1985 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

If

2x+1=8,

then

4x+1=

15

16

17

18

19

Concepts:linear equation substitution

Difficulty rating: 840

Small hint:

First solve the given equation for

x

Big hint:

Substitute

x=\frac{7}{2}

into

4x+1

Solution:

The given equation gives

2x=7,

so

x=\frac{7}{2}.

Therefore

4x+1=4(\frac{7}{2})+1=15.

Thus the correct answer is A.

2.

In an arcade game, the “monster” is the shaded sector of a circle of radius

1

cm, as shown in the figure. The missing piece (the mouth) has central angle

60^\circ.

What is the perimeter of the monster in cm?

\pi+2

2\pi

\frac53\pi

\frac56\pi+2

\frac53\pi+2

Concepts:arc circle circumference perimeter

Difficulty rating: 1000

Small hint:

The remaining circular arc has central angle

300^\circ

Big hint:

Add the length of that arc to the two exposed radii

Solution:

The curved part is

\frac{300^\circ}{360^\circ}=\frac{5}{6}

of a unit circle, so its length is

(\frac{5}{6})(2\pi)=\frac{5\pi}{3}.

The two straight sides of the mouth are radii of total length

2.

Hence the perimeter is

\frac{5\pi}{3}+2.

Thus the correct answer is E.

3.

In right

\triangle ABC

with legs

5

and

12,

arcs of circles are drawn, one with center

A

and radius

12,

the other with center

B

and radius

5.

They intersect the hypotenuse in

M

and

N.

Then

MN

has length

2

\frac{13}{5}

3

4

\frac{24}{5}

Concepts:circle distance formula Pythagorean Theorem

Difficulty rating: 1330

Small hint:

First find the hypotenuse of the

5

-

12

-right triangle

Big hint:

Measure the positions of both arc intersections from

A

Solution:

The hypotenuse has length

13.

Since

AM=12,

point

M

is

12

units from

A.

Also

BN=5,

so

AN=AB-BN=13-5=8.

Therefore

MN=AM-AN

=12-8=4.

Thus the correct answer is D.

4.

A large bag of coins contains pennies, dimes and quarters. There are twice as many dimes as pennies and three times as many quarters as dimes. An amount of money which could be in the bag is

\$306

\$333

\$342

\$348

\$360

Concepts:divisibility money ratio and proportion

Difficulty rating: 1270

Small hint:

Let the number of pennies be

p

Big hint:

Express the total value in cents as a multiple of

p

Solution:

If there are

p

pennies, there are

2p

dimes and

6p

quarters. Their total value is

p+10(2p)+25(6p)=171p

cents. Of the listed dollar amounts,

34200=171\cdot200

cents is possible.

Thus the correct answer is C.

5.

Which terms must be removed from the sum

\frac12+\frac14+\frac16+\frac18+\frac1{10}+\frac1{12}

if the sum of the remaining terms is to equal

1?

\frac14

and

\frac18

\frac14

and

\frac1{12}

\frac18

and

\frac1{12}

\frac16

and

\frac1{10}

\frac18

and

\frac1{10}

Concepts:fraction whole number operations

Difficulty rating: 1470

Small hint:

Compute how much larger the full sum is than

1

Big hint:

Use a common denominator of

120

to compare the candidate pairs

Solution:

The full sum is

\begin{aligned} \frac{\substack{60+30+20\\{}+15+12+10}}{120} &=\frac{147}{120}. \end{aligned}

The removed terms must therefore total

\frac{27}{120}=\frac{9}{40}.

Since

\frac{1}{8}+\frac{1}{10}

=\frac{5}{40}+\frac{4}{40}

=\frac{9}{40},

those are the required terms.

Thus the correct answer is E.

6.

