2018 AMC 10A Exam Solutions

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All of the real AMC 8 and AMC 10 problems in our complete solution collection are used with official permission of the Mathematical Association of America (MAA).

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1.

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

58\dfrac{5}{8}

117\dfrac{11}{7}

85\dfrac{8}{5}

1811\dfrac{18}{11}

158\dfrac{15}{8}

Solution:

We can simplify this as follows. (((2+1)1+1)1+1)1+1=((13+1)1+1)1+1=(34+1)1+1=47+1=117 \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%50\% more soda than Jacqueline, and Alice has 25%25\% more soda than Jacqueline. What is the relationship between the amounts of soda that Liliane and Alice have?

Liliane has 20%20\% more soda than Alice.

Liliane has 25%25\% more soda than Alice.

Liliane has 45%45\% more soda than Alice.

Liliane has 75%75\% more soda than Alice.

Liliane has 100%100\% more soda than Alice.

Solution:

Let xx be the number of gallons of soda that Jacqueline has. Then Alice has 1.25x1.25x gallons, and Liliane has 1.5x1.5x gallons.

Therefore, the relationship can be found by dividing the amount of soda that each has to yield 1.5x1.25x=1.2,\dfrac{1.5x}{1.25x} = 1.2, which mean Liliane has 20%20\% more soda.

Thus, A is the correct answer.

3.

A unit of blood expires after 10!=1098110! = 10 \cdot 9 \cdot 8 \cdots 1 seconds. Yasin donates a unit of blood at noon of January 1.1. On what day does his unit of blood expire?

January 22

January 1212

January 2222

February 1111

February 1212

Solution:

We can divide 10!10! by 60,60, 60,60, and 2424 to get the number of days that it takes for a unit of blood to expire.

The first division cancels a 66 and 10.10. The second division cancels 3,4,3, 4, and 5.5. The final division cancels 88 and turns the 99 into a 3.3.

This leaves 2,7,2, 7, and a 3,3, which multiply to 42.42. There are 3131 days in January, so by February 1,1, the blood only has 4231=1142 - 31 = 11 days left.

1111 days from February 11 would make the blood expire on February 12.12.

Thus, E is the correct answer.

4.

How many ways can a student schedule 33 mathematics courses — algebra, geometry, and number theory — in a 66-period day if no two mathematics courses can be taken in consecutive periods?

(What courses the student takes during the other 33 periods is of no concern here.)

33

66

1212

1818

2424

Solution:

The 33 classes can occupy the following periods: (1,3,5),(1, 3, 5),(1,3,6),(1, 3, 6),(1,4,6), (1, 4, 6),(2,4,6). (2, 4, 6).

This means that there are 44 ways to choose which periods the mathematics courses occur.

For each configuration, there are 3!3! ways to determine the order of the courses, for a total of 64=246 \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 66 miles away," Bob replied, "We are at most 55 miles away." Charlie then remarked, "Actually the nearest town is at most 44 miles away."

It turned out that none of the three statements were true. Let dd be the distance in miles to the nearest town. Which of the following intervals is the set of all possible values of d?d?

(0,4)(0,4)

(4,5)(4,5)

(4,6)(4,6)

(5,6)(5,6)

(5,)(5,\infty)

Solution:

Alice's statement tells us that d<6.d \lt 6. Bob's statement tells us that d>5.d \gt 5. Charlie's statement tells us that d>4.d \gt 4.

Combining all of these tells us that 5<d5 \lt d and d<6,d \lt 6, which means dd is in the interval (5,6).(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 0,0, and the score increases by 11 for each like vote and decreases by 11 for each dislike vote.

At one point Sangho saw that his video had a score of 90,90, and that 65%65\% of the votes cast on his video were like votes. How many votes had been cast on Sangho's video at that point?

200200

300300

400400

500500

600600

Solution:

If 65%65\% of votes were like votes, then 35%35\% of votes are dislike votes. Then Sangho's score is 65%35%=30%65\% - 35\% = 30\% the total number of votes.

We know that Sangho's score is 90,90, so the total number of votes is 90÷30%=300.90 \div 30\% = 300.

Thus, B is the correct answer.

7.

For how many (not necessarily positive) integer values of nn is the following value an integer? 4000(25)n4000 \cdot \left(\dfrac{2}{5}\right)^n

33

44

66

88

99

Solution:

We can rewrite the expression as (2553)(25)n=25+n53n. (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 5+n0n53n0n3. \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=95 + 3 + 1 = 9 values for n.n.

Thus, E is the correct answer.

8.

Joe has a collection of 2323 coins, consisting of 55-cent coins, 1010-cent coins, and 2525-cent coins. He has 33 more 1010-cent coins than 55-cent coins, and the total value of his collection is 320320 cents. How many more 2525-cent coins does Joe have than 55-cent coins?

00

11

22

33

44

Solution:

Let xx be the number of 55-cent coins that Joe has. Then the number of 1010-cent coins he has is x+3.x + 3.

Therefore, Joe has 23x(x+3)=202x 23 - x - (x + 3) = 20 - 2x 2525-cent coins.

The total value of all these coins is 5x+10(x+3)+25(202x) 5x + 10(x + 3) + 25(20 - 2x) =53035x.= 530 - 35x.

We know that 53035x=320x=6. 530 - 35x = 320 \Rightarrow x = 6.

This means that Joe has 2026=820 - 2 \cdot 6 = 8 2525-cent coins. Therefore, he has 86=28 - 6 = 2 more 2525-cent coins than 55-cent coins.

Thus, C is the correct answer.

9.

All of the triangles in the diagram below are similar to isosceles triangle ABC,ABC, in which AB=AC.AB=AC. Each of the 77 smallest triangles has area 1,1, and ABC\triangle ABC has area 40.40. What is the area of trapezoid DBCE?DBCE?

1616

1818

2020

2222

2424

Solution:

We know that the side length of the smaller triangles is 140\sqrt{\frac{1}{40}} times the length of the larger triangle from similar triangles.

Then the side length of ADE\triangle ADE is 41404\sqrt{\frac{1}{40}} times the length of the side length of the larger triangle.

This makes the ratio of the areas (4140)2=16140=25. \left(4\sqrt{\dfrac{1}{40}}\right)^2 = 16 \cdot \dfrac{1}{40} = \dfrac{2}{5}.

Therefore, the area of ADE\triangle ADE is 2540=16.\frac{2}{5} \cdot 40 = 16. The area of the trapezoid is then 4016=24.40 - 16 = 24.

Thus, E is the correct answer.

10.

Suppose that real number xx satisfies 49x225x2=3.\sqrt{49-x^2}-\sqrt{25-x^2}=3. What is the value of 49x2+25x2?\sqrt{49-x^2}+\sqrt{25-x^2}?

88

33+8\sqrt{33}+8

99

210+42\sqrt{10}+4

1212

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 49x225+x2=24. 49 - x^2 - 25 + x^2 = 24.

This means that the product of the values of the expressions is equal to 24.24. The desired value is therefore 24÷3=8.24 \div 3 = 8.

Thus, A is the correct answer.

11.

When 77 fair standard 66-sided dice are thrown, the probability that the sum of the numbers on the top faces is 1010 can be written as n67,\dfrac{n}{6^{7}}, where nn is a positive integer. What is n?n?

4242

4949

5656

6363

8484

Solution:

We can use stars and bars to find n.n. It is the same as finding the number of ways to put 1010 balls into 77 boxes, where each box has at least one ball.

The formula for such a scenario is (n1k1), \binom{n - 1}{k - 1}, where nn is the number of balls and kk is the number of boxes.

The desired answer is therefore (96)=(93)=84. \binom{9}{6} = \binom{9}{3} = 84.

Thus, E is the correct answer.

12.

How many ordered pairs of real numbers (x,y)(x,y) satisfy the following system of equations? { x+3y=3 xy=1\begin{cases} ~x+3y&=3 \\ ~\big||x|-|y|\big|&=1 \end{cases}

11

22

33

44

88

Solution:

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 3,4,3,4, and 55 inches, as shown, is folded so that point AA falls on point B.B. What is the length in inches of the crease?

1+1221+\dfrac{1}{2} \sqrt{2}

3\sqrt{3}

74\dfrac{7}{4}

158\dfrac{15}{8}

22

Solution:

Note that the crease will the perpendicular bisector of AB.\overline{AB}. Let DE\overline{DE} be the crease.

By AAAA similarity, we know that ADEACB.\triangle ADE \sim \triangle ACB. Therefore, BCAC=DEAD \dfrac{BC}{AC} = \dfrac{DE}{AD} Plugging in values: 34=DE52. \dfrac{3}{4} = \dfrac{DE}{\frac{5}{2}}.

Simplifying gets us that DE=158.DE = \dfrac{15}{8}.

Thus, D is the correct answer.

14.

What is the greatest integer less than or equal to 3100+2100396+296?\dfrac{3^{100}+2^{100}}{3^{96}+2^{96}}?

8080

8181

9696

9797

625625

Solution:

Let a=396a = 3^{96} and b=296.b = 2^{96}. Then the expression can be rewritten as 81a+16ba+b=16a+16ba+b+65aa+b \dfrac{81a + 16b}{a + b} = \dfrac{16a + 16b}{a + b} + \dfrac{65a}{a + b} =16+65aa+b. = 16 + \dfrac{65a}{a + b}.

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

Therefore, the answer is less than 65+16=81.65 + 16 = 81. 8080 is the only choice that satisfies this condition.

Thus, A is the correct answer.

15.

Two circles of radius 55 are externally tangent to each other and are internally tangent to a circle of radius 1313 at points AA and B,B, as shown in the diagram. The distance ABAB can be written in the form mn,\frac{m}{n}, where mm and nn are relatively prime positive integers. What is m+n?m+n?

2121

2929

5858

6969

9393

Solution:

Let YY and ZZ be the centers of the small circles and XX be the center of the large circle.

Then AX=BX=13, AX = BX = 13, XY=XZ=138=5, XY = XZ = 13 - 8 = 5, and YZ=25=10. YZ = 2 \cdot 5 = 10.

We also know that XYZXAB.\triangle XYZ \sim \triangle XAB. Therefore, XYYZ=XAAB \dfrac{XY}{YZ} = \dfrac{XA}{AB} Plugging in values:813=10AB. \dfrac{8}{13} = \dfrac{10}{AB}.

Multiplying yields AB=654.AB = \dfrac{65}{4}. m+n=65+4=69 m + n = 65 + 4 = 69 Thus, D is the correct answer.

16.

Right triangle ABCABC has leg lengths AB=20AB=20 and BC=21.BC=21. Including AB\overline{AB} and BC,\overline{BC}, how many line segments with integer length can be drawn from vertex BB to a point on hypotenuse AC?\overline{AC}?

55

88

1212

1313

1515

Solution:

Let PP be the foot of the altitude from BB to AC.\overline{AC}. We also get that AC=29.AC = 29.

This tells us that 29PB2=20212 \dfrac{29 \cdot PB}{2} = \dfrac{20 \cdot 21}{2}PB=202129, PB = \dfrac{20 \cdot 21}{29}, by calculating the area in two ways. This value is between 1414 and 15.15.

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

The integer values it covers therefore goes from 2020 to 15.15. Similarly, as the line segment moves from PB\overline{PB} to CB,\overline{CB}, its takes on the values from 1515 to 21.21.

This gives us 1313 unique line segments that have an integer value length.

Thus, D is the correct answer.

17.

Let SS be a set of 66 integers taken from {1,2,,12}\{1,2,\dots,12\} with the property that if aa and bb are elements of SS with a<b,a < b, then bb is not a multiple of a.a. What is the least possible value of an element in S?S?

22

33

44

55

77

Solution:

We proceed by casing on possible values for S:S:

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

22 cannot be the smallest element since we would have to include every odd number except 1.1. This would make 33 and 99 violate the rule.

Let 33 be the smallest element. Then we can include 77 and 11.11. We can finally include either 44 or 88 and 55 or 10.10.

Either way, the maximum number of elements that we can include is 5,5, so 33 cannot be the smallest element.

Starting with 4,4, we can include 6,7,96, 7, 9 and 11.11. Finally, we can add either 55 or 10,10, creating a 66-element set.

Thus, C is the correct answer.

18.

How many nonnegative integers can be written in the form a737+a636+a535a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a434+a333+a232+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a131+a030,+a_1\cdot3^1+a_0\cdot3^0, where ai{1,0,1}a_i\in \{-1,0,1\} for 0i7?0\le i \le 7?

512512

729729

10941094

32813281

59,04859,048

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 11 s to 1-1 s and 1-1 s to 11 s).

The only way for the sum to be 00 is if all the coefficients are 0.0.

The total number of numbers is 38=6561.3^8 = 6561. Because each power of 33 is larger than the sum of all previous powers of three, each combination of coefficients yields different numbers.

Therefore, there are 656112+1=3281 \dfrac{6561 - 1}{2} + 1 = 3281 distinct nonnegative integers.

Thus, D is the correct answer.

19.

A number mm is randomly selected from the set {11,13,15,17,19},\{11,13,15,17,19\}, and a number nn is randomly selected from {1999,2000,2001,,2018}.\{1999,2000,2001,\ldots,2018\}. What is the probability that mnm^n has a units digit of 1?1?

15\dfrac{1}{5}

14\dfrac{1}{4}

310\dfrac{3}{10}

720\dfrac{7}{20}

25\dfrac{2}{5}

Solution:

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

m=1m = 1

Any value of nn works. This occurs with a 15\frac{1}{5} probability.

m=3m = 3

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

This means that nn must be a multiple of 4.4. There are 55 such values. This means that nn works 520=14\frac{5}{20} = \frac{1}{4} of the time. The total probability is 1514=120. \dfrac{1}{5} \cdot \dfrac{1}{4} = \dfrac{1}{20}.

m=5m = 5

Powers of 55 always end in 5,5, which means that this case will never work.

m=7m = 7

The units digits repeat in this pattern: 7,9,3,1, 7, 9, 3, 1,\ldots This means that nn must be a multiple of 44 to work. As when m=3,m = 3, this case works with a probability of 120.\frac{1}{20}.

m=9m = 9

The units digit alternates between 11 and 9.9. This means that nn has to be even. This happens with a 12\frac{1}{2} chance. The total probability is then 1512=110. \dfrac{1}{5} \cdot \dfrac{1}{2} = \dfrac{1}{10}.

The total probability is therefore 15+2120+110=25. \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 7×77 \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 4949 squares.

A scanning code is called symmetric if its look does not change when the entire square is rotated by a multiple of 9090^{\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?

510510

10221022

81908190

81928192

65,53465,534

Solution:

Note that all the line of symmetry split the grid into 88 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 77 regions.

There are 1010 sections in this figure, and we have 22 options for each of them, for a total of 210=10242^{10} = 1024 colorings.

We have to subtract 22 since all the colors cannot be the same (10242=10221024 - 2 = 1022).

Thus, B is the correct answer.

21.

Which of the following describes the set of values of aa for which the curves x2+y2=a2x^2+y^2=a^2 and y=x2ay=x^2-a in the real xyxy-plane intersect at exactly 33 points?

a=14a = \dfrac14

14<a<12\dfrac14 \lt a \lt \dfrac12

a>14a \gt \dfrac14

a=12a = \dfrac12

a>12a \gt \dfrac12

Solution:

We can substitute yy from the second curve into the first curve.

x2+(x2a)2=a2x2+x42ax2=0x2(x2(2a1))=0 \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=0x = 0 always yields a solution (which is in fact a double root).

For the other two roots to exist and be unique, we need x2(2a1)=0 x^2 - (2a - 1) = 0 to have a positive discriminant.

For this to happen, 4(2a1)>0 4(2a - 1) \gt 0 or equivalentlya>12.a \gt \dfrac{1}{2}. Thus, E is the correct answer.

22.

Let a,b,c,a, b, c, and dd be positive integers such that gcd(a,b)=24,\gcd(a, b)=24, gcd(b,c)=36,\gcd(b, c)=36, gcd(c,d)=54,\gcd(c, d)=54, and 70<gcd(d,a<100.70 < \gcd(d, a < 100. Which of the following must be a divisor of a?a?

55

77

1111

1313

1717

Solution:

The problem statement tells us that 2424 divides aa and b,b, 3636 divides bb and c,c, and 5454 divides cc and d.d.

Note that these are the following prime factorizations: 24=23336=223254=233. \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,a, b, and cc as follows: a=233wb=2332xc=2232yd=233z. \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)=23gcd(w,z). \gcd(a, d) = 2 \cdot 3 \cdot \gcd(w, z).

Note that ww cannot have a factor of 33 since that would mean gcd(a,b)\gcd(a, b) has an extra factor of 3.3.

Similarly, we can see that zz does not have a factor of 2.2.

This means that gcd(w,z)\gcd(w, z) only has prime factors that are greater than or equal to 5.5.

The only number of the form 6p,6p, where pp satisfies the above criteria, between 7070 and 100100 is 78=613.78 = 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 33 and 44 units. In the corner where those sides meet at a right angle, he leaves a small unplanted square SS so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from SS to the hypotenuse is 22 units. What fraction of the field is planted?

2527\dfrac{25}{27}

2627\dfrac{26}{27}

7375\dfrac{73}{75}

145147\dfrac{145}{147}

7475\dfrac{74}{75}

Solution:

Let xx be the side length of S.S. Then we can split the field up into the following shapes.

We can express the area of the field in two ways: 342=x2+x(3x)2 \dfrac{3 \cdot 4}{2} = x^2 + \dfrac{x(3 - x)}{2}+x(4x)2+252. + \dfrac{x(4 - x)}{2} + \dfrac{2 \cdot 5}{2}.

Simplifying yields 6=7x2+5 6 = \dfrac{7x}{2} + 5 x=27. x = \dfrac{2}{7}.

The desired fraction is 6x26=64496=145147. \dfrac{6 - x^2}{6} = \dfrac{6 - \frac{4}{49}}{6} = \dfrac{145}{147}.

Thus, D is the correct answer.

24.

Triangle ABCABC with AB=50AB=50 and AC=10AC=10 has area 120.120. Let DD be the midpoint of AB,\overline{AB}, and let EE be the midpoint of AC.\overline{AC}. The angle bisector of BAC\angle BAC intersects DE\overline{DE} and BC\overline{BC} at FF and G,G, respectively. What is the area of quadrilateral FDBG?FDBG?

6060

6565

7070

7575

8080

Solution:

Let BC=a,BG=x,GC=y,BC = a, BG = x, GC = y, and hh be the length of the altitude through A.A.

By the angle bisector theorem, we get that 50x=10y, \dfrac{50}{x} = \dfrac{10}{y}, where y=ax.y = a - x. Substituting yields BG=5a6.BG = \frac{5a}{6}. We also know that DF=5a12DF = \frac{5a}{12} due to similar triangles.

Note that the height of the trapezoid is 12h,\frac{1}{2}h, and ah2=120.\frac{ah}{2} = 120. The area of the trapezoid is 5a8h2=58ah2=75. \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 nn and nonzero digits a,a, b,b, and c,c, let AnA_n be the nn-digit integer each of whose digits is equal to aa; let BnB_n be the nn-digit integer each of whose digits is equal to bb; and let CnC_n be the 2n2n-digit (not nn-digit) integer each of whose digits is equal to c.c. What is the greatest possible value of a+b+ca + b + c for which there are at least two values of nn such that CnBn=An2?C_n - B_n = A_n^2?

1212

1414

1616

1818

2020

Solution:

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

An=a(1111)=a(1+10+102++10n1)=a10n19 \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 Bn=b10n19 B_n = b \cdot \dfrac{10^n - 1}{9} and Cn=c102n19.C_n = c \cdot \dfrac{10^{2n} - 1}{9}.

We can substitute these expressions into our condition to get c102n19b10n19 c \cdot \dfrac{10^{2n} - 1}{9} - b \cdot \dfrac{10^n - 1}{9} =a2(10n19)2. = a^2 \left(\dfrac{10^n - 1}{9}\right)^2.

Simplifying yields c(10n+1)b=a210n199c(10n+1)9b=a2(10n1)(9ca2)10n=9b9ca2. \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 9ca29c - a^2 and 9b9ca29b - 9c - a^2 are constants.

For there to be at least 22 unique values of nn that satisfy the equation, both sides must equal zero.

We can see this by realizing that this equation is linear with respect to 10n.10^n. If both sides are non-zero, then there cannot exist 22 unique solutions to a linear equation.

This tell us that 9ca2=0 9c - a^2 = 0 and 9b9ca2=0. 9b - 9c - a^2 = 0.

The first equation gives us c=a29.c = \dfrac{a^2}{9}. Plugging this into the second equation gives us b=2a29.b = \dfrac{2a^2}{9}.

This tells us that aa must be divisible by 3.3. This gives us the following triples:(3,2,1),(6,8,4),(9,18,9). (3, 2, 1), (6, 8, 4), (9, 18, 9).

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