2002 AMC 10B — Problems & Online Timed Exam

1.

What is the value of the ratio

\dfrac{2^{2001} \cdot 3^{2003}}{6^{2002}}?

\dfrac{1}{6}

\dfrac{1}{3}

\dfrac{1}{2}

\dfrac{2}{3}

\dfrac{3}{2}

Answer: E

Concepts:exponent

Difficulty rating: 770

Small hint:

Write

6^{2002}

as

2^{2002}\cdot 3^{2002}

Big hint:

Cancel the common powers of

2

and

3

from top and bottom

Solution:

Since

6^{2002} = 2^{2002}\cdot 3^{2002},

the ratio becomes

\dfrac{2^{2001}\cdot 3^{2003}}{2^{2002}\cdot 3^{2002}} = \dfrac{3}{2}.

Thus, the correct answer is E.

2.

For the nonzero numbers

a,

b,

and

c,

define

(a, b, c) = \dfrac{abc}{a+b+c}.

What is

(2, 4, 6)?

1

2

4

6

24

Answer: C

Concepts:custom operation fraction

Difficulty rating: 720

Small hint:

Substitute

a=2,

b=4,

c=6

into the definition

Big hint:

The numerator is

2\cdot4\cdot6

and the denominator is

2+4+6

Solution:

Substituting directly,

(2, 4, 6) = \dfrac{2\cdot4\cdot6}{2+4+6} = \dfrac{48}{12} = 4.

Thus, the correct answer is C.

3.

The arithmetic mean of the nine numbers in the set

\{9, 99, 999, 9999, \ldots, 999999999\}

is a

9

-digit number

M,

all of whose digits are distinct. Which digit does the number

M

not contain?

0

2

4

6

8

Answer: A

Concepts:mean digits place value

Difficulty rating: 960

Small hint:

Each term is

9

times a repunit, so divide the sum by

9

first

Big hint:

\dfrac19(9 + 99 + \cdots)

= 1 + 11 + 111 + \cdots

+ 111111111

Solution:

The mean is

\begin{aligned} &\dfrac{1}{9} \\ &\quad {}\cdot \small \left(9 + 99 + 999 + \cdots + 999999999\right) \\ &= 1 + 11 + 111 + \cdots \\ &\quad {}+ 111111111. \end{aligned}

Adding these nine repunits column by column gives

M = 123456789.

The only digit missing from

M

is

0.

Thus, the correct answer is A.

4.

What is the value of

\begin{aligned} &(3x - 2)(4x + 1) \\ &\quad {}- (3x - 2)4x + 1 \end{aligned}

when

x = 4?

0

1

10

11

12

Answer: D

Concepts:distributive property factoring

Difficulty rating: 900

Small hint:

Factor

3x - 2

out of the first two terms before plugging in

Big hint:

The bracket

(4x + 1) - 4x

collapses to

1

Solution:

Factoring

3x-2

from the first two terms,

\begin{aligned} &(3x - 2)\big[(4x + 1) - 4x\big] + 1 \\ &= (3x - 2)(1) + 1 \\ &= 3x - 1. \end{aligned}

At

x = 4,

this equals

3\cdot4 - 1 = 11.

Thus, the correct answer is D.

5.

Circles of radius

2

and

3

are externally tangent and are circumscribed by a third circle, as shown in the figure. What is the area of the shaded region?

3\pi

4\pi

6\pi

9\pi

12\pi

Answer: E

Concepts:tangent circles circle area

Difficulty rating: 1020

Small hint:

The two small circles lie along a diameter of the large circle

Big hint:

The shaded area is the large disk minus the two smaller disks

Solution:

The two small circles line up along a diameter of the big circle, so that diameter is

2\cdot3 + 2\cdot2 = 10

and the large radius is

5.

The shaded region is the large disk with the two small disks removed:

\begin{aligned} &\pi(5^2) - \pi(3^2) - \pi(2^2) \\ &= \pi(25 - 9 - 4) \\ &= 12\pi. \end{aligned}

Thus, the correct answer is E.

6.

