1987 AIME — Problems & Online Timed Exam

1.

An ordered pair

(m,n)

of nonnegative integers is called “simple” if adding

m+n

in base

10

requires no carrying. Find the number of simple ordered pairs that sum to

1492.

Answer

Answer: 300

Concepts:digits place value multiplication principle

Difficulty rating: 1490

Small hint:

Treat the four decimal places independently

Big hint:

A target digit

d

can be split into an ordered pair of digits in

d+1

ways without carrying

Solution:

For a target digit

d,

there are

d+1

ordered pairs of nonnegative digits with sum

d.

The four digits

1,

4,

9,

2

therefore give

2,

5,

10,

3

choices independently. The answer is

2\cdot5\cdot10\cdot3=300.

2.

What is the largest possible distance between two points, one on the sphere of radius

19

centered at

(-2,-10,5)

and the other on the sphere of radius

87

centered at

(12,8,-16)?

Answer

Answer: 137

Concepts:sphere distance formula triangle inequality

Difficulty rating: 1340

Small hint:

First find the distance between the two centers

Big hint:

The maximum occurs along the line of centers, on the two outward sides

Solution:

The squared distance between the centers is

14^2+18^2+(-21)^2=961,

so they are

\sqrt{961}=31

apart. By the triangle inequality, the greatest point-to-point distance is obtained on the line of centers and equals

19+31+87=137.

3.

A proper divisor of a natural number is a positive integral divisor other than

1

and the number itself. A natural number greater than

1

is called “nice” if it equals the product of its distinct proper divisors. What is the sum of the first ten nice numbers?

Answer

Answer: 182

Concepts:factor counting prime factorization

Difficulty rating: 1830

Small hint:

Express the product of all positive divisors in terms of the number and its divisor count

Big hint:

The nice numbers are exactly those having four positive divisors

Solution:

If

N

has

d

positive divisors, their product is

N^{\frac{d}{2}}.

Removing

1

and

N

leaves product

N^{\frac{d}{2}-1},

which equals

N

exactly when

d=4.

Thus nice numbers are precisely

p^3

and

pq

for distinct primes. The first ten are

6,

8,

10,

14,

15,

21,

22,

26,

27,

33,

whose sum is

182.

4.

Find the area of the region enclosed by the graph of

|x-60|+|y|=|\frac{x}{4}|.

Answer

Answer: 480

Concepts:absolute value coordinate geometry kite

Difficulty rating: 1810

Small hint:

Solve for

|y|

and determine where its right-hand side is nonnegative

Big hint:

The boundary is a kite whose vertices occur at the breakpoints of the absolute values

Solution:

We need

|y|=\frac{|x|}{4}-|x-60|\geq0.

This forces

48\leq x\leq80.

At

x=48,

x=60,

x=80,

the boundary values are respectively

y=0,

y=\pm15,

y=0.

Hence the region is a kite with perpendicular diagonals

80-48=32

and

30,

so its area is

\frac12(32)(30)=480.

5.

Find

3x^2y^2

if

x

and

y

are integers such that

y^2+3x^2y^2=30x^2+517.

Answer

Answer: 588

Concepts:factoring Diophantine Equation divisibility

Difficulty rating: 2070

Small hint:

Move a suitable multiple of

3x^2+1

to create a product

Big hint:

Factor

507

and use that

3x^2+1\equiv1\pmod3

Solution:

Rearranging gives

(3x^2+1)(y^2-10)=507,

where

507=3\cdot13^2.

The positive divisors of

507

congruent to

1\pmod3

are

1,

13,

169.

If

3x^2+1=1,

then

x^2=0

but

y^2=517,

which is not a square. If

3x^2+1=13,

then

x^2=4

and

y^2=49.

Finally,

3x^2+1=169

gives

x^2=56,

not a square. Therefore

3x^2y^2=3\cdot4\cdot49=588.

6.

Rectangle

ABCD

is divided into four parts of equal area by five segments as shown, where

XY=YB+BC+CZ,

YB+BC+CZ=ZW,

ZW=WD+DA+AX,

and

PQ\parallel AB.

