2018 AMC 10A Solutions | LIVE by Po-Shen Loh

1.

What is the value of

\left(\left((2+1)^{-1}+1\right)^{-1}+1\right)^{-1}+1?

\dfrac{5}{8}

\dfrac{11}{7}

\dfrac{8}{5}

\dfrac{18}{11}

\dfrac{15}{8}

Concepts:fraction order of operations

Difficulty rating: 560

Small hint:

Work from the innermost reciprocal outward

Big hint:

After each reciprocal step, simplify before adding the next

1

Solution:

We can simplify this as follows.

\begin{aligned} &\left(\left((2+1)^{-1}+1\right)^{-1}+1\right)^{-1}+1 \\ &=\left(\left(\dfrac{1}{3}+1\right)^{-1}+1\right)^{-1}+1 \\ &=\left(\dfrac{3}{4}+1\right)^{-1}+1 \\ &=\dfrac{4}{7} + 1 \\ &=\dfrac{11}{7} \end{aligned}

Thus, B is the correct answer.

2.

Liliane has

50\%

more soda than Jacqueline, and Alice has

25\%

more soda than Jacqueline. What is the relationship between the amounts of soda that Liliane and Alice have?

Liliane has

20\%

more soda than Alice.

Liliane has

25\%

more soda than Alice.

Liliane has

45\%

more soda than Alice.

Liliane has

75\%

more soda than Alice.

Liliane has

100\%

more soda than Alice.

Concepts:percentage ratio and proportion

Difficulty rating: 770

Small hint:

Use Jacqueline’s amount as the base variable

Big hint:

Compare Liliane’s amount directly to Alice’s amount

Solution:

Let

x

be the number of gallons of soda that Jacqueline has. Then Alice has

1.25x

gallons, and Liliane has

1.5x

gallons.

Therefore, the relationship can be found by dividing the amount of soda that each has to yield

\dfrac{1.5x}{1.25x}=1.2,

which means Liliane has

20\%

more soda than Alice.

Thus, A is the correct answer.

3.

A unit of blood expires after

10! = 10 \cdot 9 \cdot 8 \cdots 1

seconds. Yasin donates a unit of blood at noon on January

1.

On what day does his unit of blood expire?

January

2

January

12

January

22

February

11

February

12

Concepts:factorial unit conversion date and time

Difficulty rating: 960

Small hint:

Convert

10!

seconds into days

Big hint:

Use

60\cdot60\cdot24

seconds per day

Solution:

We can divide

10!

by

60,

60,

and

24

to get the number of days that it takes for a unit of blood to expire.

The first division cancels a

6

and

10.

The second division cancels

3, 4,

and

5.

The final division cancels

8

and turns the

9

into a

3.

This leaves

2, 7,

and a

3,

which multiply to

42.

There are

31

days in January, so by February

1,

the blood only has

42 - 31 = 11

days left.

11

days from February

1

would make the blood expire on February

12.

Thus, E is the correct answer.

4.

How many ways can a student schedule

3

mathematics courses — algebra, geometry, and number theory — in a

6

-period day if no two mathematics courses can be taken in consecutive periods?

(What courses the student takes during the other

3

periods is of no concern here.)

3

6

12

18

24

Concepts:arrangements with restrictions permutations systematic listing

Difficulty rating: 1220

Small hint:

First choose the three nonconsecutive periods

Big hint:

After the periods are chosen, arrange the three named courses

Solution:

The

3

classes can occupy the following periods:

(1, 3, 5),

(1, 3, 6),

(1, 4, 6),

(2, 4, 6).

This means that there are

4

ways to choose which periods the mathematics courses occur.

For each configuration, there are

3!

ways to determine the order of the courses, for a total of

6 \cdot 4 = 24

schedules.

Thus, E is the correct answer.

5.

Alice, Bob, and Charlie were on a hike and were wondering how far away the nearest town was. When Alice said, “We are at least

6

miles away,” Bob replied, “We are at most

5

miles away.” Charlie then remarked, “Actually the nearest town is at most

4

miles away.” It turned out that none of the three statements was true. Let

d

be the distance in miles to the nearest town. Which of the following intervals is the set of all possible values of

d?

(0,4)

(4,5)

(4,6)

(5,6)

(5,\infty)

Concepts:logical deduction inequality

Difficulty rating: 900

Small hint:

Negate each person’s statement

Big hint:

Intersect the three resulting inequalities for

d

Solution:

Alice’s statement tells us that

d \lt 6.