For how many positive integers

n

is

n^2 - 3n + 2

a prime number?

none

one

two

more than two, but finitely many

infinitely many

Answer: B

Concepts:factoring prime

Difficulty rating: 1070

Small hint:

Factor

n^2 - 3n + 2

Big hint:

For

(n-1)(n-2)

to be prime, one factor must equal

1

Solution:

Factor as

n^2 - 3n + 2 = (n-1)(n-2).

For

n \ge 4,

both factors exceed

1,

so the product is composite. For

n = 1

and

n = 2

the value is

0,

and for

n = 3

the value is

(2)(1) = 2,

which is prime.

So exactly one value of

n

works.

Thus, the correct answer is B.

7.

Let

n

be a positive integer such that

\dfrac12 + \dfrac13 + \dfrac17 + \dfrac1n

is an integer. Which of the following statements is not true?

2

divides

n

3

divides

n

6

divides

n

7

divides

n

n \gt 84

Answer: E

Concepts:fraction divisibility bounding to limit cases

Difficulty rating: 1170

Small hint:

The sum is positive and less than

2,

so it must equal exactly

1

Big hint:

Solve

\dfrac1n = 1 - \dfrac12 - \dfrac13 - \dfrac17

for

n

Solution:

The sum

\dfrac12 + \dfrac13 + \dfrac17 + \dfrac1n

is greater than

0

and less than

\dfrac12 + \dfrac13 + \dfrac17 + 1 \lt 2,

so as an integer it must equal

1.

Since

\dfrac12 + \dfrac13 + \dfrac17 = \dfrac{41}{42},

we need

\dfrac1n = \dfrac{1}{42},

so

n = 42.

Then

2,

3,

6,

and

7

all divide

42,

but

n = 42

is not greater than

84.

So the false statement is

n \gt 84.

Thus, the correct answer is E.

8.

Suppose July of year

N

has five Mondays. Which of the following must occur five times in August of year

N?

(Note: both months have

31

days.)

Monday

Tuesday

Wednesday

Thursday

Friday

Answer: D

Concepts:date and time modular arithmetic casework

Difficulty rating: 1240

Small hint:

A

31

-day month has exactly three weekdays that occur five times

Big hint:

Those are the weekdays of days

1,

2,

and

3

; find where a five-Monday July forces them

Solution:

A

31

-day month is

4

weeks plus

3

extra days, so exactly the weekdays of the

1

st,

2

nd, and

3

rd of the month occur five times.

For July to have five Mondays, Monday must be one of July

1,

2,

or

3.

In all three cases August

1

lands on a Tuesday, Wednesday, or Thursday, and the common weekday among the resulting five-time days is Thursday.

More directly, since July has

31

days, August

1

is the same weekday as July

4.

With Monday on July

1,

2,

or

3,

the three five-time weekdays of August always include Thursday.

Thus, the correct answer is D.

9.

Using the letters

A,

M,

O,

S,

and

U,

we can form

120

five-letter “words.” If these “words” are arranged in alphabetical order, then the “word”

USAMO

occupies which position?

112

113

114

115

116

Answer: D

Concepts:permutations systematic listing

Difficulty rating: 1280

Small hint:

Words starting with

A,

M,

O,

or

S

come first; count them

Big hint:

Among the words starting with

U,

find how many precede

USAMO

Solution:

The alphabetical order of the letters is

A, M, O, S, U.

Words beginning with

A,

M,

O,

or

S

fill positions

1

through

96

(four choices of first letter,

24

each).

Words beginning with

U

occupy positions

97

–

120.

Listing them alphabetically,

USAMO

is the

19

th such word, so it occupies position

96 + 19 = 115.

Thus, the correct answer is D.

10.

Suppose that

a

and

b

are nonzero real numbers, and that the equation

x^2 + ax + b = 0

has solutions

a

and

b.

What is the pair

(a, b)?

(-2, 1)

(-1, 2)

(1, -2)

(2, -1)

(4, 4)

Answer: C

Concepts:Vieta’s Formulas system of equations

Difficulty rating: 1280

Small hint:

By Vieta’s formulas, the sum of the roots is

-a

and the product is

b

Big hint:

So

a + b = -a

and

ab = b

; use

b \ne 0

Solution:

Since the roots are

a

and

b,

Vieta’s formulas give

a + b = -a

and

ab = b.