One student in a class of boys and girls is to be chosen to represent the class. Each student is equally likely to be chosen and the probability that a boy is chosen is

\frac23

of the probability that a girl is chosen. The ratio of the number of boys to the total number of boys and girls is

\frac13

\frac25

\frac12

\frac35

\frac23

Concepts:basic probability ratio and proportion

Difficulty rating: 1100

Small hint:

Equal likelihood makes the selection probabilities proportional to the class counts

Big hint:

A boys-to-girls ratio of

2:3

uses five equal parts in all

Solution:

Because every student is equally likely to be selected, the number of boys is

\frac{2}{3}

of the number of girls. Thus the boys-to-girls ratio is

2:3,

and boys make up

\frac{2}{2+3}=\frac{2}{5}

of the class.

Thus the correct answer is B.

7.

In some computer languages (such as APL), when there are no parentheses in an algebraic expression, the operations are grouped from right to left. Thus,

a\times b-c

in such languages means the same as

a(b-c)

in ordinary algebraic notation. If

a\div b-c+d

is evaluated in such a language, the result in ordinary algebraic notation would be

\frac ab-c+d

\frac ab-c-d

\frac{d+c-b}{a}

\frac{a}{b-c+d}

\frac{a}{b-c-d}

Concepts:algebraic manipulation order of operations

Difficulty rating: 1360

Small hint:

Begin with the rightmost operation,

c+d

Big hint:

Next subtract that result from

b,

and only then divide

a

Solution:

Grouping from the right first gives

c+d.

Subtracting this result from

b

gives

b-(c+d)=b-c-d.

Finally, dividing

a

by that quantity gives

\frac{a}{b-c-d}.

Thus the correct answer is E.

8.

Let

a,

a',

b,

b'

be real numbers with

a

and

a'

nonzero. The solution to

ax+b=0

is less than the solution to

a'x+b'=0

if and only if

a'b\lt ab'

ab'\lt a'b

ab\lt a'b'

\frac ba\lt\frac{b'}{a'}

\frac{b'}{a'}\lt\frac ba

Concepts:inequality linear equation

Difficulty rating: 1360

Small hint:

Write each equation’s solution without cross-multiplying

Big hint:

Multiplying an inequality by

-1

reverses its direction

Solution:

The two solutions are

-\frac{b}{a}

and

-\frac{b'}{a'}.

Thus the given comparison is

-\frac ba\lt-\frac{b'}{a'}.

Multiplying both sides by

-1

reverses the inequality and gives

\frac{b}{a}\gt \frac{b'}{a'},

equivalently

\frac{b'}{a'}\lt \frac{b}{a}.

No multiplication by the unknown-sign quantity

aa'

is valid.

Thus the correct answer is E.

9.

The odd positive integers,

1,

3,

5,

7,

\ldots,

are arranged in five columns continuing with the pattern shown. Counting from the left, the column in which

1985

appears is the

\begin{array}{ccccc} &1&3&5&7\\ 15&13&11&9&\\ &17&19&21&23\\ 31&29&27&25&\\ &33&35&37&39\\ 47&45&43&41&\\ &49&51&53&55 \end{array}

first

second

third

fourth

fifth

Concepts:arithmetic sequence modular arithmetic pattern recognition

Difficulty rating: 1310

Small hint:

Look at the entries that begin each left-to-right row

Big hint:

Relate

1985

to a nearby multiple of

16

Solution:

The last number in each right-to-left row is

16k-1,

and the next row begins with

16k+1

in the second column. Since

1985=16\cdot124+1,

it begins such a row and lies in the second column.

Thus the correct answer is B.

10.

An arbitrary circle can intersect the graph of

y=\sin x

in

at most

2

points

at most

4

points

at most

6

points

at most

8

points

more than

16

points

Concepts:circle counting intersections trigonometry

Difficulty rating: 2240

Small hint:

Consider a circle tangent to the

x

-axis at the origin

Big hint:

A very large radius keeps the lower arc close to the axis across many sine waves

Solution:

Take a circle tangent to the

x

-axis at the origin with its center high above the axis. As its radius grows, its lower arc stays positive but arbitrarily close to the

x

-axis over an arbitrarily long interval. On each positive arch of

y=\sin x,

the sine curve is

0

at the endpoints and rises well above this circular arc in between, producing two crossings. Choosing a sufficiently large radius therefore produces more than

16

intersections.

Thus the correct answer is E.

11.

How many distinguishable rearrangements of the letters in CONTEST have both the vowels first? (For instance, OETCNST is one such arrangement, but OTETSNC is not.)

60

120

240

720

2520

Concepts:multiplication principle multiset permutations permutations

Difficulty rating: 1480

Small hint:

Order the two vowels in the first two positions

Big hint:

Permute the remaining five letters while accounting for the repeated

T

Solution:

The vowels

O,E

can be ordered first in

2!

ways. The remaining letters

C,N,T,S,T

can be arranged in

\frac{5!}{2!}=60

distinguishable ways because the two

T

’s agree. The total is

2\cdot60=120.

Thus the correct answer is B.

12.

Let

p,

q

and

r

be distinct prime numbers, where

1

is not considered a prime. Which of the following is the smallest positive perfect cube having

n=pq^2r^4

as a divisor?

p^8q^8r^8

(pq^2r^2)^3

(p^2q^2r^2)^3

(pqr^2)^3

4p^3q^3r^3

Concepts:divisibility perfect power prime factorization

Difficulty rating: 1660

Small hint:

Every prime exponent in a perfect cube is a multiple of

3

Big hint:

Raise the exponents

1,

2,

and

4

to the least larger multiples of

3

Solution:

A perfect cube has prime exponents divisible by

3.

The least multiples of

3

that are at least

1,

2,

and

4

are

3,

3,

and

6,

respectively. Thus the smallest possible cube is

p^3q^3r^6=(pqr^2)^3.

Therefore the correct answer is D.

13.

Pegs are put in a board

1

unit apart both horizontally and vertically. A rubber band is stretched over

4

pegs as shown in the figure, forming a quadrilateral. Its area in square units is

4

4.5

5

5.5

6

Concepts:coordinate geometry lattice point shoelace formula

Difficulty rating: 1420

Small hint:

Assign integer coordinates to the four corner pegs

Big hint:

Use the shoelace formula on

(1,3),

(4,1),

(3,0),

and

(0,1)

Solution:

With the lower-left peg as

(0,0),

the quadrilateral’s vertices in order are

(1,3),

(4,1),

(3,0),

and

(0,1).

The shoelace formula gives

\begin{aligned} &\frac12\left|(1+0+3+0)\right.\\ &\quad\left.{}-(12+3+0+1)\right|=6. \end{aligned}

Therefore the correct answer is E.

14.

Exactly three of the interior angles of a convex polygon are obtuse. What is the maximum number of sides of such a polygon?

4

5

6

7

8

Concepts:angle sum extremal argument inequality

Difficulty rating: 1960

Small hint:

Bound the three obtuse angles by

180^\circ

and every other angle by

90^\circ

Big hint:

Compare that upper bound with the interior-angle sum

(n-2)180^\circ

Solution:

For an

n

-gon, the three obtuse angles have total less than

3\cdot180^\circ,

while the other

n-3

angles have total at most

(n-3)90^\circ.

Hence

\begin{aligned} (n-2)180^\circ &\lt3(180^\circ)\\ &\quad{}+(n-3)90^\circ, \end{aligned}

which simplifies to

n\lt7.

Six sides are attainable, for example with interior angles

150^\circ,

150^\circ,

150^\circ,

90^\circ,

90^\circ,

and

90^\circ.

Thus the maximum is

6.

Therefore the correct answer is C.

15.

If

a

and

b

are positive numbers such that

a^b=b^a

and

b=9a,

then the value of

a

is

9

\frac19

\sqrt[9]{9}

\sqrt[3]{9}

\sqrt[4]{3}

Concepts:exponent radical substitution

Difficulty rating: 1960

Small hint:

Substitute

b=9a

into

a^b=b^a

Big hint:

Because

a\gt0,

take the

a

-th root and simplify

Solution:

Substitution gives

a^{9a}=(9a)^a.

Taking the positive

a

-th root yields

a^9=9a,

so

a^8=9.

Therefore

a=9^{\frac{1}{8}}=3^{\frac{1}{4}}=\sqrt[4]{3}.

Thus the correct answer is E.

16.

If

A=20^\circ

and

B=25^\circ,

then the value of

(1+\tan A)(1+\tan B)

is

\sqrt3

2

1+\sqrt2

2(\tan A+\tan B)

none of these

Concepts:trigonometric identity trigonometry

Difficulty rating: 1920

Small hint:

Use

A+B=45^\circ

in the tangent addition formula

Big hint:

Express

\tan A+\tan B

in terms of

\tan A\tan B

Solution:

Because

\tan(A+B)=\tan45^\circ=1,

\frac{\tan A+\tan B}{1-\tan A\tan B}=1,

so

\tan A+\tan B

=1-\tan A\tan B.

Therefore

\begin{aligned} &(1+\tan A)(1+\tan B)\\ &\quad=1+(1-\tan A\tan B)\\ &\qquad+\tan A\tan B\\ &\quad=2. \end{aligned}

Thus the correct answer is B.

17.

Diagonal

DB

of rectangle

ABCD

is divided into three segments of length

1

by parallel lines

L

and

L'

that pass through

A

and

C

and are perpendicular to

DB.

The area of

ABCD,

rounded to one decimal place, is

4.1

4.2

4.3

4.4

4.5

Concepts:area diagonal rectangle vector

Difficulty rating: 2240

Small hint:

Let the rectangle’s side lengths be

w,h;

then its diagonal has length

3

Big hint:

Project the vertical and horizontal sides onto the diagonal to obtain

\frac{h^2}{3}=1

and

\frac{w^2}{3}=2

Solution:

Let

D=(0,0),

B=(w,h),

A=(0,h)

and

C=(w,0).

Since

DB=3,

we have

w^2+h^2=9.

The projection of

DA

onto

DB

has length

\frac{h^2}{3},

and the first marked segment has length

1,

so

h^2=3.

Similarly, the projection of

DC

reaches the second division point, so

\frac{w^2}{3}=2

and

w^2=6.

Hence the area is

wh=\sqrt{18}=3\sqrt2\approx4.2426,

which rounds to

4.2.

Therefore the correct answer is B.

18.

Six bags of marbles contain

18,

19,

21,

23,

25

and

34

marbles, respectively. One bag contains chipped marbles only. The other

5

bags contain no chipped marbles. Jane takes three of the bags and George takes two of the others. Only the bag of chipped marbles remains. If Jane gets twice as many marbles as George, how many chipped marbles are there?

18

19

21

23

25

Concepts:divisibility modular arithmetic partitions and compositions

Difficulty rating: 1790

Small hint:

Add all six bag sizes, then remove the chipped bag

Big hint:

Jane’s and George’s totals are in the ratio

2:1,

so their combined total is divisible by

3

Solution:

The six bags total

140.

If the chipped bag contains

c

marbles, then the other bags total three times George’s amount, so

140-c

is divisible by

3.

Of the choices, only

c=23

has this property. It is attainable: George can take

18+21=39,

while Jane takes

19+25+34=78.

Thus the correct answer is D.

19.

Consider the graphs of

y=Ax^2

and

y^2+3=x^2+4y,

where

A

is a positive constant and

x

and

y

are real variables. In how many points do the two graphs intersect?

exactly

4

exactly

2

at least

1,

but the number varies for different positive values of

A

0

for at least one positive value of

A

none of these

Concepts:algebraic manipulation counting intersections quadratic

Difficulty rating: 2350

Small hint:

Use

x^2=\frac{y}{A}

to eliminate

x^2

Big hint:

Show that the resulting quadratic has two distinct positive

y

-roots

Solution:

Substituting

x^2=\frac{y}{A}

into the second equation gives

Ay^2-(4A+1)y+3A=0.