Bob’s statement tells us that

d \gt 5.

Charlie’s statement tells us that

d \gt 4.

Combining all of these tells us that

5 \lt d

and

d \lt 6,

which means

d

is in the interval

(5, 6).

Thus, D is the correct answer.

6.

Sangho uploaded a video to a website where viewers can vote that they like or dislike a video. Each video begins with a score of

0,

and the score increases by

1

for each like vote and decreases by

1

for each dislike vote.

At one point Sangho saw that his video had a score of

90,

and that

65\%

of the votes cast on his video were like votes. How many votes had been cast on Sangho’s video at that point?

200

300

400

500

600

Concepts:percentage linear equation

Difficulty rating: 900

Small hint:

Let

N

be the total number of votes

Big hint:

The score is likes minus dislikes

Solution:

If

65\%

of votes were like votes, then

35\%

of votes are dislike votes. Then Sangho’s score is

65\% - 35\% = 30\%

the total number of votes.

We know that Sangho’s score is

90,

so the total number of votes is

90 \div 30\% = 300.

Thus, B is the correct answer.

7.

For how many (not necessarily positive) integer values of

n

is the following value an integer?

4000 \cdot \left(\dfrac{2}{5}\right)^n

3

4

6

8

9

Concepts:exponent prime factorization counting integers in a range

Difficulty rating: 1070

Small hint:

Write

4000

as

2^5\cdot5^3

Big hint:

The exponents of

2

and

5

must both be nonnegative

Solution:

We can rewrite the expression as

(2^5 \cdot 5^3) \cdot \left(\dfrac{2}{5}\right)^n = 2^{5 + n} \cdot 5^{3 - n}.

For this to be an integer, both exponents must be nonnegative. This means that

\begin{aligned} 5 + n \geq 0 &\Rightarrow n \geq -5 \\ 3 - n \geq 0 &\Rightarrow n \leq 3. \end{aligned}

This gives us

5 + 3 + 1 = 9

values for

n.

Thus, E is the correct answer.

8.

Joe has a collection of

23

coins, consisting of

5

-cent coins,

10

-cent coins, and

25

-cent coins. He has

3

more

10

-cent coins than

5

-cent coins, and the total value of his collection is

320

cents. How many more

25

-cent coins does Joe have than

5

-cent coins?

0

1

2

3

4

Concepts:money system of equations

Difficulty rating: 1220

Small hint:

Let the number of

5

-cent coins be

x

Big hint:

Use both the total number of coins and total value equations

Solution:

Let

x

be the number of

5

-cent coins that Joe has. Then the number of

10

-cent coins he has is

x + 3.

Therefore, Joe has

23 - x - (x + 3) = 20 - 2x

25

-cent coins.

The total value of all these coins is

5x + 10(x + 3) + 25(20 - 2x)

= 530 - 35x.

We know that

530 - 35x = 320 \Rightarrow x = 6.

This means that Joe has

20 - 2 \cdot 6 = 8

25

-cent coins. Therefore, he has

8 - 6 = 2

more

25

-cent coins than

5

-cent coins.

Thus, C is the correct answer.

9.

All of the triangles in the diagram below are similar to isosceles triangle

ABC,

in which

AB=AC.

Each of the

7

smallest triangles has area

1,

and

\triangle ABC

has area

40.

What is the area of trapezoid

DBCE?

16

18

20

22

24

Concepts:similarity area ratio

Difficulty rating: 1420

Small hint:

Area scales as the square of side length for similar triangles

Big hint:

Find the area of the top triangle and subtract from

40

Solution:

We know that the side length of the smaller triangles is

\sqrt{\frac{1}{40}}

times the length of the larger triangle from similar triangles.

Then the side length of

\triangle ADE

is

4\sqrt{\frac{1}{40}}

times the length of the side length of the larger triangle.

This makes the ratio of the areas

\left(4\sqrt{\dfrac{1}{40}}\right)^2 = 16 \cdot \dfrac{1}{40} = \dfrac{2}{5}.

Therefore, the area of

\triangle ADE

is

\frac{2}{5} \cdot 40 = 16.

The area of the trapezoid is then

40 - 16 = 24.

Thus, E is the correct answer.

10.

Suppose that real number

x

satisfies

\sqrt{49-x^2}-\sqrt{25-x^2}=3.

What is the value of

\sqrt{49-x^2}+\sqrt{25-x^2}?

