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

$\dfrac{5}{8}$

$\dfrac{11}{7}$

$\dfrac{8}{5}$

$\dfrac{18}{11}$

$\dfrac{15}{8}$

###### Answer: B

###### 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.

###### Answer: A

###### 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 $12$

###### Answer: E

###### 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.)

$3$

$6$

$12$

$18$

$24$

###### Answer: E

###### 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 $d?$

$(0,4)$

$(4,5)$

$(4,6)$

$(5,6)$

$(5,\infty)$

###### Answer: D

###### 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?

$200$

$300$

$400$

$500$

$600$

###### Answer: B

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

$3$

$4$

$6$

$8$

$9$

###### Answer: E

###### 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?

$0$

$1$

$2$

$3$

$4$

###### Answer: C

###### 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 $DBCE?$

$16$

$18$

$20$

$22$

$24$

###### Answer: E

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

$8$

$\sqrt{33}+8$

$9$

$2\sqrt{10}+4$

$12$

###### Answer: A

###### 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 $n?$

$42$

$49$

$56$

$63$

$84$

###### Answer: E

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

$1$

$2$

$3$

$4$

$8$

###### Answer: C

###### 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?

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

$\sqrt{3}$

$\dfrac{7}{4}$

$\dfrac{15}{8}$

$2$

###### Answer: D

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

$80$

$81$

$96$

$97$

$625$

###### Answer: A

###### 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 $m+n?$

$21$

$29$

$58$

$69$

$93$

###### Answer: D

###### 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 $\overline{AC}?$

$5$

$8$

$12$

$13$

$15$

###### Answer: D

###### 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 $S?$

$2$

$3$

$4$

$5$

$7$

###### Answer: C

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

$512$

$729$

$1094$

$3281$

$59,048$

###### Answer: D

###### 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 $1?$

$\dfrac{1}{5}$

$\dfrac{1}{4}$

$\dfrac{3}{10}$

$\dfrac{7}{20}$

$\dfrac{2}{5}$

###### Answer: E

###### 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?

$510$

$1022$

$8190$

$8192$

$65,534$

###### Answer: B

###### 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?

$a = \dfrac14$

$\dfrac14 \lt a \lt \dfrac12$

$a \gt \dfrac14$

$a = \dfrac12$

$a \gt \dfrac12$

###### Answer: E

###### 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 $a?$

$5$

$7$

$11$

$13$

$17$

###### Answer: D

###### 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?

$\dfrac{25}{27}$

$\dfrac{26}{27}$

$\dfrac{73}{75}$

$\dfrac{145}{147}$

$\dfrac{74}{75}$

###### Answer: D

###### 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 $FDBG?$

$60$

$65$

$70$

$75$

$80$

###### Answer: D

###### 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 $C_n - B_n = A_n^2?$

$12$

$14$

$16$

$18$

$20$

###### Answer: D

###### 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.