Its discriminant is

4A^2+8A+1\gt0.

Its roots have positive product

3

and positive sum

\frac{4A+1}{A},

so both roots are positive and distinct. For each root

y,

the equation

x^2=\frac{y}{A}

gives two distinct values of

x.

Thus there are exactly four intersection points.

Therefore the correct answer is A.

20.

A wooden cube with edge length

n

units (where

n

is an integer

\gt2

) is painted black all over. By slices parallel to its faces, the cube is cut into

n^3

smaller cubes each of unit edge length. If the number of smaller cubes with just one face painted black is equal to the number of smaller cubes completely free of paint, what is

n?

5

6

7

8

none of these

Concepts:algebraic manipulation basic counting cube geometry

Difficulty rating: 1660

Small hint:

Count the unpainted cubes in the interior cube of side length

n-2

Big hint:

On each face, exactly

(n-2)^2

cubes avoid the edges

Solution:

There are

(n-2)^3

completely unpainted cubes. Each of the six faces contributes

(n-2)^2

cubes with exactly one painted face, so equality gives

6(n-2)^2=(n-2)^3.

Since

n\gt2,

division by

(n-2)^2

gives

n-2=6,

hence

n=8.

Therefore the correct answer is D.

21.

How many integers

x

satisfy the equation

(x^2-x-1)^{x+2}=1?

2

3

4

5

none of these

Concepts:casework exponent quadratic

Difficulty rating: 2070

Small hint:

An integer power can equal

1

when the base is

1,

when the base is

-1

with even exponent, or when a nonzero base has exponent

0

Big hint:

Solve those three cases separately and check every candidate

Solution:

If the base is

1,

then

x^2-x-1=1,

giving

x=-1

and

x=2.

If the base is

-1,

then

x(x-1)=0;

only

x=0

makes the exponent

x+2

even. If the exponent is

0,

then

x=-2,

and its base is

5,

so it also works. Thus the four solutions are

-2,

-1,

0,

and

2.

Therefore the correct answer is C.

22.

In a circle with center

O,

AD

is a diameter,

ABC

is a chord,

BO=5

and

\angle ABO=\overset{\frown}{CD}=60^\circ.

Then the length of

BC

is

3

3+\sqrt3

5-\frac{\sqrt3}{2}

5

none of the above

Concepts:circle inscribed angle special right triangle

Difficulty rating: 2130

Small hint:

The

60^\circ

arc

CD

makes

\angle CAD=30^\circ

Big hint:

Use the resulting

30^\circ

-

60^\circ

-

90^\circ

triangles to compare

AB

and

AC

Solution:

Arc

CD

gives

\angle CAD=30^\circ.

Since

A,B,C

are collinear and

A,O,D

are collinear,

\angle BAO=30^\circ.

With

\angle ABO=60^\circ,

triangle

ABO

is a

30^\circ

-

60^\circ

-

90^\circ

triangle. Because

BO=5,

we get

AB=10

and

AO=5\sqrt3.

Also

\angle ACD=90^\circ

because

AD

is a diameter. In the

30^\circ

-

60^\circ

-

90^\circ

triangle

ACD,

the hypotenuse is

AD=10\sqrt3,

so

AC=15.

Therefore

BC=AC-AB=5.

Thus the correct answer is D.

23.

If

\begin{aligned} x&=\frac{-1+i\sqrt3}{2},\\ y&=\frac{-1-i\sqrt3}{2}, \end{aligned}

where

i^2=-1,

then which of the following is not correct?

x^5+y^5=-1

x^7+y^7=-1

x^9+y^9=-1

x^{11}+y^{11}=-1

x^{13}+y^{13}=-1

Concepts:complex number modular arithmetic roots of unity

Difficulty rating: 2130

Small hint:

Recognize

x

and

y

as the two nonreal cube roots of unity

Big hint:

Their powers depend only on the exponent modulo

3

Solution:

The numbers are the two nonreal cube roots of unity, so

x^n+y^n=2\cos\left(\frac{2\pi n}{3}\right).

This equals

-1

when

n

is not divisible by

3,

but equals

2

when

n

is divisible by

3.

Of the listed exponents, only

9

is divisible by

3,

so its displayed equation is the one that is not correct.

Therefore the correct answer is C.

24.

A non-zero digit is chosen in such a way that the probability of choosing digit

d

is

\log_{10}(d+1)-\log_{10}d.

The probability that the digit

2

is chosen is exactly

\frac{1}{2}

the probability that the digit chosen is in the set

\{2,3\}

\{3,4\}

\{4,5,6,7,8\}

\{5,6,7,8,9\}

\{4,5,6,7,8,9\}

Concepts:basic probability logarithm telescoping

Difficulty rating: 1800

Small hint:

Twice the probability of digit

2

is

2\log_{10}(\frac{3}{2})

Big hint:

Sum the probabilities over a consecutive set by telescoping its logarithms

Solution:

The probability of

2

is

\log_{10}(\frac{3}{2}),

so twice that probability is

\log_{10}(\frac{9}{4}).

For the digits

4

through

8,

the sum telescopes:

\begin{aligned} &\sum_{d=4}^{8} \bigl(\log_{10}(d+1)\\ &\qquad{}-\log_{10}d\bigr) =\log_{10}\frac94. \end{aligned}

Therefore that set has exactly twice the probability of digit

2.

Thus the correct answer is C.

25.

The volume of a certain rectangular solid is

8\text{ cm}^3,

its total surface area is

32\text{ cm}^2,

and its three dimensions are in geometric progression. The sum of the lengths in cm of all the edges of this solid is

28

32

36

40

44

Concepts:geometric sequence rectangular prism surface area volume

Difficulty rating: 1920

Small hint:

Write the three dimensions as

\frac{t}{r},t,tr

Big hint:

Use the volume to find

t,

then use the surface area to find

r+\frac{1}{r}

Solution:

Write the dimensions as

\frac{2}{r},2,2r,

since their product is

8.

Half the surface area is the sum of the three pairwise products, so

\frac4r+4+4r=16,

giving

r+\frac{1}{r}=3.

The sum of the dimensions is

2(r+1+\frac{1}{r})=8.

Since each dimension occurs on four edges, the sum of all edge lengths is

4\cdot8=32.

Thus the correct answer is B.

26.

Find the least positive integer

n

for which

\frac{n-13}{5n+6}

is a non-zero reducible fraction.

45

68

155

226

none of these

Concepts:greatest common divisor modular arithmetic

Difficulty rating: 2130

Small hint:

Any common divisor of

n-13

and

5n+6

also divides a suitable linear combination

Big hint:

Compute

(5n+6)-5(n-13)

Solution:

Euclid’s algorithm gives

\begin{aligned} &\gcd(n-13,5n+6)\\ &\quad=\gcd(n-13,71). \end{aligned}

Since

71

is prime, the nonzero fraction is reducible exactly when

n-13

is a nonzero multiple of

71.

The least positive possibility is

n=13+71=84,

which is not among the four numerical choices.

Thus the correct answer is E.

27.

Consider a sequence

x_1,

x_2,

x_3,

\ldots,

defined by

\begin{aligned} x_1&=\sqrt[3]{3},\\ x_2&=(\sqrt[3]{3})^{\sqrt[3]{3}}. \end{aligned}

and in general

x_n=(x_{n-1})^{\sqrt[3]{3}}

for

n\gt1.

What is the smallest value of

n

for which

x_n

is an integer?

2

3

4

9

27

Concepts:exponent inequality radical recursion

Difficulty rating: 2380

Small hint:

Let

c=\sqrt[3]{3}

and track the exponent of

3

Big hint:

Show that

x_n=3^{\frac{c^{\,n-1}}{3}}

and examine the first four terms

Solution:

Let

c=\sqrt[3]{3}.

Repeated application of the recurrence gives

x_n=3^{\frac{c^{\,n-1}}{3}},

so

x_4=3^{\frac{c^3}{3}}=3.