8

\sqrt{33}+3

9

2\sqrt{10}+4

12

Concepts:radical difference of squares algebraic manipulation

Difficulty rating: 1310

Small hint:

Treat the two radicals as conjugate-like quantities

Big hint:

Multiply the given difference by the desired sum

Solution:

Note that the left hand side of the equation and the desired expression are conjugates. Multiplying them would remove the square roots.

Multiplying them yields

49 - x^2 - 25 + x^2 = 24.

This means that the product of the values of the expressions is equal to

24.

The desired value is therefore

24 \div 3 = 8.

Thus, A is the correct answer.

11.

When

7

fair standard

6

-sided dice are thrown, the probability that the sum of the numbers on the top faces is

10

can be written as

\dfrac{n}{6^{7}},

where

n

is a positive integer. What is

n?

42

49

56

63

84

Concepts:dice (probability)stars and bars

Difficulty rating: 1420

Small hint:

List the unordered ways seven positive die rolls can sum to

10

Big hint:

Count the orderings for each listed pattern

Solution:

We can use stars and bars to find

n.

It is the same as finding the number of ways to put

10

balls into

7

boxes, where each box has at least one ball.

The formula for such a scenario is

\binom{n - 1}{k - 1},

where

n

is the number of balls and

k

is the number of boxes.

No die can exceed

4

in a sum of

10

from seven positive rolls, so the upper bound of

6

creates no additional restriction. The desired answer is therefore

\binom{9}{6} = \binom{9}{3} = 84.

Thus, E is the correct answer.

12.

How many ordered pairs of real numbers

(x,y)

satisfy the following system of equations?

\begin{cases} ~x+3y&=3 \\ ~\big||x|-|y|\big|&=1 \end{cases}

1

2

3

4

8

Concepts:absolute value system of equations casework

Difficulty rating: 1370

Small hint:

Replace the absolute-value equation by four linear cases

Big hint:

Distinct solutions can repeat across cases, so count unique ordered pairs

Solution:

The second equation says

|x|-|y|=1

or

|x|-|y|=-1

, so it is enough to check the four linear possibilities

x=y\pm1

and

x=-y\pm1

.

Combining these with

x+3y=3

gives

(x,y)=\left(\dfrac32,\dfrac12\right)

,

(0,1)

,

(0,1)

again, and

(-3,2)

.

These are three distinct ordered pairs, and each satisfies the original absolute-value equation. Thus, C is the correct answer.

13.

A paper triangle with sides of lengths

3,

4,

and

5

inches, as shown, is folded so that point

A

falls on point

B.

What is the length in inches of the crease?

1+\dfrac{1}{2} \sqrt{2}

\sqrt{3}

\dfrac{7}{4}

\dfrac{15}{8}

2

Concepts:paper folding perpendicular bisector similarity

Difficulty rating: 1420

Small hint:

The crease is the perpendicular bisector of

AB

Big hint:

Use similarity with the original

3

-

4

-

5

triangle

Solution:

The crease is the perpendicular bisector of

\overline{AB}.

Let

\overline{DE}

be the crease.

By

AA

similarity,

\triangle ADE\sim\triangle ACB.

Therefore,

\dfrac{BC}{AC}=\dfrac{DE}{AD}.

Plugging in the side lengths gives

\dfrac34=\dfrac{DE}{\frac{5}{2}},

so

DE=\dfrac{15}{8}.

Thus, D is the correct answer.

14.

What is the greatest integer less than or equal to

\dfrac{3^{100}+2^{100}}{3^{96}+2^{96}}?

80

81

96

97

625

Concepts:exponent floor and ceiling functions bounding to limit cases

Difficulty rating: 1540

Small hint:

Compare the expression to

81

first

Big hint:

Then prove it is still greater than

80

Solution:

Let

a=3^{96}

and

b=2^{96}

. The expression is

\dfrac{81a+16b}{a+b}=16+\dfrac{65a}{a+b}

, so it is less than

16+65=81

.

To show the floor is

80

, we also need the expression to be greater than

80

. This is equivalent to

81a+16b>80a+80b

, or

a>64b

.

Because

\left(\dfrac32\right)^2=\dfrac94>2,

we have

\dfrac{a}{b}=\left(\dfrac32\right)^{96}>2^{48}>64.

Hence the expression is greater than

80

and less than

81.

Thus, A is the correct answer.

15.

Two circles of radius

5

are externally tangent to each other and are internally tangent to a circle of radius

13

at points

A

and

B,

as shown in the diagram. The distance

AB

can be written in the form

\frac{m}{n},

where

m

and

n

are relatively prime positive integers. What is

m+n?

