### 2018 AMC 10A Exam Problems

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Used with permission of the Mathematical Association of America.

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1.

What is the value of $\left(\left((2+1)^{-1}+1\right)^{-1}+1\right)^{-1}+1?$

$$\dfrac{5}{8}$$

$$\dfrac{11}{7}$$

$$\dfrac{8}{5}$$

$$\dfrac{18}{11}$$

$$\dfrac{15}{8}$$

###### Solution(s):

We can simplify this as follows. \begin{align*} &\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{align*}

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.

###### Solution(s):

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 mean Liliane has $$20\%$$ more soda.

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 of January $$1.$$ On what day does his unit of blood expire?

January $$2$$

January $$12$$

January $$22$$

February $$11$$

February $$12$$

###### Solution(s):

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.)

$$3$$

$$6$$

$$12$$

$$18$$

$$24$$

###### Solution(s):

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 were 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 $$d?$$

$$(0,4)$$

$$(4,5)$$

$$(4,6)$$

$$(5,6)$$

$$(5,\infty)$$

###### Solution(s):

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?

$$200$$

$$300$$

$$400$$

$$500$$

$$600$$

###### Solution(s):

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

$$3$$

$$4$$

$$6$$

$$8$$

$$9$$

###### Solution(s):

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, the exponents must be positive. This means that \begin{align*} 5 + n \geq 0 &\Rightarrow n \geq -5 \\ 3 - n \geq 0 &\Rightarrow n \leq 3. \end{align*}

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?

$$0$$

$$1$$

$$2$$

$$3$$

$$4$$

###### Solution(s):

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 $$DBCE?$$ $$16$$

$$18$$

$$20$$

$$22$$

$$24$$

###### Solution(s):

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

$$8$$

$$\sqrt{33}+8$$

$$9$$

$$2\sqrt{10}+4$$

$$12$$

###### Solution(s):

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 $$n?$$

$$42$$

$$49$$

$$56$$

$$63$$

$$84$$

###### Solution(s):

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.

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

$$1$$

$$2$$

$$3$$

$$4$$

$$8$$

###### Solution(s):

We can graph these equations to easily find out where the intersection points are. From the graph, we see that the line intersects the other graph three times.

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

$$\sqrt{3}$$

$$\dfrac{7}{4}$$

$$\dfrac{15}{8}$$

$$2$$

###### Solution(s): Note that the crease will the perpendicular bisector of $$\overline{AB}.$$ Let $$\overline{DE}$$ be the crease.

By $$AA$$ similarity, we know that $$\triangle ADE \sim \triangle ACB.$$ Therefore, $\dfrac{BC}{AC} = \dfrac{DE}{AD}$Plugging in values: $\dfrac{3}{4} = \dfrac{DE}{\frac{5}{2}}.$

Simplifying gets us that $DE = \dfrac{15}{8}.$

Thus, D is the correct answer.

14.

What is the greatest integer less than or equal to $\dfrac{3^{100}+2^{100}}{3^{96}+2^{96}}?$

$$80$$

$$81$$

$$96$$

$$97$$

$$625$$

###### Solution(s):

Let $$a = 3^{96}$$ and $$b = 2^{96}.$$ Then the expression can be rewritten as $\dfrac{81a + 16b}{a + b} = \dfrac{16a + 16b}{a + b} + \dfrac{65a}{a + b}$ $= 16 + \dfrac{65a}{a + b}.$

Note that $\dfrac{65a}{a + b} \lt \dfrac{65a}{a} = 65.$

Therefore, the answer is less than $$65 + 16 = 81.$$ $$80$$ is the only choice that satisfies this condition.

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 $$m+n?$$ $$21$$

$$29$$

$$58$$

$$69$$

$$93$$

###### Solution(s): Let $$Y$$ and $$Z$$ be the centers of the small circles and $$X$$ be the center of the large circle.

Then $AX = BX = 13,$ $XY = XZ = 13 - 8 = 5,$ and $YZ = 2 \cdot 5 = 10.$

We also know that $$\triangle XYZ \sim \triangle XAB.$$ Therefore, $\dfrac{XY}{YZ} = \dfrac{XA}{AB}$Plugging in values:$\dfrac{8}{13} = \dfrac{10}{AB}.$

Multiplying yields $$AB = \dfrac{65}{4}.$$ $m + n = 65 + 4 = 69$ 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 $$\overline{AC}?$$

$$5$$

$$8$$

$$12$$

$$13$$

$$15$$

###### Solution(s): Let $$P$$ be the foot of the altitude from $$B$$ to $$\overline{AC}.$$ We also get that $$AC = 29.$$

This tells us that $\dfrac{29 \cdot PB}{2} = \dfrac{20 \cdot 21}{2}$$PB = \dfrac{20 \cdot 21}{29},$ by calculating the area in two ways. This value is between $$14$$ and $$15.$$

Note that as we move the line segment from $$\overline{AB}$$ to $$\overline{PB},$$ the line segment's length ranges from $$AB$$ to $$PB.$$

The integer values it covers therefore goes from $$20$$ to $$15.$$ Similarly, as the line segment moves from $$\overline{PB}$$ to $$\overline{CB},$$ its takes on the values from $$15$$ to $$21.$$

This gives us $$13$$ unique line segments that have an integer value length.

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 $$S?$$

$$2$$

$$3$$

$$4$$

$$5$$

$$7$$

###### Solution(s):

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 $$0\le i \le 7?$$

$$512$$

$$729$$

$$1094$$

$$3281$$

$$59,048$$

###### Solution(s):

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 different numbers.

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 $$1?$$

$$\dfrac{1}{5}$$

$$\dfrac{1}{4}$$

$$\dfrac{3}{10}$$

$$\dfrac{7}{20}$$

$$\dfrac{2}{5}$$

###### Solution(s):

Since we only care about the units digit, we can turn the set $\{11,13,15,17,19\}$ into $\{1,3,5,7,9\}.$ Then we can case on the value of $$m.$$

$$m = 1$$

Any value of $$n$$ works. This occurs with a $$\frac{1}{5}$$ probability.

$$m = 3$$

Looking at powers of $$3,$$ we see that this sequence of units digits repeats: $3, 9, 7, 1, \ldots$

This means that $$n$$ must be a multiple of $$4.$$ There are $$5$$ such values. This means that $$n$$ works $$\frac{5}{20} = \frac{1}{4}$$ of the time. The total probability is $\dfrac{1}{5} \cdot \dfrac{1}{4} = \dfrac{1}{20}.$

$$m = 5$$

Powers of $$5$$ always end in $$5,$$ which means that this case will never work.

$$m = 7$$

The units digits repeat in this pattern: $7, 9, 3, 1,\ldots$ This means that $$n$$ must be a multiple of $$4$$ to work. As when $$m = 3,$$ this case works with a probability of $$\frac{1}{20}.$$

$$m = 9$$

The units digit alternates between $$1$$ and $$9.$$ This means that $$n$$ has to be even. This happens with a $$\frac{1}{2}$$ chance. The total probability is then $\dfrac{1}{5} \cdot \dfrac{1}{2} = \dfrac{1}{10}.$

The total probability is therefore $\dfrac{1}{5} + 2 \cdot \dfrac{1}{20} + \dfrac{1}{10} = \dfrac{2}{5}.$

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?

$$510$$

$$1022$$

$$8190$$

$$8192$$

$$65,534$$

###### Solution(s):

Note that all the line of symmetry split the grid into $$8$$ congruent regions that like look the figure below. If we analyze one of these pieces, we can see that what we color this region determines the coloring of the other $$7$$ regions.

There are $$10$$ sections in this figure, and we have $$2$$ options for each of them, for a total of $$2^{10} = 1024$$ colorings.

We have to subtract $$2$$ since all the colors cannot be the same ($$1024 - 2 = 1022$$).

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?

$$a = \dfrac14$$

$$\dfrac14 \lt a \lt \dfrac12$$

$$a \gt \dfrac14$$

$$a = \dfrac12$$

$$a \gt \dfrac12$$

###### Solution(s):

We can substitute $$y$$ from the second curve into the first curve.

\begin{align*} x^2 + (x^2 - a)^2 &= a^2 \\ x^2 + x^4 - 2ax^2 &= 0 \\ x^2(x^2 - (2a - 1)) &= 0 \end{align*}

From this, we see that $$x = 0$$ always yields a solution (which is in fact a double root).

For the other two roots to exist and be unique, we need $x^2 - (2a - 1) = 0$ to have a positive discriminant.

For this to happen, $4(2a - 1) \gt 0$or equivalently$a \gt \dfrac{1}{2}.$ 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 $$a?$$

$$5$$

$$7$$

$$11$$

$$13$$

$$17$$

###### Solution(s):

The problem statement tells us that $$24$$ divides $$a$$ and $$b,$$ $$36$$ divides $$b$$ and $$c,$$ and $$54$$ divides $$c$$ and $$d.$$

Note that these are the following prime factorizations: $\begin{gather*} 24 = 2^3 \cdot 3 \\ 36 = 2^2 \cdot 3^2 \\ 54 = 2 \cdot 3^3. \end{gather*}$

This means that we can express $$a, b,$$ and $$c$$ as follows: $\begin{gather*} a = 2^3 \cdot 3 \cdot w \\ b = 2^3 \cdot 3^2 \cdot x \\ c = 2^2 \cdot 3^2 \cdot y \\ d = 2 \cdot 3^3 \cdot z. \end{gather*}$

Then we get that $\gcd(a, d) = 2 \cdot 3 \cdot \gcd(w, z).$

Note that $$w$$ cannot have a factor of $$3$$ since that would mean $$\gcd(a, b)$$ has an extra factor of $$3.$$

Similarly, we can see that $$z$$ does not have a factor of $$2.$$

This means that $$\gcd(w, z)$$ only has prime factors that are greater than or equal to $$5.$$

The only number of the form $$6p,$$ where $$p$$ satisfies the above criteria, between $$70$$ and $$100$$ is $$78 = 6 \cdot 13.$$

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? $$\dfrac{25}{27}$$

$$\dfrac{26}{27}$$

$$\dfrac{73}{75}$$

$$\dfrac{145}{147}$$

$$\dfrac{74}{75}$$

###### Solution(s):

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 $$FDBG?$$

$$60$$

$$65$$

$$70$$

$$75$$

$$80$$

###### Solution(s): Let $$BC = a, BG = x, GC = y,$$ and $$h$$ be the length of the altitude through $$A.$$

By the angle bisector theorem, we get that $\dfrac{50}{x} = \dfrac{10}{y},$ where $$y = a - x.$$ Substituting yields $$BG = \frac{5a}{6}.$$ We also know that $$DF = \frac{5a}{12}$$ due to similar triangles.

Note that the height of the trapezoid is $$\frac{1}{2}h,$$ and $$\frac{ah}{2} = 120.$$ The area of the trapezoid is $\dfrac{5a}{8} \cdot \dfrac{h}{2} = \dfrac{5}{8} \cdot \dfrac{ah}{2} = 75.$ 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 $C_n - B_n = A_n^2?$

$$12$$

$$14$$

$$16$$

$$18$$

$$20$$

###### Solution(s):

We can use the formula for the sum of a geometric sequence to rewrite $$A_n, B_n,$$ and $$C_n.$$

$\begin{gather*} A_n = a(11\cdots11) \\ =a(1 + 10 + 10^2 + \cdots + 10^{n - 1}) \\ =a \cdot \dfrac{10^n - 1}{9} \end{gather*}$

Similarly, we get that $B_n = b \cdot \dfrac{10^n - 1}{9}$ and $C_n = c \cdot \dfrac{10^{2n} - 1}{9}.$

We can substitute these expressions into our condition to get $c \cdot \dfrac{10^{2n} - 1}{9} - b \cdot \dfrac{10^n - 1}{9}$ $= a^2 \left(\dfrac{10^n - 1}{9}\right)^2.$

Simplifying yields \begin{align*} c(10^n + 1) - b &= a^2 \cdot \dfrac{10^n - 1}{9} \\ 9c(10^n + 1) - 9b &= a^2 \cdot (10^n - 1) \\ (9c - a^2)10^n &= 9b - 9c - a^2. \end{align*}

From the last line, we see that $$9c - a^2$$ and $$9b - 9c - a^2$$ are constants.

For there to be at least $$2$$ unique values of $$n$$ that satisfy the equation, both sides must equal zero.

We can see this by realizing that this equation is linear with respect to $$10^n.$$ If both sides are non-zero, then there cannot exist $$2$$ unique solutions to a linear equation.

This tell us that $9c - a^2 = 0$ and $9b - 9c - a^2 = 0.$

The first equation gives us $c = \dfrac{a^2}{9}.$ Plugging this into the second equation gives us $b = \dfrac{2a^2}{9}.$

This tells us that $$a$$ must be divisible by $$3.$$ This gives us the following triples:$(3, 2, 1), (6, 8, 4), (9, 18, 9).$

The last triple is not allowed, so the maximum sum is $6 + 8 + 4 = 18.$ Thus, D is the correct answer.