It remains to rule out the first three terms. The sequence is strictly increasing. Also

c\lt\frac{3}{2},

so

x_2=c^c \lt\left(\frac32\right)^{\frac{3}{2}} \lt2.

Finally

x_3=3^{\frac{1}{c}}.

Since

2^c\lt2^{\frac{3}{2}}\lt3,

we have

x_3\gt2,

while

\frac{1}{c}\lt1

gives

x_3\lt3.

Thus

x_1,x_2,x_3

are not integers, and the first integral term is

x_4.

Therefore the correct answer is C.

28.

In

\triangle ABC,

we have

\angle C=3\angle A,

a=27

and

c=48.

What is

b?

33

35

37

39

not uniquely determined

Concepts:law of sines trigonometric identity

Difficulty rating: 2460

Small hint:

Use the sine law to write

\frac{48}{27}=\frac{\sin3A}{\sin A}

Big hint:

After finding

\cos A,

use

B=180^\circ-4A

and

\sin4A=4\sin A\cos A\cos2A

Solution:

By the sine law and the triple-angle identity,

\frac{48}{27} =\frac{\sin3A}{\sin A} =3-4\sin^2A.

Hence

\sin^2A=\frac{11}{36}.

Since

A\lt60^\circ,

\cos A=\frac{5}{6},

and therefore

\cos2A=\frac{7}{18}.

Also

B=180^\circ-4A,

so another application of the sine law gives

\begin{aligned} b &=27\frac{\sin B}{\sin A}\\ &=27\frac{\sin4A}{\sin A}\\ &=27\cdot4\cos A\cos2A\\ &=35. \end{aligned}

Thus the correct answer is B.

29.

In their base

10

representations, the integer

a

consists of a sequence of

1985

eights and the integer

b

consists of a sequence of

1985

fives. What is the sum of the digits of the base

10

representation of the integer

9ab?

15880

17856

17865

17874

19851

Concepts:algebraic manipulation digits pattern recognition place value

Difficulty rating: 2530

Small hint:

Write a string of

k

identical digits using

R=\frac{10^k-1}{9}

Big hint:

After removing the terminal zero, express the product as a string of

2k

fours minus a string of

k

eights

Solution:

Let

k=1985

and

R=\frac{10^k-1}{9}.

Then

a=8R,

b=5R,

and

9ab

ends in

0.

Removing that zero does not change the digit sum and leaves

\begin{aligned} N=\frac{9ab}{10} &=36R^2\\ &=\frac49(10^{2k}-1)\\ &\quad{}-\frac89(10^k-1). \end{aligned}

Thus

N

is a string of

2k

fours minus a string of

k

eights. The subtraction produces

k-1

fours, then

3,

then

k-1

fives, then

6.

Its digit sum is

\begin{aligned} &4(k-1)+3+5(k-1)+6\\ &\quad=9k=17865. \end{aligned}

Therefore the correct answer is C.

30.

Let

\lfloor x\rfloor

be the greatest integer less than or equal to

x.

Then the number of real solutions to

4x^2-40\lfloor x\rfloor+51=0

is

0

1

2

3

4

Concepts:casework floor and ceiling functions inequality quadratic

Difficulty rating: 2380

Small hint:

Set

n=\lfloor x\rfloor

and solve the equation for

x^2

Big hint:

Enforce

n\le x\lt n+1

to determine which integer values of

n

work

Solution:

Let

n=\lfloor x\rfloor.

A solution must be positive and satisfy

\begin{aligned} x&=\sqrt{10n-\frac{51}{4}},\\ n&\le x\lt n+1. \end{aligned}

The lower inequality is equivalent to

4n^2-40n+51\le0,

so

n=2,3,\ldots,8.

The upper inequality is equivalent to

4n^2-32n+55\gt0.

Among

2,\ldots,8,

this holds exactly for

n=2,6,7,8.

Each interval contributes one root, so there are four real solutions.

Therefore the correct answer is E.