21

29

58

69

93

Concepts:tangent circles similarity

Difficulty rating: 1820

Small hint:

Join the three circle centers

Big hint:

The triangle through tangent points is similar to the triangle through centers

Solution:

Let

X

be the center of the large circle and let

Y,Z

be the centers of the two smaller circles. Then

XY=XZ=13-5=8

and

YZ=10

.

The radii to tangent points make

XA

collinear with

XY

and

XB

collinear with

XZ

, so

\triangle XAB\sim \triangle XYZ

. Thus

\dfrac{AB}{YZ}=\dfrac{XA}{XY}=\dfrac{13}{8}

.

Hence

AB=10\cdot\dfrac{13}{8}=\dfrac{65}{4}

, so

m+n=65+4=69

. Thus, D is the correct answer.

16.

Right triangle

ABC

has leg lengths

AB=20

and

BC=21.

Including

\overline{AB}

and

\overline{BC},

how many line segments with integer length can be drawn from vertex

B

to a point on hypotenuse

\overline{AC}?

5

8

12

13

15

Concepts:right triangle altitude counting integers in a range

Difficulty rating: 1540

Small hint:

Find the altitude from

B

to the hypotenuse

Big hint:

Integer-length segments occur in symmetric pairs around the altitude

Solution:

Let

P

be the foot of the altitude from

B

to

\overline{AC}.

The Pythagorean Theorem gives

AC=29.

Computing the area in two ways gives

\dfrac{29\cdot PB}{2}=\dfrac{20\cdot21}{2},

so

PB=\dfrac{420}{29},

which lies between

14

and

15.

As the endpoint moves from

A

to

P,

its distance from

B

decreases continuously from

20

to

PB.

Thus there is one segment of each integer length

15,16,17,18,19,20.

As the endpoint moves from

P

to

C,

the distance increases continuously from

PB

to

21,

giving one segment of each integer length

15,16,17,18,19,20,21.

These are

6+7=13

distinct segments.

Thus, D is the correct answer.

17.

Let

S

be a set of

6

integers taken from

\{1,2,\dots,12\}

with the property that if

a

and

b

are elements of

S

with

a < b,

then

b

is not a multiple of

a.

What is the least possible value of an element in

S?

2

3

4

5

7

Concepts:divisibility extremal argument casework

Difficulty rating: 1970

Small hint:

Group numbers so at most one from each group can be chosen

Big hint:

Try forcing six choices from the divisor chains

Solution:

We proceed by casing on possible values for

S:

1

cannot be the smallest element since that would mean that no other number can be in the set.

2

cannot be the smallest element since we would have to include every odd number except

1.

This would make

3

and

9

violate the rule.

Let

3

be the smallest element. Then we can include

7

and

11.

We can finally include either

4

or

8

and

5

or

10.

Either way, the maximum number of elements that we can include is

5,

so

3

cannot be the smallest element.

Starting with

4,

we can include

6, 7, 9

and

11.

Finally, we can add either

5

or

10,

creating a

6

-element set.

Thus, C is the correct answer.

18.

How many nonnegative integers can be written in the form

a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5

+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2

+a_1\cdot3^1+a_0\cdot3^0,

where

a_i\in \{-1,0,1\}

for

0\le i \le 7?

512

729

1094

3281

59{,}048

Concepts:number base symmetry pairing and grouping

Difficulty rating: 1770

Small hint:

Think of the expression as balanced ternary

Big hint:

Positive and negative values pair off symmetrically around

0

Solution:

Note that every number formed by this sum is either positive, negative, or zero.

The number of positive numbers equals the number of negative numbers due to symmetry (flip the

1

s to

-1

s and

-1

s to

1

s).

The only way for the sum to be

0

is if all the coefficients are

0.

The total number of numbers is

3^8 = 6561.

Because each power of

3

is larger than the sum of all previous powers of three, each combination of coefficients yields a different value. More explicitly, at the highest place where two combinations differ, the difference has magnitude at least

3^k,

while all lower places together can cancel at most

2(1+3+\cdots+3^{k-1})=3^k-1.

Therefore, there are

\dfrac{6561 - 1}{2} + 1 = 3281

distinct nonnegative integers.

Thus, D is the correct answer.

19.

A number

m

is randomly selected from the set

\{11,13,15,17,19\},

and a number

n

is randomly selected from

\{1999,2000,2001,\ldots,2018\}.

What is the probability that

m^n

has a units digit of

1?