Find

AB

(in cm) if

BC=19

cm and

PQ=87

cm.

Answer

Answer: 193

Concepts:rectangle system of equations trapezoid

Difficulty rating: 2270

Small hint:

Call the common boundary length

t

and the rectangle width

L

Big hint:

Equal areas above and below

PQ

place

PQ

halfway up the rectangle

Solution:

Put

AB=L

and let the four equal boundary lengths be

t.

Writing

AX=x

and

DW=w,

the left boundary condition gives

x+w=t-19.

Substituting

XY=ZW=t

into the right condition gives

2L+38=4t,

so

t=\frac{L+19}{2}.

The upper and lower central regions are trapezoids with the same bases

t

and

87.

Since their areas are equal,

PQ

lies halfway up the

19

-cm rectangle. Each central region therefore has area

\frac{19(t+87)}{4}.

This is one quarter of the rectangle,

\frac{19L}{4},

so

t+87=L.

Combining with

t=\frac{L+19}{2}

yields

L=193.

7.

Let

[r,s]

denote the least common multiple of positive integers

r,

s.

Find the number of ordered triples

(a,b,c)

for which

[a,b]=1000,

[b,c]=2000,

and

[c,a]=2000.

Answer

Answer: 70

Concepts:least common multiple prime factorization multiplication principle

Difficulty rating: 2230

Small hint:

Treat the exponents of

2

and

5

independently

Big hint:

For each prime, translate every least common multiple into a condition on pairwise maxima

Solution:

For the exponent of

5,

all three pairwise maxima equal

3.

Thus at least two exponents are

3:

there is one all-

3

triple and

3\cdot3

triples with exactly two

3

’s, for

10

choices.

For the exponent of

2,

the first pair has maximum

3

while the other two have maximum

4.

The exponent of

c

must be

4,

and the exponents of

a

and

b

lie in

\{0,1,2,3\}

with maximum

3,

giving

4^2-3^2=7

choices. Independence gives

10\cdot7=70.

8.

What is the largest positive integer

n

for which there is a unique integer

k

such that

\frac8{15}<\frac{n}{n+k}<\frac7{13}?

Answer

Answer: 112

Concepts:inequality fraction bounding to limit cases

Difficulty rating: 1860

Small hint:

Solve both inequalities for

k

Big hint:

Study the integers in the open interval

(\frac{6n}{7},\frac{7n}{8})

Solution:

The inequalities are equivalent to

\frac{6n}{7}<k<\frac{7n}{8}.

This interval has length

\frac{n}{56}.

At

n=112

it is

(96,98),

containing only

97.

For

n>112

its length exceeds

2,

so it contains at least two integers. Thus the largest possible

n

is

112.

9.

Triangle

ABC

has a right angle at

B

and contains a point

P

for which

PA=10,

PB=6,

and

\angle APB=\angle BPC=\angle CPA.

Find

PC.

Answer

Answer: 33

Concepts:vector right triangle

Difficulty rating: 2380

Small hint:

The three equal angles around

P

are each

120^\circ

Big hint:

Use vectors from

P

and translate the right angle at

B

into a dot product

Solution:

Let

\mathbf a,

\mathbf b,

and

\mathbf c

be the vectors from

P

to

A,

B,

C,

and put

|\mathbf c|=x.

Their pairwise angles are

120^\circ,

so

\mathbf a\cdot\mathbf b=-30,

\mathbf b\cdot\mathbf c=-3x,

and

\mathbf a\cdot\mathbf c=-5x.

Since

AB\perp BC,

(\mathbf a-\mathbf b)\cdot(\mathbf c-\mathbf b)=0.

Expanding gives

-5x+30+3x+36=0,

so

x=33.

10.

Al walks down an escalator that is moving up and counts

150

steps. Bob walks up and counts

75

steps. If Al’s walking speed is three times Bob’s, how many steps are visible at a given time? Assume this is constant.