From

ab = b

with

b \ne 0,

we get

a = 1.

Then

a + b = -a

gives

1 + b = -1,

so

b = -2.

Thus

(a, b) = (1, -2),

and the correct answer is C.

11.

The product of three consecutive positive integers is

8

times their sum. What is the sum of their squares?

50

77

110

149

194

Answer: B

Concepts:algebraic manipulation quadratic

Difficulty rating: 1140

Small hint:

Call the integers

n - 1,

n,

n + 1

; their sum is

3n

Big hint:

The product is

(n-1)n(n+1) = n(n^2 - 1),

which equals

8\cdot 3n

Solution:

Let the integers be

n - 1,

n,

n + 1.

Their product is

n(n^2 - 1)

and their sum is

3n,

so

n(n^2 - 1) = 8(3n) = 24n.

Since

n \ne 0,

we get

n^2 - 1 = 24,

so

n^2 = 25

and

n = 5.

The three integers are

4,

5,

and

6,

and

4^2 + 5^2 + 6^2 = 16 + 25 + 36

= 77.

Thus, the correct answer is B.

12.

For which of the following values of

k

does the equation

\dfrac{x - 1}{x - 2} = \dfrac{x - k}{x - 6}

have no solution for

x?

1

2

3

4

5

Answer: E

Concepts:rational equation linear equation

Difficulty rating: 1370

Small hint:

Cross multiply and expand both sides

Big hint:

The equation becomes linear; it has no solution when the coefficient of

x

vanishes

Solution:

Cross multiplying gives

(x - 1)(x - 6) = (x - 2)(x - k),

which expands to

\begin{aligned} x^2 - 7x + 6 &= x^2 - (2 + k)x \\ &\quad {}+ 2k. \end{aligned}

Cancelling

x^2

leaves

(k - 5)x = 2k - 6.

For

k

equal to

1,

2,

3,

or

4,

this gives a valid value of

x

that is neither excluded denominator value

2

nor

6.

For

k=5,

the equation instead becomes

0\cdot x=4,

which has no solution.

Thus, the correct answer is E.

13.

What value of

x

makes

8xy - 12y + 2x - 3 = 0

true for all values of

y?

\dfrac23

\dfrac32

or

-\dfrac14

-\dfrac23

or

-\dfrac14

\dfrac32

-\dfrac32

or

-\dfrac14

Answer: D

Concepts:factoring algebraic manipulation

Difficulty rating: 1220

Small hint:

Group terms to factor the left side

Big hint:

8xy - 12y + 2x - 3

= (4y + 1)(2x - 3)

; it must vanish for every

y

Solution:

Grouping and factoring,

\begin{aligned} &8xy - 12y + 2x - 3 \\ &= 4y(2x - 3) + (2x - 3) \\ &= (4y + 1)(2x - 3). \end{aligned}

For this to equal

0

for all

y,

the factor that depends on

y

cannot be forced to zero, so we need

2x - 3 = 0,

giving

x = \dfrac32.

Thus, the correct answer is D.

14.

The number

25^{64} \cdot 64^{25}

is the square of a positive integer

N.

In decimal representation, what is the sum of the digits of

N?

7

14

21

28

35

Answer: B

Concepts:exponent digits

Difficulty rating: 1420

Small hint:

Take the square root by writing everything as powers of

2

and

5

Big hint:

N = 5^{64}\cdot 2^{75} = 10^{64}\cdot 2^{11}

; only the digits of

2^{11}

matter

Solution:

Since

25 = 5^2

and

64 = 2^6,

we have

25^{64}\cdot 64^{25} = 5^{128}\cdot 2^{150},

so

N = \sqrt{5^{128}\cdot 2^{150}} = 5^{64}\cdot 2^{75}.

Writing

2^{75} = 2^{64}\cdot 2^{11},

we get

N = (5\cdot 2)^{64}\cdot 2^{11} = 10^{64}\cdot 2048.

So

N

is

2048

followed by

64

zeros, and its digit sum is

2 + 0 + 4 + 8 = 14.

Thus, the correct answer is B.

15.