\dfrac{1}{5}

\dfrac{1}{4}

\dfrac{3}{10}

\dfrac{7}{20}

\dfrac{2}{5}

Concepts:units digit modular exponentiation basic probability casework

Difficulty rating: 1540

Small hint:

Only units-digit cycles matter

Big hint:

Average the success counts for bases ending in

1,3,5,7,9

Solution:

Only the units digit of

m

matters. Among the

20

consecutive possible exponents, each residue modulo

4

occurs

5

times, and

10

exponents are even. A base ending in

1

succeeds for all

20

exponents; bases ending in

3

or

7

succeed for the

5

exponents divisible by

4;

a base ending in

5

never succeeds; and a base ending in

9

succeeds for the

10

even exponents.

Thus

20+5+0+5+10=40

of the

5\cdot20=100

equally likely pairs work, and the probability is

\dfrac{40}{100}=\dfrac25.

Thus, E is the correct answer.

20.

A scanning code consists of a

7 \times 7

grid of squares, with some of its squares colored black and the rest colored white. There must be at least one square of each color in this grid of

49

squares.

A scanning code is called symmetric if its look does not change when the entire square is rotated by a multiple of

90^{\circ}

counterclockwise around its center, nor when it is reflected across a line joining opposite corners or a line joining midpoints of opposite sides.

What is the total number of possible symmetric scanning codes?

510

1022

8190

8192

65{,}534

Concepts:symmetry multiplication principle

Difficulty rating: 1970

Small hint:

Classify grid squares by symmetry orbits

Big hint:

Choose one color per orbit, excluding the two constant colorings

Solution:

Number rows and columns from

-3

through

3,

with the center at

(0,0).

Rotations and reflections can change signs and exchange coordinates, so the orbit of

(x,y)

is determined by the ordered pair obtained by sorting

(|x|,|y|).

These pairs are

(u,v)

with

0\le u\le v\le3,

of which there are

4+3+2+1=10.

Once one square in each orbit is colored, symmetry forces the colors of all other squares in that orbit.

There are therefore

2^{10}

symmetric colorings before the condition about using both colors. The all-black and all-white colorings are not allowed.

The total number of valid symmetric scanning codes is

2^{10}-2=1022

. Thus, B is the correct answer.

21.

Which of the following describes the set of values of

a

for which the curves

x^2+y^2=a^2

and

y=x^2-a

in the real

xy

-plane intersect at exactly

3

points?

a = \dfrac14

\dfrac14 \lt a \lt \dfrac12

a \gt \dfrac14

a = \dfrac12

a \gt \dfrac12

Concepts:parabola circle substitution

Difficulty rating: 1820

Small hint:

Substitute the parabola equation into the circle equation

Big hint:

The nonzero roots appear only when

2a-1

is positive

Solution:

Substitute

y=x^2-a

into

x^2+y^2=a^2

. This gives

x^2+(x^2-a)^2=a^2

, so

x^2(x^2-(2a-1))=0

.

The factor

x^2=0

always gives the single point

(0,-a)

. The other factor gives two additional real points exactly when

2a-1>0

.

There are exactly three intersection points when

a>\dfrac12

. Thus, E is the correct answer.

22.

Let

a,

b,

c,

and

d

be positive integers such that

\gcd(a, b)=24,

\gcd(b, c)=36,

\gcd(c, d)=54,

and

70 < \gcd(d, a) < 100.

Which of the following must be a divisor of

a?

5

7

11

13

17

Concepts:greatest common divisor prime factorization bounding to limit cases

Difficulty rating: 2010

Small hint:

Track the forced powers of

2

and

3

in the gcds

Big hint:

The remaining factor of

\gcd(d,a)

must lie between

12

and

17

Solution:

From

\gcd(a,b)=24=2^3\cdot3

and

\gcd(b,c)=36=2^2\cdot3^2

, the number

a

is divisible by

2^3\cdot3

but not by

3^2

.

From

\gcd(b,c)=36

and

\gcd(c,d)=54=2\cdot3^3

, the number

d

is divisible by

2\cdot3^3

but not by

2^2

. Therefore

\gcd(d,a)=2\cdot3\cdot n

, where

n

has no factor

2

or

3

.

Since

70<6n<100,

the integer

n

satisfies

12\le n\le16.

The only value in this range with no factor

2

or

3

is

13,

so

13

must divide

a.

Thus, D is the correct answer.

23.