Answer

Answer: 120

Concepts:distance rate and time system of equations

Difficulty rating: 1770

Small hint:

Let Bob’s speed be

b,

the escalator’s upward speed be

e,

and the visible count be

N

Big hint:

Write

N

as net speed times travel time for each person

Solution:

Bob’s time is

\frac{75}{b},

so

N=(b+e)(\frac{75}{b}),

or

N=75(1+\frac{e}{b}).

Al’s time is

\frac{150}{3b}=\frac{50}{b},

so

N=(3b-e)(\frac{50}{b}),

or

N=150-\frac{50e}{b}.

Equating gives

\frac{125e}{b}=75,

hence

\frac{e}{b}=\frac{3}{5}

and

N=75(1+\frac{3}{5})=120.

11.

Find the largest possible

k

for which

3^{11}

is expressible as the sum of

k

consecutive positive integers.

Answer

Answer: 486

Concepts:arithmetic sequence divisibility exponent

Difficulty rating: 2070

Small hint:

Write the sum as

\frac{k(2a+k-1)}{2}

Big hint:

The only possible lengths divide

2\cdot3^{11}

; then enforce a positive first term

Solution:

If the first term is

a,

then

2\cdot3^{11}=k(2a+k-1).

Thus

k

is

3^j

or

2\cdot3^j.

The largest viable even choice is

k=2\cdot3^5=486,

for which

2a+k-1=3^6=729

and

a=122>0.

The next candidates,

3^6

and

2\cdot3^6,

force a nonpositive first term, as do all larger choices. Hence

k=486.

12.

Let

m

be the smallest integer whose cube root has the form

n+r,

where

n

is a positive integer and

0<r<\frac{1}{1000}.

Find

n.

Answer

Answer: 19

Concepts:perfect power estimation inequality

Difficulty rating: 2110

Small hint:

For fixed

n,

the smallest possible integer is

m=n^3+1

Big hint:

Compare

n^3+1

with

(n+0.001)^3

Solution:

For a given

n,

the closest integer cube-root candidate above

n

is

m=n^3+1.

We need

n^3+1<(n+0.001)^3,

or

1<0.003n^2+0.000003n+10^{-9}.

This fails at

n=18

and holds at

n=19.

The right-hand side is increasing for positive

n,

so every smaller

n

fails; every larger

n

has a larger least candidate

m=n^3+1.

Hence the smallest

m

occurs with

n=19.

13.

A given sequence

r_1,

r_2,

\ldots,

r_n

of distinct real numbers can be put in ascending order by means of one or more “bubble passes.” A bubble pass through a given sequence consists of comparing the second term with the first term, and exchanging them if and only if the second term is smaller, then comparing the third term with the second term and exchanging them if and only if the third term is smaller, and so on in order, through comparing the last term,

r_n,

with its current predecessor and exchanging them if and only if the last term is smaller.

The example below shows how the sequence

1,

9,

8,

7

is transformed into the sequence

1,

8,

7,

9

by one bubble pass. The numbers compared at each step are underlined.

\begin{aligned} \underline{1}\quad\underline{9}\quad8\quad7\\ 1\quad\underline{9}\quad\underline{8}\quad7\\ 1\quad8\quad\underline{9}\quad\underline{7}\\ 1\quad8\quad7\quad9 \end{aligned}

Suppose that

n=40,

and that the terms of the initial sequence

r_1,

r_2,

\ldots,

r_{40}

are distinct from one another and are in random order. Let

\frac{p}{q},

in lowest terms, be the probability that the number that begins as

r_{20}

will end up, after one bubble pass, in the

30

th place. Find

p+q.

Answer

Answer: 931

Concepts:basic probability permutations process simulation

Difficulty rating: 2450

Small hint:

After the comparison reaching position

j,

that position holds the maximum of the first

j

original terms

Big hint:

Characterize the relative ranks of

r_{20}

and

r_{31}

among the first

31

terms

Solution:

For

r_{20}

to move right to position

30,

it must exceed every other term among

r_1,\ldots,r_{30}.

It stops at position

30

exactly when

r_{31}>r_{20}.

