23.

The polynomial $x^3 - 2004x^2 + mx + n$ has integer coefficients and three distinct positive zeros. Exactly one of these is an integer, and it is the sum of the other two. How many values of $n$ are possible?

$250{,}000$

$250{,}250$

$250{,}500$

$250{,}750$

$251{,}000$

Concepts:Vieta’s Formulas polynomial counting integers in a range

Difficulty rating: 2280

Small Hint:

The integer zero is the sum of the other two, which must be conjugates $\dfrac{a}{2} \pm r$

Big Hint:

The integer zero is half of $2004,$ so it is $1002;$ then count valid $r^2$

Solution:

Let the integer zero be $a.$ The other two zeros are irrational conjugates $\dfrac{a}{2} \pm r,$ whose sum $a$ equals the integer zero. Vieta’s formula on the $x^2$ coefficient gives $a + a = 2004,$ so $a = 1002$ and the conjugate pair is $501 \pm r.$

The coefficients are integers exactly when $r^2$ is a positive integer, and the zeros are positive and distinct when $1 \le r^2 \le 501^2 - 1 = 251{,}000.$ Since $r$ cannot be an integer, we exclude the $500$ perfect-square values $r^2 = 1^2, \ldots, 500^2,$ leaving $251{,}000 - 500 = 250{,}500$ values of $n.$

Thus, the correct answer is C.

24.

In $\triangle ABC,$ $AB = BC,$ and $\overline{BD}$ is an altitude. Point $E$ is on the extension of $\overline{AC}$ such that $BE = 10.$ The values of $\tan \angle CBE,$ $\tan \angle DBE,$ and $\tan \angle ABE$ form a geometric progression, and the values of $\cot \angle DBE,$ $\cot \angle CBE,$ $\cot \angle DBC$ form an arithmetic progression. What is the area of $\triangle ABC?$

$16$

$\dfrac{50}{3}$

$10\sqrt{3}$

$8\sqrt{5}$

$18$

Concepts:trigonometric identity geometric sequence isosceles triangle

Difficulty rating: 2390

Small Hint:

Let $\angle DBE = \alpha$ and $\angle DBC = \beta;$ then $\angle CBE = \alpha - \beta$ and $\angle ABE = \alpha + \beta$

Big Hint:

The geometric progression gives $\tan(\alpha - \beta)\tan(\alpha + \beta) = \tan^2\alpha,$ which forces $\alpha = 45^\circ$

Solution:

Let $\angle DBE = \alpha$ and $\angle DBC = \beta.$ Since $\overline{BD}$ is the altitude of the isosceles triangle, $\angle CBE = \alpha - \beta$ and $\angle ABE = \alpha + \beta.$ The geometric progression gives $\tan(\alpha - \beta)\tan(\alpha + \beta) = \tan^2\alpha,$ which simplifies to $\tan^2\beta(\tan^4\alpha - 1) = 0,$ so $\tan\alpha = 1$ and $\alpha = 45^\circ.$

Writing $DC = a$ and $BD = b,$ the arithmetic progression $\cot\angle DBE,$ $\cot\angle CBE,$ $\cot\angle DBC$ becomes $1, \dfrac{b + a}{b - a}, \dfrac{b}{a},$ forcing $b = 3a.$ With $BE = 10$ and $\angle DBE = 45^\circ,$ we get $b = \dfrac{BE}{\sqrt2} = 5\sqrt2,$ so $a = \dfrac{5\sqrt2}{3}.$

The area of $\triangle ABC$ is $\tfrac12 (AC)(BD) = ab$ $= 5\sqrt2 \cdot \dfrac{5\sqrt2}{3}$ $= \dfrac{50}{3}.$

Thus, the correct answer is B.

25.

Given that $2^{2004}$ is a $604$ -digit number whose first digit is $1,$ how many elements of the set $S = \{2^0, 2^1, 2^2, \ldots, 2^{2003}\}$ have a first digit of $4?$

$194$

$195$

$196$

$197$

$198$

Concepts:digits pattern recognition

Difficulty rating: 2360

Small Hint:

The number of digits of $2^k$ increases by $1$ exactly when its leading digit forces a carry

Big Hint:

A power of $2$ has leading digit $8$ or $9$ precisely when the previous power has leading digit $4$

Solution:

The smallest power of $2$ with any given digit-count lies between $10^k$ and $2\cdot10^k,$ so it has leading digit $1.$ Because $2^{2004}$ is a $604$ -digit number beginning with $1,$ it is the first $604$ -digit power. Thus the powers in $S$ contain exactly one leading- $1$ number for each digit-count from $1$ through $603,$ or $603$ in all.

After each leading- $1$ power, the next power leads with $2$ or $3,$ and the following power leads with $4, 5, 6,$ or $7.$ Hence $603$ elements lead with $2$ or $3,$ another $603$ lead with $4$ through $7,$ and the remaining $2004 - 3(603) = 195$ lead with $8$ or $9.$

Finally, halving a power that leads with $8$ or $9$ produces the preceding power with leading digit $4,$ and doubling any leading- $4$ power reverses this. This is a bijection, so there are $195$ elements with first digit $4.$

Thus, the correct answer is B.