The positive integers

A,

B,

A - B,

and

A + B

are all prime numbers. The sum of these four primes is

even

divisible by

3

divisible by

5

divisible by

7

prime

Answer: E

Concepts:prime parity

Difficulty rating: 1480

Small hint:

A - B

and

A + B

have the same parity, and both are prime

Big hint:

This forces

B = 2,

so

A - 2,

A,

and

A + 2

are three primes in a row

Solution:

The numbers

A - B

and

A + B

differ by

2B,

so they have the same parity. Being prime, they must both be odd, which forces

A

and

B

to have opposite parity.

Since

2

is the only even prime, either

A=2

or

B=2.

The first case is impossible because the positive prime

B

would make

A-B<0.

Hence

B=2.

Now

A - 2,

A,

and

A + 2

are three primes. One of any three integers spaced

2

apart is divisible by

3,

so that member must itself be

3.

The only positive possibility is

3,

5,

7.

The four primes are

2,

3,

5,

7,

and their sum is

17,

which is prime.

Thus, the correct answer is E.

16.

For how many integers

n

is

\dfrac{n}{20 - n}

the square of an integer?

1

2

3

4

10

Answer: D

Concepts:Diophantine Equation divisibility perfect square

Difficulty rating: 1580

Small hint:

Set

\dfrac{n}{20 - n} = k^2

and solve for

n

in terms of

k

Big hint:

n = \dfrac{20k^2}{k^2 + 1}

; since

k^2

and

k^2 + 1

are coprime,

20

is divisible by

k^2 + 1

Solution:

Suppose

\dfrac{n}{20 - n} = k^2

for some integer

k \ge 0.

Solving,

n = \dfrac{20k^2}{k^2 + 1}.

Since

k^2

and

k^2 + 1

share no common factor,

k^2 + 1

must divide

20.

This happens only for

k = 0,

1,

2,

3,

giving

k^2 + 1 = 1,

2,

5,

10.

The corresponding values

n = 0,

10,

16,

18

are all integers, so there are

4

such

n.

Thus, the correct answer is D.

17.

A regular octagon

ABCDEFGH

has sides of length two. What is the area of

\triangle ADG?

4 + 2\sqrt{2}

6 + \sqrt{2}

4 + 3\sqrt{2}

3 + 4\sqrt{2}

8 + \sqrt{2}

Answer: C

Concepts:regular polygon coordinate geometry triangle area

Difficulty rating: 1660

Small hint:

Place the octagon on a grid; each slanted side moves

\sqrt2

horizontally and vertically

Big hint:

DG

is a horizontal chord; use it as the base and the vertical distance from

A

as the height

Solution:

Set the octagon on coordinate axes with the axis-aligned sides of length

2

and each slanted side spanning

\sqrt2

horizontally and

\sqrt2

vertically. Then

\begin{aligned} A &= (\sqrt2, 0), \\ D &= (2 + 2\sqrt2, 2 + \sqrt2), \\ G &= (0, 2 + \sqrt2). \end{aligned}

Since

D

and

G

share the height

2 + \sqrt2,

segment

DG

is horizontal with length

2 + 2\sqrt2,

and the height from

A

up to that level is

2 + \sqrt2.

Therefore

\begin{aligned} [\triangle ADG] &= \tfrac12(2 + 2\sqrt2)(2 + \sqrt2) \\ &= (1 + \sqrt2)(2 + \sqrt2) \\ &= 4 + 3\sqrt2. \end{aligned}

Thus, the correct answer is C.

18.

Four distinct circles are drawn in a plane. What is the maximum number of points where at least two of the circles intersect?

8

9

10

12

16

Answer: D

Concepts:counting intersections counting pairs

Difficulty rating: 1280

Small hint:

Two distinct circles meet in at most

2

points

Big hint:

Count the pairs of circles, then multiply by

2

Solution:

Any two distinct circles intersect in at most

2

points. There are

\binom{4}{2} = 6

pairs of circles, giving at most

6\cdot 2 = 12

intersection points.

This maximum is achievable by a configuration where every pair of circles crosses twice, so the answer is

12.

Thus, the correct answer is D.

19.

Suppose that

\{a_n\}

is an arithmetic sequence with

a_1 + a_2 + \cdots + a_{100} = 100

and

a_{101} + a_{102} + \cdots + a_{200} = 200.

