2002 AMC 12A — Problems & Online Timed Exam

Timed

1:15:00

1.

Compute the sum of all the roots of $\begin{aligned} &(2x+3)(x-4) \\ &\quad {}+(2x+3)(x-6)=0. \end{aligned}$

$\dfrac{7}{2}$

$4$

$5$

$7$

$13$

Answer: A

Concepts:factoring Vieta’s Formulas

Difficulty rating: 890

Small Hint:

Both terms share the factor $2x+3$

Big Hint:

The equation becomes $(2x+3)(2x-10)=0$

Solution:

Factoring out $2x+3$ gives $(2x+3)\big[(x-4)+(x-6)\big]$ $= (2x+3)(2x-10)$ $= 0.$

The roots are $-\dfrac{3}{2}$ and $5,$ whose sum is $-\dfrac{3}{2} + 5 = \dfrac{7}{2}.$

Thus, the correct answer is A.

2.

Cindy was asked by her teacher to subtract $3$ from a certain number and then divide the result by $9.$ Instead, she subtracted $9$ and then divided the result by $3,$ giving an answer of $43.$ What would her answer have been had she worked the problem correctly?

$15$

$34$

$43$

$51$

$138$

Answer: A

Concepts:work backwards linear equation

Difficulty rating: 1020

Small Hint:

Her computation was $\dfrac{x-9}{3}=43$

Big Hint:

Solve for $x,$ then evaluate $\dfrac{x-3}{9}$

Solution:

Let $x$ be the number. Cindy computed $\dfrac{x-9}{3} = 43,$ so $x - 9 = 129$ and $x = 138.$

The correct computation gives $\dfrac{138-3}{9} = \dfrac{135}{9} = 15.$

Thus, the correct answer is A.

3.

According to the standard convention for exponentiation, $2^{2^{2^2}} = 2^{\left(2^{\left(2^2\right)}\right)} = 2^{16} = 65{,}536.$ If the order in which the exponentiations are performed is changed, how many other values are possible?

$0$

$1$

$2$

$3$

$4$

Answer: B

Concepts:order of operations exponent casework

Difficulty rating: 1270

Small Hint:

There are five ways to parenthesize a tower of four $2$ ’s

Big Hint:

Every parenthesization evaluates to either $2^{16}$ or $2^{8}$

Solution:

The five parenthesizations of $2^{2^{2^2}}$ give $\begin{aligned} &\left(\left(2^2\right)^2\right)^2 = 2^8, \\ &\left(2^{2^2}\right)^2 = 2^8, \\ &\left(2^2\right)^{2^2} = 2^8, \end{aligned}$ $2^{\left(2^2\right)^2} = 2^{16},\quad 2^{2^{2^2}} = 2^{16}.$

So the only values are $2^{16} = 65{,}536$ and $2^8 = 256.$ Besides the standard value there is exactly $1$ other.

Thus, the correct answer is B.

4.

Find the degree measure of an angle whose complement is $25\%$ of its supplement.

$48$

$60$

$75$

$120$

$150$

Answer: B

Concepts:angle sum linear equation

Difficulty rating: 1130

Small Hint:

The complement is $90 - x$ and the supplement is $180 - x$

Big Hint:

Solve $90 - x = \dfrac14(180 - x)$

Solution:

Let the angle be $x.$ Then $90 - x = \dfrac14(180 - x),$ so $360 - 4x = 180 - x.$

This gives $3x = 180,$ so $x = 60.$

Thus, the correct answer is B.

5.

Each of the small circles in the figure has radius one. The innermost circle is tangent to the six circles that surround it, and each of those circles is tangent to the large circle and to its small-circle neighbors. Find the area of the shaded region.

$\pi$

$1.5\pi$

$2\pi$

$3\pi$

$3.5\pi$

Answer: C

Concepts:circle area tangent circles

Difficulty rating: 1270

Small Hint:

Each surrounding circle has its center $2$ units from the center, so the large circle has radius $3$

Big Hint:

Subtract the areas of the seven unit circles from the area of the large circle

Solution:

Each of the six outer unit circles is tangent to the central unit circle, so its center is $2$ units from the center. Adding one more radius, the large circle has radius $3$ and area $9\pi.$

The seven unit circles have total area $7\pi,$ so the shaded region has area $9\pi - 7\pi = 2\pi.$

Thus, the correct answer is C.

6.

For how many positive integers $m$ does there exist at least one positive integer $n$ such that $m \cdot n \le m + n?$

$4$

$6$

$9$

$12$

infinitely many

Answer: E

Concepts:inequality small cases

Difficulty rating: 1350

Small Hint:

Consider what happens when $n = 1$

Big Hint:

For $n = 1,$ the inequality reduces to $m \le m + 1$

Solution:

Taking $n = 1,$ the inequality becomes $m \le m + 1,$ which holds for every positive integer $m.$

So every positive integer $m$ works, and there are infinitely many.

Thus, the correct answer is E.

7.

If an arc of $45^\circ$ on circle $A$ has the same length as an arc of $30^\circ$ on circle $B,$ then the ratio of the area of circle $A$ to the area of circle $B$ is

$\dfrac{4}{9}$

$\dfrac{2}{3}$

$\dfrac{5}{6}$

$\dfrac{3}{2}$

$\dfrac{9}{4}$

Answer: A

Concepts:arc circle area ratio and proportion

Difficulty rating: 1350

Small Hint:

Equal arc lengths: $\dfrac{45}{360}\cdot 2\pi R_A = \dfrac{30}{360}\cdot 2\pi R_B$

Big Hint:

Find $\dfrac{R_A}{R_B};$ the area ratio is its square

Solution:

Equal arc lengths give $\dfrac{45}{360}\cdot 2\pi R_A = \dfrac{30}{360}\cdot 2\pi R_B,$ so $\dfrac{R_A}{R_B} = \dfrac{30}{45} = \dfrac{2}{3}.$

The ratio of areas is $\left(\dfrac{R_A}{R_B}\right)^2 = \dfrac{4}{9}.$

Thus, the correct answer is A.

8.

Betsy designed a flag using blue triangles, small white squares, and a red center square, as shown. Let $B$ be the total area of the blue triangles, $W$ the total area of the white squares, and $R$ the area of the red square. Which of the following is correct?

$B = W$

$W = R$

$B = R$

$3B = 2R$

$2R = W$

Answer: A

Concepts:area decomposition congruence (geometry)

Difficulty rating: 1380

Small Hint:

Cover the whole flag with congruent triangles

Big Hint:

Count how many of those triangles lie in the blue region and in the white region

Solution:

Drawing the diagonals shown, the entire flag is tiled by congruent triangles. Counting, there are $24$ triangles in the blue region, $24$ in the white region, and $16$ in the red square.

Since the blue and white regions contain the same number of triangles, $B = W.$

Thus, the correct answer is A.

9.

Jamal wants to store $30$ computer files on floppy disks, each of which has a capacity of $1.44$ megabytes (mb). Three of his files require $0.8$ mb of memory each, $12$ more require $0.7$ mb each, and the remaining $15$ require $0.4$ mb each. No file can be split between floppy disks. What is the minimal number of floppy disks that will hold all the files?

$12$

$13$

$14$

$15$

$16$

Answer: B

Concepts:optimization bounding to limit cases

Difficulty rating: 1570

Small Hint:

The total memory needed is $3(0.8)+12(0.7)+15(0.4)$ mb

Big Hint:

A disk holding a $0.8$ file has room for only one $0.4$ file, wasting $0.24$ mb; add up this waste

Solution:

The files need $3(0.8)+12(0.7)$ $+15(0.4) = 16.8$ mb, so at least $\dfrac{16.8}{1.44} = 11\tfrac{2}{3}$ disks by volume alone.

A disk containing a $0.8$ -mb file has room for only one more $0.4$ -mb file, leaving at least $0.24$ mb unused. Across the three $0.8$ -mb files this wastes at least $3(0.24) = 0.72$ mb, over half a disk, forcing at least $13$ disks.

Thirteen suffice: six disks each hold two $0.7$ -mb files, three disks each hold one $0.8$ -mb file plus one $0.4$ -mb file, and four disks each hold three $0.4$ -mb files.

Thus, the correct answer is B.

10.

Sarah pours four ounces of coffee into an eight-ounce cup and four ounces of cream into a second cup of the same size. She then transfers half the coffee from the first cup to the second and, after stirring thoroughly, transfers half the liquid in the second cup back to the first. What fraction of the liquid in the first cup is now cream?

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{3}{8}$

$\dfrac{2}{5}$

$\dfrac{1}{2}$

Answer: D

Concepts:mixture fraction

Difficulty rating: 1440

Small Hint:

After the first transfer, cup $2$ holds $2$ oz coffee and $4$ oz cream

Big Hint:

Transferring back half of cup $2$ moves its coffee and cream in equal proportion; find the cream in cup $1$

Solution:

After transferring half the coffee, cup $1$ has $2$ oz coffee and cup $2$ has $2$ oz coffee and $4$ oz cream, a total of $6$ oz.

Transferring half of cup $2$ back moves $1$ oz coffee and $2$ oz cream. Cup $1$ then holds $2 + 1 = 3$ oz coffee and $2$ oz cream.

The fraction that is cream is $\dfrac{2}{2+3} = \dfrac{2}{5}.$

Thus, the correct answer is D.

11.

Mr. Earl E. Bird leaves his house for work at exactly $8{:}00$ A.M. every morning. When he averages $40$ miles per hour, he arrives at his workplace three minutes late. When he averages $60$ miles per hour, he arrives three minutes early. At what average speed, in miles per hour, should Mr. Bird drive to arrive at his workplace precisely on time?

$45$

$48$

$50$

$55$

$58$

Answer: B

Concepts:distance rate and time system of equations

Difficulty rating: 1380

Small Hint:

Let $t$ be the on-time travel time in hours; three minutes is $0.05$ hours

Big Hint:

Set $40(t+0.05) = 60(t-0.05),$ solve for $t,$ then find the distance

Solution:

Let $t$ be the time in hours to arrive on time. Since three minutes is $0.05$ hours, $40(t+0.05) = 60(t-0.05).$

This gives $40t + 2 = 60t - 3,$ so $t = 0.25.$ The distance is $40(0.25+0.05) = 12$ miles, and the required speed is $\dfrac{12}{0.25} = 48$ miles per hour.

Thus, the correct answer is B.

12.

Both roots of the quadratic equation $x^2 - 63x + k = 0$ are prime numbers. The number of possible values of $k$ is

$0$

$1$

$2$

$4$

more than four

Answer: B

Concepts:Vieta’s Formulas prime parity

Difficulty rating: 1350

Small Hint:

If the roots are primes $p$ and $q,$ then $p + q = 63$

Big Hint:

An odd sum forces one of the primes to be $2$

Solution:

If the roots are primes $p$ and $q,$ then by Vieta’s formulas $p + q = 63$ and $pq = k.$

Since $63$ is odd, one prime must be even, namely $2,$ and the other is $61.$ Both are prime, so $k = 2\cdot 61 = 122$ is the only possible value.

Thus, the correct answer is B.

13.

Two different positive numbers $a$ and $b$ each differ from their reciprocals by $1.$ What is $a + b?$

$1$

$2$

$\sqrt{5}$

$\sqrt{6}$

$3$

Answer: C

Concepts:rational equation quadratic

Difficulty rating: 1500

Small Hint:

A number $x$ with $\left|x - \dfrac1x\right| = 1$ satisfies $x^2 - x - 1 = 0$ or $x^2 + x - 1 = 0$

Big Hint:

The two positive solutions are reciprocals of each other; add them using the quadratic formula

Solution:

A positive number $x$ differs from its reciprocal by $1$ when $x - \dfrac1x = 1$ or $x - \dfrac1x = -1,$ i.e. $x^2 - x - 1 = 0$ or $x^2 + x - 1 = 0.$

The positive roots are $\dfrac{1+\sqrt5}{2}$ and $\dfrac{-1+\sqrt5}{2},$ which are reciprocals of each other. Their sum is $a + b = \sqrt5.$

Thus, the correct answer is C.

14.

For all positive integers $n,$ let $f(n) = \log_{2002} n^2.$ Let $N = f(11) + f(13) + f(14).$ Which of the following relations is true?

$N \gt 1$

$N = 1$

$1 \lt N \lt 2$

$N = 2$

$N \gt 2$

Answer: D

Concepts:logarithm

Difficulty rating: 1500

Small Hint:

$f(n) = \log_{2002} n^2 = 2\log_{2002} n$

Big Hint:

Combine into $N = \log_{2002}(11\cdot 13\cdot 14)^2$ and evaluate the product inside

Solution:

Using $\log a^2 = 2\log a$ and adding logs, $\begin{aligned} N &= \log_{2002} 11^2 + \log_{2002} 13^2 \\ &\quad {}+ \log_{2002} 14^2 \\ &= \log_{2002}(11\cdot 13\cdot 14)^2. \end{aligned}$

Since $11\cdot 13\cdot 14 = 2002,$ this is $\log_{2002} 2002^2 = 2.$

Thus, the correct answer is D.

15.

The mean, median, unique mode, and range of a collection of eight integers are all equal to $8.$ The largest integer that can be an element of this collection is

$11$

$12$

$13$

$14$

$15$

Answer: D

Concepts:mean range extremal argument

Difficulty rating: 1660

Small Hint:

The eight integers sum to $8\cdot 8 = 64,$ and the mode $8$ must repeat

Big Hint:

Test the largest value $15;$ the smallest would then be $7,$ and check the remaining sum

Solution:

The collection $6, 6, 6, 8, 8, 8, 8, 14$ has mean, median, unique mode, and range all equal to $8,$ so $14$ is attainable.

Suppose the largest were $15.$ The range $8$ forces the smallest to be $7.$ Because $8$ is the mode, it occurs in the sorted list; together with median $8,$ this forces the two middle values to be $8, 8.$ Then $7 + 8 + 8 + 15 = 38,$ so the remaining four values sum to $64 - 38 = 26,$ averaging $6.5.$ At least one would be below $7,$ contradicting the minimum. So $15$ is impossible.

Thus, the correct answer is D.

16.

Tina randomly selects two distinct numbers from the set $\{1, 2, 3, 4, 5\},$ and Sergio randomly selects a number from the set $\{1, 2, \ldots, 10\}.$ The probability that Sergio’s number is larger than the sum of the two numbers chosen by Tina is

$\dfrac{2}{5}$

$\dfrac{9}{20}$

$\dfrac{1}{2}$

$\dfrac{11}{20}$

$\dfrac{24}{25}$

Answer: A

Concepts:basic probability casework

Difficulty rating: 1630

Small Hint:

Tina has $\binom{5}{2} = 10$ equally likely pairs

Big Hint:

For a pair with sum $s,$ Sergio exceeds it with probability $\dfrac{10 - s}{10}$

Solution:

Tina’s ten pairs have sums $3, 4, 5, 5, 6, 6, 7, 7, 8, 9.$ For a sum $s,$ Sergio’s number exceeds it with probability $\dfrac{10 - s}{10}.$

The corresponding values of $10 - s$ are $7, 6, 5, 5, 4, 4, 3, 3, 2, 1,$ totaling $40.$ The overall probability is $\dfrac{40}{10\cdot 10} = \dfrac{2}{5}.$

Thus, the correct answer is A.

17.

Several sets of prime numbers, such as $\{7, 83, 421, 659\},$ use each of the nine nonzero digits exactly once. What is the smallest possible sum such a set of primes could have?

$193$

$207$

$225$

$252$

$477$

Answer: B

Concepts:prime digits optimization

Difficulty rating: 1800

Small Hint:

The digits $4, 6, 8$ cannot end a prime past one digit, so each must be a tens digit

Big Hint:

Those three contribute at least $40 + 60 + 80,$ and the remaining six digits at least their own sum as units digits; then check a valid set exists

Solution:

The even digits $4, 6, 8$ cannot be the units digit of a multi-digit prime, so each must appear in a tens place or higher, contributing at least $40 + 60 + 80 = 180.$ The other six digits contribute at least $1 + 2 + 3 + 5 + 7 + 9 = 27,$ so the sum is at least $207.$

This bound is achieved, for example by $\{2, 3, 5, 41, 67, 89\},$ whose sum is $207.$

Thus, the correct answer is B.

18.

Let $C_1$ and $C_2$ be circles defined by $(x - 10)^2 + y^2 = 36$ and $(x + 15)^2 + y^2 = 81,$ respectively. What is the length of the shortest line segment $\overline{PQ}$ that is tangent to $C_1$ at $P$ and to $C_2$ at $Q?$

$15$

$18$

$20$

$21$

$24$

Answer: C

Concepts:tangent line similarity Pythagorean Theorem

Difficulty rating: 1660

Small Hint:

The shortest such segment is the internal tangent; it crosses $\overline{AB}$ between the centers

Big Hint:

The crossing point divides $\overline{AB}$ in the ratio of the radii $6 : 9;$ use similar right triangles

Solution:

The centers are $A = (10, 0)$ and $B = (-15, 0),$ with radii $6$ and $9,$ so $AB = 25.$ The shortest tangent is the internal one, meeting $\overline{AB}$ at a point $D$ that splits it in the ratio $6 : 9,$ giving $D = (0, 0).$

The right triangles $APD$ and $BQD$ are similar with ratio $2 : 3.$ Then $PD = \sqrt{10^2 - 6^2} = 8$ and $QD = \sqrt{15^2 - 9^2} = 12,$ so $PQ = 8 + 12 = 20.$

Thus, the correct answer is C.

19.

The graph of the function $f$ is shown below. How many solutions does the equation $f(f(x)) = 6$ have?

$2$

$4$

$5$

$6$

$7$

Answer: D

Concepts:function data and graph interpretation

Difficulty rating: 1660

Small Hint:

$f(f(x)) = 6$ means $f(x)$ must be an input where $f$ equals $6$

Big Hint:

From the graph $f(t) = 6$ at $t = -2$ and $t = 1;$ count solutions of $f(x) = -2$ and $f(x) = 1$

Solution:

The graph reaches $6$ at $x = -2$ and $x = 1,$ so $f(f(x)) = 6$ requires $f(x) = -2$ or $f(x) = 1.$

The horizontal line $y = -2$ meets the graph twice, and $y = 1$ meets it four times, giving $2 + 4 = 6$ solutions.

Thus, the correct answer is D.

20.

Suppose that $a$ and $b$ are digits, not both nine and not both zero, and the repeating decimal $0.\overline{ab}$ is expressed as a fraction in lowest terms. How many different denominators are possible?

$3$

$4$

$5$

$8$

$9$

Answer: C

Concepts:repeating decimal factor

Difficulty rating: 1630

Small Hint:

$0.\overline{ab} = \dfrac{\overline{ab}}{99}$ where $\overline{ab}$ is the two-digit value

Big Hint:

In lowest terms the denominator is a divisor of $99 = 3^2\cdot 11;$ exclude $1$ since $a, b$ are not both $9$

Solution:

Since $0.\overline{ab} = \dfrac{\overline{ab}}{99},$ the reduced denominator divides $99 = 3^2\cdot 11.$ The divisors are $1, 3, 9, 11, 33, 99.$

The denominator $1$ would require $\overline{ab} = 99,$ i.e. $a = b = 9,$ which is excluded. Each is achievable: numerators $33, 11, 9, 3,$ and $1$ reduce to denominators $3, 9, 11, 33,$ and $99,$ respectively. Thus there are $5$ possible denominators.

Thus, the correct answer is C.

21.

Consider the sequence of numbers $4,$ $7,$ $1,$ $8,$ $9,$ $7,$ $6,$ $\ldots$ For $n \gt 2,$ the $n$ th term of the sequence is the units digit of the sum of the two previous terms. Let $S_n$ denote the sum of the first $n$ terms of this sequence. The smallest value of $n$ for which $S_n \gt 10{,}000$ is

$1992$

$1999$

$2001$

$2002$

$2004$

Answer: B

Concepts:units digit summation pattern recognition

Difficulty rating: 1840

Small Hint:

Write out terms until the pattern repeats

Big Hint:

The first $12$ terms sum to $60$ and then repeat; find how many full blocks stay under $10{,}000$

Solution:

Continuing the sequence gives $4, 7, 1, 8, 9, 7, 6,$ $3, 9, 2, 1, 3, 4, 7, 1, \ldots,$ which repeats with period $12.$ Each block of $12$ terms sums to $60.$

The largest $k$ with $60k \le 10{,}000$ is $k = 166,$ giving $S_{12\cdot 166} = 9960.$ Adding the next terms $4, 7, 1, 8, 9, 7, 6$ contributes $42,$ pushing the total past $10{,}000.$ So $n = 12\cdot 166 + 7 = 1999.$

Thus, the correct answer is B.

22.

Triangle $ABC$ is a right triangle with $\angle ACB$ as its right angle, $m\angle ABC = 60^\circ,$ and $AB = 10.$ Let $P$ be randomly chosen inside $\triangle ABC,$ and extend $\overline{BP}$ to meet $\overline{AC}$ at $D.$ What is the probability that $BD \gt 5\sqrt{2}?$

$\dfrac{2 - \sqrt{2}}{2}$

$\dfrac{1}{3}$

$\dfrac{3 - \sqrt{3}}{3}$

$\dfrac{1}{2}$

$\dfrac{5 - \sqrt{5}}{5}$

Answer: C

Concepts:geometric probability area ratio special right triangle

Difficulty rating: 1990

Small Hint:

$\triangle ABC$ is $30$ - $60$ - $90$ with $BC = 5$ and $AC = 5\sqrt{3}$

Big Hint:

Choose $E$ on $\overline{AC}$ with $CE = 5,$ so $BE = 5\sqrt{2};$ then $BD \gt 5\sqrt{2}$ iff $P$ lies inside $\triangle ABE$

Solution:

Since $AB = 10$ and $\angle ABC = 60^\circ,$ the $30$ - $60$ - $90$ triangle has $BC = 5$ and $AC = 5\sqrt3.$

Place $E$ on $\overline{AC}$ with $CE = 5;$ then $BE = \sqrt{5^2 + 5^2} = 5\sqrt2.$ As $D$ moves along $\overline{AC},$ $BD = \sqrt{25 + CD^2}$ exceeds $5\sqrt2$ exactly when $CD \gt 5,$ i.e. when $D$ lies beyond $E,$ which happens iff $P$ is inside $\triangle ABE.$

The probability is $\begin{aligned} \dfrac{[ABE]}{[ABC]} &= \dfrac{EA}{CA} \\ &= \dfrac{5\sqrt3 - 5}{5\sqrt3} \\ &= \dfrac{3 - \sqrt3}{3}. \end{aligned}$

Thus, the correct answer is C.

23.

In triangle $ABC,$ side $\overline{AC}$ and the perpendicular bisector of $\overline{BC}$ meet in point $D,$ and $\overline{BD}$ bisects $\angle ABC.$ If $AD = 9$ and $DC = 7,$ what is the area of triangle $ABD?$

$14$

$21$

$28$

$14\sqrt{5}$

$28\sqrt{5}$

Answer: D

Concepts:angle bisector theorem perpendicular bisector law of cosines Heron’s Formula

Difficulty rating: 2110

Small Hint:

$D$ is on the perpendicular bisector of $\overline{BC},$ so $DB = DC = 7$

Big Hint:

The angle bisector gives $\dfrac{AB}{BC} = \dfrac{AD}{DC} = \dfrac{9}{7};$ apply the Law of Cosines with $\cos\theta$ read from $\triangle BDC$

Solution:

Since $D$ lies on the perpendicular bisector of $\overline{BC},$ $DB = DC = 7.$ The angle bisector $\overline{BD}$ gives $\dfrac{AB}{BC} = \dfrac{AD}{DC} = \dfrac{9}{7},$ so write $AB = 9x$ and $BC = 7x.$

Let $\theta = \angle ABD = \angle DBC.$ In isosceles $\triangle BDC,$ the foot of the perpendicular is the midpoint $M$ of $\overline{BC},$ so $\cos\theta = \dfrac{BM}{BD} = \dfrac{\frac{7x}{2}}{7} = \dfrac{x}{2}.$

Applying the Law of Cosines in $\triangle ABD:$ $9^2 = (9x)^2 + 7^2 - 2(9x)(7)\cdot\dfrac{x}{2},$ which simplifies to $81 = 18x^2 + 49,$ so $x = \dfrac43$ and $AB = 12.$

Now $\triangle ABD$ has sides $9, 7, 12.$ By Heron’s formula with $s = 14,$ the area is $\sqrt{14\cdot 5\cdot 7\cdot 2} = \sqrt{980} = 14\sqrt5.$

Thus, the correct answer is D.

24.

Find the number of ordered pairs of real numbers $(a, b)$ such that $(a + bi)^{2002} = a - bi.$

$1001$

$1002$

$2001$

$2002$

$2004$

Answer: E

Concepts:complex number roots of unity

Difficulty rating: 2170

Small Hint:

Write $z = a + bi,$ so the equation is $z^{2002} = \overline{z};$ take magnitudes

Big Hint:

$|z|^{2002} = |z|$ forces $|z| = 0$ or $|z| = 1;$ when $|z| = 1,$ $\overline{z} = \dfrac1z$

Solution:

Let $z = a + bi.$ The equation is $z^{2002} = \overline{z}.$ Taking magnitudes, $|z|^{2002} = |z|,$ so $|z|\big(|z|^{2001} - 1\big) = 0,$ giving $|z| = 0$ or $|z| = 1.$

If $|z| = 0,$ then $(a, b) = (0, 0),$ one solution. If $|z| = 1,$ then $\overline{z} = \dfrac1z,$ so $z^{2002} = \dfrac1z,$ i.e. $z^{2003} = 1,$ which has $2003$ distinct roots.

Altogether there are $1 + 2003 = 2004$ ordered pairs.

Thus, the correct answer is E.

25.

The nonzero coefficients of a polynomial $P$ with real coefficients are all replaced by their mean to form a polynomial $Q.$ Which of the following could be a graph of $y = P(x)$ and $y = Q(x)$ over the interval $-4 \le x \le 4?$

Answer: B

Concepts:polynomial data and graph interpretation

Difficulty rating: 2270

Small Hint:

The sum of the coefficients of $P$ equals the sum of the coefficients of $Q$

Big Hint:

Evaluating at $x = 1$ gives $P(1) = Q(1),$ so the two graphs must intersect at $x = 1$

Solution:

Replacing the nonzero coefficients by their mean keeps the total of the coefficients unchanged, so $P$ and $Q$ have the same coefficient sum. Since $P(1)$ and $Q(1)$ each equal that sum, $P(1) = Q(1).$

Therefore the graphs of $y = P(x)$ and $y = Q(x)$ must cross at $x = 1.$ The only choice showing an intersection at $x = 1$ is graph B. (There, $P(x) = 2x^4 - 3x^2 - 3x - 4$ and $Q(x) = -2x^4 - 2x^2 - 2x - 2.$ )

Thus, the correct answer is B.