Farmer Pythagoras has a field in the shape of a right triangle. The right triangle’s legs have lengths

3

and

4

units. In the corner where those sides meet at a right angle, he leaves a small unplanted square

S

so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from

S

to the hypotenuse is

2

units. What fraction of the field is planted?

\dfrac{25}{27}

\dfrac{26}{27}

\dfrac{73}{75}

\dfrac{145}{147}

\dfrac{74}{75}

Concepts:area decomposition right triangle

Difficulty rating: 2150

Small hint:

Let

x

be the side length of the unplanted square

Big hint:

Decompose the triangle into the square, two right triangles, and a triangle of height

2

Solution:

Let

x

be the side length of

S.

Then we can split the field up into the following shapes.

We can express the area of the field in two ways:

\dfrac{3 \cdot 4}{2} = x^2 + \dfrac{x(3 - x)}{2}

+ \dfrac{x(4 - x)}{2} + \dfrac{2 \cdot 5}{2}.

Simplifying yields

6 = \dfrac{7x}{2} + 5

x = \dfrac{2}{7}.

The desired fraction is

\dfrac{6 - x^2}{6} = \dfrac{6 - \frac{4}{49}}{6} = \dfrac{145}{147}.

Thus, D is the correct answer.

24.

Triangle

ABC

with

AB=50

and

AC=10

has area

120.

Let

D

be the midpoint of

\overline{AB},

and let

E

be the midpoint of

\overline{AC}.

The angle bisector of

\angle BAC

intersects

\overline{DE}

and

\overline{BC}

at

F

and

G,

respectively. What is the area of quadrilateral

FDBG?

60

65

70

75

80

Concepts:angle bisector theorem midpoint trapezoid area ratio

Difficulty rating: 2040

Small hint:

Use the angle bisector theorem to locate

G

Big hint:

The midsegment

DE

makes a similar half-height trapezoid

Solution:

Let

BC = a, BG = x, GC = y,

and

h

be the length of the altitude through

A.

By the Angle Bisector Theorem,

BG:GC=AB:AC=5:1,

so

BG=\dfrac{5a}{6}.

Because

DE

is a midsegment,

DE=\dfrac a2.

Applying the same theorem in

\triangle ADE

gives

DF:FE=AD:AE=5:1,

so

DF=\dfrac56DE=\dfrac{5a}{12}.

The trapezoid’s height is

\frac{h}{2},

and

\frac{ah}{2}=120.

Its average base length is

\dfrac12\left(\dfrac{5a}{12}+\dfrac{5a}{6}\right)=\dfrac{5a}{8}.

Therefore, its area is

\dfrac{5a}{8}\cdot\dfrac h2=\dfrac58\cdot\dfrac{ah}{2}=75.

Thus, D is the correct answer.

25.

For a positive integer

n

and nonzero digits

a,

b,

and

c,

let

A_n

be the

n

-digit integer each of whose digits is equal to

a

; let

B_n

be the

n

-digit integer each of whose digits is equal to

b

; and let

C_n

be the

2n

-digit (not

n

-digit) integer each of whose digits is equal to

c.

What is the greatest possible value of

a + b + c

for which there are at least two values of

n

such that

C_n - B_n = A_n^2?

12

14

16

18

20

Concepts:digits geometric sequence algebraic manipulation

Difficulty rating: 2390

Small hint:

Write the repeated-digit numbers as geometric sums

Big hint:

Having two different

n

values forces the linear equation in

10^n

to be identically true

Solution:

The repeated-digit numbers can be written as

\begin{aligned} A_n&=a\dfrac{10^n-1}{9},\\ B_n&=b\dfrac{10^n-1}{9},\\ C_n&=c\dfrac{10^{2n}-1}{9}. \end{aligned}

Substituting these expressions into

C_n-B_n=A_n^2

and dividing by

\frac{10^n-1}{9}

gives

c(10^n+1)-b=a^2\dfrac{10^n-1}{9}.

Rearranging yields

(9c-a^2)10^n=9b-9c-a^2.

If this holds for two different values of

n,

subtracting the two equations shows that

9c-a^2=0.

The displayed equation then also forces

9b-9c-a^2=0.

Hence

c=\frac{a^2}{9}

and

b=\frac{2a^2}{9}.

Because

a,b,c

are nonzero digits, the candidates are

(a,b,c)=(3,2,1),

(6,8,4),

and

(9,18,9).

The last triple is invalid because

18

is not a digit. The greatest valid sum is

6+8+4=18.

Thus, D is the correct answer.