What is the value of

a_2 - a_1?

0.0001

0.001

0.01

0.1

1

Answer: C

Concepts:arithmetic sequence summation

Difficulty rating: 1460

Small hint:

Each term

a_{k+100}

exceeds

a_k

by

100d,

where

d = a_2 - a_1

Big hint:

Subtract the two block sums to isolate

10000d

Solution:

Let

d = a_2 - a_1.

Then

a_{k+100} = a_k + 100d,

so the second block sum is the first plus

100\cdot 100 d:

\begin{aligned} &a_{101} + \cdots + a_{200} \\ &= (a_1 + \cdots + a_{100}) \\ &\quad {}+ 10000d. \end{aligned}

Therefore

200 = 100 + 10000d,

giving

d = \dfrac{100}{10000} = 0.01.

Thus, the correct answer is C.

20.

Let

a,

b,

and

c

be real numbers such that

a - 7b + 8c = 4

and

8a + 4b - c = 7.

What is

a^2 - b^2 + c^2?

0

1

4

7

8

Answer: B

Concepts:algebraic manipulation system of equations

Difficulty rating: 1790

Small hint:

Rearrange to

a + 8c = 4 + 7b

and

8a - c = 7 - 4b

Big hint:

Square both equations and add; the cross terms with

b

cancel nicely

Solution:

Rewrite the equations as

a + 8c = 4 + 7b

and

8a - c = 7 - 4b.

Squaring both and adding,

\begin{aligned} &(a + 8c)^2 + (8a - c)^2 \\ &= (4 + 7b)^2 \\ &\quad {}+ (7 - 4b)^2. \end{aligned}

The left side expands to

65a^2 + 65c^2

(the

ac

terms cancel), and the right side expands to

65 + 65b^2

(the

b

terms cancel). So

65(a^2 + c^2) = 65(1 + b^2),

giving

a^2 - b^2 + c^2 = 1.

Thus, the correct answer is B.

21.

Andy’s lawn has twice as much area as Beth’s lawn and three times as much area as Carlos’ lawn. Carlos’ lawn mower cuts half as fast as Beth’s mower and one third as fast as Andy’s mower. If they all start to mow their lawns at the same time, who will finish first?

Andy

Beth

Carlos

Andy and Carlos tie for first.

All three tie.

Answer: B

Concepts:rate ratio and proportion

Difficulty rating: 1370

Small hint:

Time to mow equals area divided by mowing rate

Big hint:

Let Andy’s area be

A

and Carlos’ rate be

R,

then write each person’s time

Solution:

Let Andy’s lawn have area

A,

so Beth’s is

\dfrac{A}{2}

and Carlos’ is

\dfrac{A}{3}.

Let Carlos mow at rate

R,

so Beth mows at

2R

and Andy at

3R.

The times are

\begin{aligned} &\text{Andy: } \dfrac{A}{3R}, \\ &\text{Beth: } \dfrac{\frac{A}{2}}{2R} = \dfrac{A}{4R}, \\ &\text{Carlos: } \dfrac{\frac{A}{3}}{R} = \dfrac{A}{3R}. \end{aligned}

Since

\dfrac{A}{4R}

is the smallest, Beth finishes first.

Thus, the correct answer is B.

22.

Let

\triangle XOY

be a right-angled triangle with

m\angle XOY = 90^\circ.

Let

M

and

N

be the midpoints of legs

OX

and

OY,

respectively. Given that

XN = 19

and

YM = 22,

what is

XY?

24

26

28

30

32

Answer: B

Concepts:Pythagorean Theorem median (geometry)system of equations

Difficulty rating: 1690

Small hint:

Let

OM = a

and

ON = b,

so

OX = 2a

and

OY = 2b

Big hint:

Write the Pythagorean relations for

XN

and

YM,

then add them to find

a^2 + b^2

Solution:

Let

OM = a

and

ON = b,

so

OX = 2a

and

OY = 2b.

The right angle at

O

gives

19^2 = (2a)^2 + b^2

and

22^2 = a^2 + (2b)^2.

Adding these,

5(a^2 + b^2) = 19^2 + 22^2 = 845,

so

a^2 + b^2 = 169

and

MN = \sqrt{a^2 + b^2} = 13.

Since

\triangle XOY \sim \triangle MON

with ratio

2,

we have

XY = 2\cdot MN = 26.

Thus, the correct answer is B.

23.

Let

\{a_k\}

be a sequence of integers such that

a_1 = 1

and

a_{m+n} = a_m + a_n + mn

for all positive integers

m

and

n.

What is

a_{12}?

45

56

67

78

89

Answer: D

Concepts:functional equation telescoping triangular number

Difficulty rating: 1510

Small hint:

Set

n = 1

to get

a_{m+1} = a_m + a_1 + m

Big hint:

So

a_{m+1} - a_m = m + 1

; sum these differences up to

a_{12}

Solution:

Setting

n = 1,

we get

a_{m+1} = a_m + a_1 + m

= a_m + (m + 1),

so

a_{m+1} - a_m = m + 1.

Summing from

m = 1

to

11,

\begin{aligned} a_{12} - a_1 &= 2 + 3 + \cdots + 12 \\ &= \dfrac{12\cdot 13}{2} - 1 \\ &= 77. \end{aligned}

Therefore

a_{12} = 1 + 77 = 78.

Thus, the correct answer is D.

24.

Riders on a Ferris wheel travel in a circle in a vertical plane. A particular wheel has radius

20

feet and revolves at the constant rate of one revolution per minute. How many seconds does it take a rider to travel from the bottom of the wheel to a point

10

vertical feet above the bottom?

5

6

7.5

10

15

Answer: D

Concepts:circle special right triangle ratio and proportion

Difficulty rating: 1580

Small hint:

Place the center at height

20

; the rider rises

10

feet from the bottom

Big hint:

The rider is now

10

feet below center; find the central angle turned from the bottom

Solution:

Put the center

O

at height

20.

The bottom

A

is at height

0,

and the rider reaches height

10,

which is

10

feet below the center.

Draw a horizontal segment from the rider to the wheel’s vertical diameter. The resulting right triangle has a vertical leg of length

10

and a hypotenuse (the radius) of length

20.

That leg is half the hypotenuse, so the radius to the rider makes

60^\circ

with the downward vertical.

The wheel turns

360^\circ

in

60

seconds, so turning

60^\circ

takes

\dfrac{60}{360}\cdot 60 = 10

seconds.

Thus, the correct answer is D.

25.

When

15

is appended to a list of integers, the mean is increased by

2.

When

1

is appended to the enlarged list, the mean of the enlarged list is decreased by

1.

How many integers were in the original list?

4

5

6

7

8

Answer: A

Concepts:mean system of equations

Difficulty rating: 1690

Small hint:

Let the list have

n

integers with mean

m,

so the sum is

mn

Big hint:

Translate each appending into an equation for the new sum and mean

Solution:

Let the original list have

n

integers with mean

m,

so its sum is

mn.

Appending

15

gives

\begin{aligned} &(m + 2)(n + 1) \\ &= mn + 15 \\ &\implies m + 2n = 13. \end{aligned}

Appending

1

to that enlarged list gives

\begin{aligned} &(m + 1)(n + 2) \\ &= mn + 16 \\ &\implies 2m + n = 14. \end{aligned}

Solving

m + 2n = 13

and

2m + n = 14

yields

m = 5

and

n = 4.

Thus, the correct answer is A.

2002 AMC 10B Problems

Answer: E

Difficulty rating: 770

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Answer: C

Difficulty rating: 720

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Answer: A

Difficulty rating: 960

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Answer: D

Difficulty rating: 900

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Answer: E

Difficulty rating: 1020

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Answer: B

Difficulty rating: 1070

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Answer: E

Difficulty rating: 1170

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Answer: D

Difficulty rating: 1240

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Answer: D

Difficulty rating: 1280

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Answer: C

Difficulty rating: 1280

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Answer: B

Difficulty rating: 1140

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Answer: E

Difficulty rating: 1370

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Answer: D

Difficulty rating: 1220

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Answer: B

Difficulty rating: 1420

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Answer: E

Difficulty rating: 1480

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Answer: D

Difficulty rating: 1580

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