Thus among the first

31

terms,

r_{31}

must be greatest and

r_{20}

second greatest. These two ordered rank assignments have probability

\frac1{31}\cdot\frac1{30}=\frac1{930}.

Therefore

p+q=1+930=931.

14.

Compute

\frac{\begin{gathered}(10^4+324)(22^4+324)\\{}\cdot(34^4+324)(46^4+324)\\{}\cdot(58^4+324)\end{gathered}}{\begin{gathered}(4^4+324)(16^4+324)\\{}\cdot(28^4+324)(40^4+324)\\{}\cdot(52^4+324)\end{gathered}}.

Answer

Answer: 373

Concepts:algebraic manipulation factoring telescoping

Difficulty rating: 2380

Small hint:

Use Sophie Germain’s identity with

324=4\cdot3^4

Big hint:

If

g(x)=x^2+6x+18,

rewrite

x^4+324

as

g(x-6)g(x)

Solution:

Let

g(x)=x^2+6x+18.

Sophie Germain’s identity gives

x^4+324=g(x-6)g(x).

The numerator therefore supplies

g(4),g(10),\ldots,g(58),

while the denominator supplies

g(-2),g(4),\ldots,g(52).

Everything cancels except

\begin{aligned}\frac{g(58)}{g(-2)}&=\frac{58^2+6(58)+18}{4-12+18}\\&=\frac{3730}{10}=373.\end{aligned}

15.

Squares

S_1,

S_2

are inscribed in right triangle

ABC

as shown. Find

AC+CB

if

\operatorname{area}(S_1)=441

and

\operatorname{area}(S_2)=440.

Answer

Answer: 462

Concepts:right triangle similarity square (geometry)

Difficulty rating: 2450

Small hint:

Let the legs be

a

and

b

and use the first square to relate

ab

to

a+b

Big hint:

For the second square, use the altitude to the hypotenuse and similar cross-sections

Solution:

Put

p=a+b,

q=ab,

and let the hypotenuse be

c.

Since

S_1

has side

21,

the standard leg-aligned-square relation gives

21=\frac{ab}{a+b},

so

q=21p.

Hence

c^2=p^2-2q=p(p-42).

The altitude to the hypotenuse is

h=\frac{q}{c}.

If the side of

S_2

is

t,

similarity gives

t=\frac{ch}{c+h}.

Substitution simplifies this to

t=\frac{21c}{p-21}.

Therefore

440=\frac{441p(p-42)}{(p-21)^2}.

Since

p(p-42)=(p-21)^2-441,

this becomes

440=441\left(1-\frac{441}{(p-21)^2}\right).

Thus

(p-21)^2=441^2,

and

p>42

gives

p=462.

1987 AIME Problems

Answer: 300

Difficulty rating: 1490

Small hint:

Big hint:

Solution:

Answer: 137

Difficulty rating: 1340

Small hint:

Big hint:

Solution:

Answer: 182

Difficulty rating: 1830

Small hint:

Big hint:

Solution:

Answer: 480

Difficulty rating: 1810

Small hint:

Big hint:

Solution:

Answer: 588

Difficulty rating: 2070

Small hint:

Big hint:

Solution:

Answer: 193

Difficulty rating: 2270

Small hint:

Big hint:

Solution:

Answer: 70

Difficulty rating: 2230

Small hint:

Big hint:

Solution:

Answer: 112

Difficulty rating: 1860

Small hint:

Big hint:

Solution:

Answer: 33

Difficulty rating: 2380

Small hint:

Big hint:

Solution:

Answer: 120

Difficulty rating: 1770

Small hint:

Big hint:

Solution:

Answer: 486

Difficulty rating: 2070

Small hint:

Big hint:

Solution:

Answer: 19

Difficulty rating: 2110

Small hint:

Big hint:

Solution:

Answer: 931

Difficulty rating: 2450

Small hint:

Big hint:

Solution:

Answer: 373

Difficulty rating: 2380

Small hint:

Big hint:

Solution:

Answer: 462

Difficulty rating: 2450

Small hint:

Big hint:

Solution: