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2018 AMC 12A Answer Key答案解析

Category: AMC美国数学竞赛, 国际竞赛, 福利干货, 竞赛资料 Date: 2018年2月9日上午10:24

2018 AMC 12A Answer Key答案解析

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Problem 1

答案：D
Solution：There are $36$ red balls; for these red balls to comprise $72 \%$ of the urn, there must be only $14$ blue balls. Since there are currently $64$ blue balls, this means we must remove $50 = \boxed{ \textbf{(D)}.}$

Problem 2

答案：c
Solution：The answer is just $3\cdot 18=54$ minus the minimum number of rocks we need to make $18$ pounds, or $54-4=\boxed{\textbf{(C)} 50.}$

Problem 3

答案：E
Solution 1：

We must place the classes into the periods such that no two balls are in the same period or in consecutive period.

Ignoring distinguishability, we can thus list out the ways that three periods can be chosen for the classes, when periods cannot be consecutive:

Periods $1, 3, 5$

Periods $1, 3, 6$

Periods $1, 4, 6$

Periods $2, 4, 6$

There are $4$ ways to place $3$ nondistinguishable classes into $6$ periods such that no two classes are in consecutive periods. For each of these ways, there are $3! = 6$ orderings of the classes among themselves.

Therefore, there are $4 \cdot 6 = \boxed{\mathrm{(E) \ } 24}$ ways to choose the classes.

Solution 2：

First draw 6 $X$ 's representing the 6 periods.

$XXXXXX$

Let the $O$ 's represent the classes that occupy each period.

$XOXOXO$

There are 6 ways to place the first class.

There are 4 ways to place the second class.

There is 1 way to place the third class.

We multiply $6 \cdot 4 \cdot 1= \boxed{(E) 24}$

—Baolan

Problem 4

答案：D
Solution：From Alice and Bob, we know that $5 < d < 6.$ From Charlie, we know that $4 < d.$ We take the union of these two intervals to yield $\boxed{\textbf{(D)} (5,6)}.$

Problem 5

答案：E
Solution：We factor $x^2-3x+2$ into $(x-1)(x-2)$ . Thus, either $1$ or $2$ is a root of $x^2-5x+k$ . If $1$ is a root, then $1^2-5\cdot1+k=0$ , so $k=4$ . If $2$ is a root, then $2^2-5\cdot2+k=0$ , so $k=6$ . The sum of all possible values of $k$ is $\boxed{\textbf{(E)}10}$ .

Problem 6

答案：B
Solution 1：The mean and median are $\frac{3m+4n+17}{6}=\frac{m+n+11}{2}=n,$ so $3m+17=2n$ and $m+11=n$ . Solving this gives $\left(m,n\right)=\left(5,16\right)$ for $m+n=\boxed{21}$ . (trumpeter)
Solution 2：

This is an alternate solution if you don't want to solve using algebra. First, notice that the median $n$ is the average of $m+10$ and $n+1$ . Therefore, $n=m+11$ , so the answer is $m+n=2m+11$ , which must be odd. This leaves two remaining options: ${(B) 21}$ and ${(D) 23}$ . Notice that if the answer is $(B)$ , then $m$ is odd, while $m$ is even if the answer is $(D)$ . Since the average of the set is an integer $n$ , the sum of the terms must be even. $4+10+1+2+2n$ is odd by definition, so we know that $3m+2n$ must also be odd, thus with a few simple calculations $m$ is odd. Because all other answers have been eliminated, $(B)$ is the only possibility left. Therefore, $m+n=\boxed{21}$ .

Problem 7

答案：E
Solution：The prime factorization of $4000$ is $2^{5}5^{3}$ . Therefore, the maximum number for $n$ is $3$ , and the minimum number for $n$ is $-5$ . The range from $-5$ to $3$ , which is $3-(-5) + 1 = 8 + 1 = \fbox{\textbf{(E) }9}$ .

Problem 8

答案：E
Solution 1：You can see that we can create a "stack" of 5 triangles congruent to the 7 small triangles shown here, arranged in a row above those 7, whose total area would be 5. Similarly, we can create another row of 3, and finally 1 more at the top, as follows. We know this cumulative area will be $7+5+3+1=16$ , so to find the area of such trapezoid $BCED$ , we just take $40-16=\boxed{24}$ , like so. --anna0kear

Solution 2：Let $x$ be the area of $ADE$ . Note that $x$ is comprised of the $7$ small isosceles triangles and a triangle similar to $ADE$ with side length ratio $3:4$ (so an area ratio of $9:16$ ). Thus, we have $x=7+\dfrac{9}{16}x$ This gives $x=16$ , so the area of $DBCE=40-x=\boxed{24}$ .
Solution 3：The area of $ADE$ is 16 times the area of the small triangle, as they are similar and their side ratio is $4:1$ . Therefore the area of the trapezoid is $40-16=\boxed{24}$ .

Problem 9

答案：E
Solution：On the interval $[0, \pi]$ sine is nonnegative; thus $\sin(x + y) = \sin x \cos y + \sin y \cos x \le \sin x + \sin y$ for all $x, y \in [0, \pi]$ . The answer is $\boxed{\textbf{(E) } 0\le y\le \pi}$ . (CantonMathGuy)

Problem 10

答案：C
Solution 1：

The graph looks something like this:

$[asy] draw((-3,0)--(3,0), Arrows); draw((0,-3)--(0,3), Arrows); draw((2,3)--(0,1)--(-2,3), blue); draw((-3,2)--(-1,0)--(-3,-2), blue); draw((-2,-3)--(0,-1)--(2,-3), blue); draw((3,-2)--(1,0)--(3,2), blue); draw((-3,2)--(3,0), red); dot((-3,2)); dot((3/2,1/2)); dot((0,1)); [/asy]$

Now it's clear that there are $\boxed{3}$ intersection points. (pinetree1)

Solution 2：

$x+3y=3$ can be rewritten to $x=3-3y$ . Substituting $3-3y$ for $x$ in the second equation will give $||3-3y|-y|=1$ . Splitting this question into casework for the ranges of y will give us the total number of solutions.

Case 1: $y>1$

$3-3y$ will be negative so $|3-3y| = 3y-3$

$|3y-3-y| = |2y-3| = 1$

   Subcase 1:

$2y-3$ is positive so $2y-3 = 1$ and $y = 2$ and $x = 3-3(2) = -3$

   Subcase 2:

$2y-3$ is negative so $|2y-3| = 3-2y = 1$ . $2y = 2$ and so there are no solutions ( $y$ can't equal to $1$ )

Case 2: $y = 1 x = 0$

Case 3: $y<1$

$3-3y$ will be positive so $|3-3y-y| = |3-4y| = 1$

   Subcase 1:

$3-4y$ will be negative so $4y-3 = 1$ --> $4y = 4$ . There are no solutions (again, $y$ can't equal to $1$ )

   Subcase 2: y<4/3

$3-4y$ will be positive so $3-4y = 1$ --> $4y = 2$ . $y = \frac{1}{2}$ and $x = \frac{3}{2}$ Solutions: $(-3,2) (0,1) (\frac{3}{2},\frac{1}{2})$ $\boxed{\textbf{(C)} \ 3}$

NOTE: Please fix this up using latex I have no idea how

Solution by Danny Li JHS

Problem 11

答案：D
Solution 1：First, we need to realize that the crease line is just the perpendicular bisector of side $AB$ , the hypotenuse of right triangle $\triangle ABC$ . Call the midpoint of $AB$ point $D$ . Draw this line and call the intersection point with $AC$ as $E$ . Now, $\triangle ABC$ is similar to $\triangle ADE$ by $AA$ similarity. Setting up the ratios, we find that $\frac{BC}{AC}=\frac{DE}{AD} \Rightarrow \frac{3}{4}=\frac{DE}{\frac{5}{2}} \Rightarrow DE=\frac{15}{8}.$ Thus, our answer is $\boxed{D}$ . ~Nivek
Solution 2 (if you are already out of time)：Simply make a 3x4x5 inch triangle and then cut it out (using fine rips). Then, make the fold and measure. It will be $\boxed{D} \frac{15}{8}$ inches in length.

Problem 12

答案：C
Solution：If we start with $1$ , we can include nothing else, so that won't work. If we start with $2$ , we would have to include every odd number except $1$ to fill out the set, but then $3$ and $9$ would violate the rule, so that won't work. Experimentation with $3$ shows it's likewise impossible. You can include $7$ , $11$ , and either $5$ or $10$ (which are always safe). But after adding either $4$ or $8$ we have nowhere else to go. Finally, starting with $4$ , we find that the sequence $4,5,6,7,9,11$ works, giving us $\boxed{\textbf{(C)} \text{ 4}}$ . (Random_Guy)

Problem 13

答案：D
Solution 1：This looks like balanced ternary, in which all the integers with absolute values less than $\frac{3^n}{2}$ are represented in $n$ digits. There are 8 digits. Plugging in 8 into the formula gives a maximum bound of $|x|=3280.5$ , which means there are 3280 positive integers, 0, and 3280 negative integers. Since we want all nonnegative integers, there are $3280+1=\boxed{3281}$ integers or $\boxed{D}$ .
Solution 2：Note that all numbers formed from this sum are either positive, negative or zero. The number of positive numbers formed by this sum is equal to the number of negative numbers formed by this sum, because of symmetry. There is only one way to achieve a sum of zero, if all $a_i=0$ . The total number of ways to pick $a_i$ from $i=1, 2, 3, ... 7$ is $3^8=6561$ . $\frac{6561-1}{2}=3280$ gives the number of possible negative integers. The question asks for the number of nonnegative integers, so subtracting from the total gives $6561-3280=\boxed{3281}$ . ~RegularHexagon
Solution 3 (Quick Solution)：Note that the number of total possibilities (ignoring the conditions set by the problem) is $3^8=6561$ . So, E is clearly unrealistic. Note that if $a_7$ is 1, then it's impossible for $a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,$ to be negative. Therefore, if $a_7$ is 1, there are $3^7=2187$ possibilities. As A, B, and C are all less than 2187, the answer must be $\boxed{(D) 3281}$

Problem 14

答案：D
Solution：

Base switch to log 2 and you have

$\frac{\log_2 4}{\log_2 3x} = \frac{\log_2 8}{\log_2 2x}$ .

$\frac{2}{\log_2 3x} = \frac{3}{\log_2 2x}$

$2*\log_2 2x = 3*\log_2 3x$

Then $\log_2 (2x)^2 = \log_2 (3x)^3$ . so $4x^2=27x^3$ and we have $x=\frac{4}{27}$ leading to $\boxed{ (D) 31}$ (jeremylu)

Problem 15

答案：B
Solution：

Draw a $7 \times 7$ square.

$\begin{tabular}{|c|c|c|c|c|c|c|} \hline a & b & c & d & c & b & a \\ \hline b & e & f & g & f & e & b \\ \hline c & f & h & i & h & f & c \\ \hline d & g & i & j & i & g & d \\ \hline c & f & h & i & h & f & c \\ \hline b & e & f & g & f & e & b \\ \hline a & b & c & d & c & b & a \\ \hline \end{tabular}$

There are only ten squares we get to actually choose, and two independent choices for each, for a total of $2^{10} = 1024$ codes. Two codes must be subtracted (due to the rule that there must be at least one square of each color) for an answer of $\fbox{\textbf{(B)} \text{ 1022}}$

Problem 16

答案：E
Solution 1：Substituting $y=x^2-a$ into $x^2+y^2=a^2$ , we get $x^2+(x^2-a)^2=a^2 \implies x^2+x^4-2ax^2=0 \implies x^2(x^2-(2a-1))=0$ Since this is a quartic, there are 4 total roots (counting multiplicity). We see that $x=0$ always at least one intersection at $(0,-a)$ (and is in fact a double root). The other two intersection points have $x$ coordinates $\sqrt{2a-1}$ . We must have $2a-1> 0,$ otherwise we are in the case where the parabola lies entirely above the circle (tangent to it at the point $(0,a)$ ). This only results in a single intersection point in the real coordinate plane. Thus, we see $a>\frac{1}{2}$ . (projecteulerlover)
Solution 2： Looking at a graph, it is obvious that the two curves intersect at (0, -a). We also see that if the parabola go's 'in' the circle, than by going out of it (as it will) it will intersect five times, an impossibility. Thus we only look for cases where the parabola becomes externally tangent to the circle. We have $x^2 - a = -\sqrt(a^2 - x^2)$ . Squaring both sides and solving yields $x^4 - (2a - 1)x^2 = 0$ . Since x = 0 is already accounted for, we only need to find 1 solution for $x^2 = 2a - 1$ , where the right hand side portion is obviously increasing. Since a = 1/2 begets x = 0 (an overcount), we have $a > 1/2 -> E$ is the right answer. Solution by JohnHankock
Solution 3： This describes a unit parabola, with a circle centered at the axis of symmetry and tangent to the vertex. As the curvature of the unit parabola at the vertex is 2, the radius of the circle that matches it has a radius of $\frac{1}{2}$ . This circle is tangent to an infinitesimally close pair of points, one on each side. Therefore, it is tangent to only 1 point. When a larger circle is used, it is tangent to 3 points because the points on either side are now separated from the vertex. Therefore, $\boxed{a > \frac{1}{2}}$ or $\boxed{E}$ is correct.
Solution 4： Notice, the equations are of that of a circle of radius a centered at the origin and a parabola translated down by a units. They always intersect at the point $(0, a)$ , and they have symmetry across the y-axis, thus, for them to intersect at exactly 3 points, it suffices to find the y solution.

First, rewrite the second equation to $y=x^2-a\implies x^2=y+a$ And substitute into the first equation: $y+a+y^2=a^2$ Since we're only interested in seeing the interval in which a can exist, we find the discriminant: $1-4a+4a^2$ . This value must not be less than 0 (It is the square root part of the quadratic formula). To find when it is 0, we find the roots: $4a^2-4a+1=0 \implies a=\frac{4\pm\sqrt{16-16}}{8}=\frac{1}{2}$ Since $\lim_{a\to \infty}(4a^2-4a+1)=\infty$ , our range is $\boxed{a>\frac{1}{2}}$

Problem 17

答案：D
Solution 1：Let the square have side length $x$ . Connect the upper-right vertex of square $S$ with the two vertices of the triangle's hypotenuse. This divides the triangle in several regions whose areas must add up to the area of the whole triangle, which is $6$ .
Square $S$ has area $x^2$ , and the two thin triangle regions have area $\dfrac{x(3-x)}{2}$ and $\dfrac{x(4-x)}{2}$ . The final triangular region with the hypotenuse as its base and height $2$ has area $5$ . Thus, we have $x^2+\dfrac{x(3-x)}{2}+\dfrac{x(4-x)}{2}+5=6$ Solving gives $x=\dfrac{2}{7}$ . The area of $S$ is $\dfrac{4}{49}$ and the desired ratio is $\dfrac{6-\dfrac{4}{49}}{6}=\boxed{\dfrac{145}{147}}$ .
Solution 2： Let the square have side length $s$ . If we were to extend the sides of the square further into the triangle until they intersect on point on the hypotenuse, we'd have a similar right triangle formed between the hypotenuse and the two new lines, and 2 smaller similar triangles that share a side of length 2. Using the side-to-side ratios of these triangles, we can find that the length of the big similar triangle is $\frac{5}{3}(2)=\frac{10}{3}$ . Now, let's extend this big similar right triangle to the left until it hits the side of length 3. Now, the length is $\frac{10}{3}+s$ , and using the ratios of the side lengths, the height is $\frac{3}{4}(\frac{10}{3}+s)=\frac{5}{2}+\frac{3s}{4}$ . Looking at the diagram, if we add the height of this triangle to the side length of the square, we'd get 3, so $\frac{5}{2}+\frac{3s}{4}+s=\frac{5}{2}+\frac{7s}{4}=3 \\ \frac{7s}{4}=\frac{1}{2} \\ s=\frac{2}{7} \implies \textrm{ area of square is } (\frac{2}{7})^2=\frac{4}{49}$ Now comes the easy part: finding the ratio of the areas: $\frac{3\cdot 4 \cdot \frac{1}{2} -\frac{4}{49}}{3\cdot 4 \cdot \frac{1}{2}}=\frac{6-\frac{4}{49}}{6}=\frac{294-4}{294}=\frac{290}{294}=\boxed{\frac{145}{147}.}$
Solution 3： We use coordinate geometry. Let the right angle be at $(0,0)$ and the hypotenuse be the line $3x+4y = 12$ for $0\le x\le 3$ . Denote the position of $S$ as $(s,s)$ , and by the point to line distance formula, we know that $\frac{|3s+4s-12|}{5} = 2$ $\Rightarrow |7s-12| = 10$ Obviously $s<\frac{22}{7}$ , so $s = \frac{2}{7}$ , and from here the rest of the solution follows to get $\boxed{\frac{145}{147}}$ .
Solution 4： Let the side length of the square be $x$ . First off, let us make a similar triangle with the segment of length $2$ and the top-right corner of $S$ . Therefore, the longest side of the smaller triangle must be $2 \cdot \frac54 = \frac52$ . We then do operations with that side in terms of $x$ . We subtract $x$ from the bottom, and $\frac{3x}{4}$ from the top. That gives us the equation of $3-\frac{7x}{4} = \frac{5}{2}$ . Solving, $12-7x = 10 \implies x = \frac{2}{7}.$ Thus, $x^2 = \frac{4}{49}$ , so the fraction of the triangle (area $6$ ) covered by the square is $\frac{2}{147}$ . The answer is then $\boxed{\dfrac{145}{147}}$ .

Problem 18

答案：D
Solution 1：
By angle bisector theorem, $BG=\frac{5a}{6}$ . By similar triangles, $DF=\frac{5a}{12}$ , and the height of this trapezoid is $\frac{h}{2}$ , where $h$ is the length of the altitude to $BC$ . Then $\frac{ah}{2}=120$ and we wish to compute $\frac{5a}{8}\cdot\frac{h}{2}=\boxed{75}$ .
Solution 2： $\overline{DE}$ is midway from $A$ to $\overline{BC}$ , and $DE = \frac{BC}{2}$ . Therefore, $\bigtriangleup ADE$ is a quarter of the area of $\bigtriangleup ABC$ , which is $30$ . Subsequently, we can compute the area of quadrilateral $BDEC$ to be $120 - 90 = 30$ . Using the angle bisector theorem in the same fashion as the previous problem, we get that $\overline{BG}$ is $5$ times the length of $\overline{GC}$ . We want the larger piece, as described by the problem. Because the heights are identical, one area is $5$ times the other, and $\frac{5}{6} \cdot 90 = \boxed{75}$ .

Problem 19

答案：C
Solution：its just $\sum_{a\ge 0}\frac1{2^a}\sum_{b\ge 0}\frac1{3^b}\sum_{c\ge 0}\frac{1}{5^c} = 2 \cdot \frac32 \cdot \frac54 = \frac{15}{4}\Rightarrow\textbf{(C)}.$ (ayushk)

Problem 20

答案：D
Solution：Observe that $\triangle{EMI}$ is isosceles right ( $M$ is the midpoint of diameter arc $EI$ ), so $MI=2,MC=\frac{3}{\sqrt{2}}$ . With $\angle{MCI}=45^\circ$ , we can use Law of Cosines to determine that $CI=\frac{3\pm\sqrt{7}}{2}$ . The same calculations hold for $BE$ also, and since $CI<BE$ , we deduce that $CI$ is the smaller root, giving the answer of $\boxed{12}$ . (trumpeter)

Problem 21

答案：B
Solution：We can see that our real solution has to lie in the open interval $(-1,0)$ . From there, note that $x^a < x^b$ if a, b are odd positive integers so $a<b$ , so hence it can only either be B or E(as all of the other polynomials will be larger than the polynomial B). Finally, we can see that plugging in the root of $2019x+2018$ into B gives a negative, and so the answer is $\fbox{B}$ . (cpma213)

Problem 22

答案：A
Solution：The roots are $\pm\left(\sqrt{10}+i\sqrt{6}\right),\pm\left(\sqrt{3}+i\right)$ (easily derivable by using DeMoivre and half-angle). From there, shoelace on $\left(0,0\right),\left(\sqrt{10},\sqrt{6}\right),\left(\sqrt{3},1\right)$ and multiplying by $4$ gives the area of $6\sqrt{2}-2\sqrt{10}$ , so the answer is $\boxed{20}$ . (trumpeter)

Problem 23

答案：E
Solution 1：

Let $P$ be the origin, and $PA$ lie on the x axis.

We can find $U=\left(\cos(36), \sin(36)\right)$ and $G=\left(10-\cos(56), \sin(56)\right)$

Then, we have $M=(5, 0)$ and $N=\left(\frac{10+\cos(36)-\cos(56)}{2}, \frac{\sin(36)+\sin(56)}{2}\right)$

Notice that the tangent of our desired points is the the absolute difference between the y coordinates of the two points divided by the absolute difference between the x coordinates of the two points.

This evaluates to $\frac{\cos(36)-\cos(56)}{\sin(36)+\sin(56)}$ Now, using sum to product identities, we have this equal to $\frac{2\cos(10)\cos(46)}{2\sin(10)\cos(46)}=\tan(80)$ so the answer is $\boxed{\textbf{(E)}.}$ (lifeisgood03)

Solution 2 (Overkill)：

Note that $X$ , the midpoint of major arc $PA$ on $(PAT)$ is the Miquel Point of $PUAG$ (Because $PU = AG$ ). Then, since $1 = \frac{UN}{NG} = \frac{PM}{MA}$ , this spiral similarity carries $M$ to $N$ . Thus, we have $\triangle XMN \sim \triangle XAG$ , so $\angle XMN = \angle XAG$ .

But, we have $\angle XAG = \angle PAG = \angle PAX = 56 - \frac{180 - \angle PXA}{2} =56 - \frac{180 - \angle T}{2} = 56 - \frac{\angle A + \angle P}{2} = 56 - \frac{56+36}{2} = 56 - 46 = 10$ ; thus $\angle XMN = 10$ .

Then, as $X$ is the midpoint of the major arc, it lies on the perpendicular bisector of $PA$ , so $\angle XMA = 90$ . Since we want the acute angle, we have $\angle NMA = \angle XMA - \angle XMN = 90 - 10 = 80$ , so the answer is $\boxed{\textbf{(E)}}$ .

(stronto)

Problem 24

答案：B
Solution 1：Plug in all the answer choices to get $\boxed{\textbf{(B)}.}$
Solution 2：Let the value we want be $x$ . The probability that Alice's number is less than Carol's number and Bob's number is greater than Carol's number is $x(\frac{2}{3}-x)$ . Similarly, the probability that Bob's number is less than Carol's number and Alice's number is greater than Carol's number is $(x-\frac{1}{2})(1-x)$ . Adding these together, the probability that Carol wins given a certain number $x$ is $-2x^2+\frac{13}{6}x-\frac{1}{2}$ . Using calculus or the fact that the extremum of a parabola occurs at $\frac{-b}{2a}$ , the maximum value occurs at $x=\frac{13}{24}$ , which is $\boxed{\textbf{(B)}.}$
Solution 3：The expected value of Alice's number is $\frac{1}{2}$ and the expected value of Bob's number is $\frac{7}{12}$ . To maximize her chance of winning, Carol would choose number exactly in between the two expected values, giving: $\frac{6+7}{12*2}=\frac{13}{24}$ . This is $\boxed{\textbf{(B)}}$ . (Random_Guy)

Problem 25

答案：D
Solution：Observe $A_n = a(1 + 10 + \dots + 10^{n - 1}) = a \cdot \tfrac{10^n - 1}{9}$ ; similarly $B_n = b \cdot \tfrac{10^n - 1}{9}$ and $C_n = c \cdot \tfrac{10^{2n} - 1}{9}$ . The relation $C_n - B_n = A_n^2$ rewrites as $c \cdot \frac{10^{2n} - 1}{9} - b \cdot \frac{10^n - 1}{9} = a^2 \cdot \left(\frac{10^n - 1}{9}\right)^2.$ Since $n > 0$ , $10^n > 1$ and we may cancel out a factor of $\tfrac{10^n - 1}{9}$ to obtain $c \cdot (10^n + 1) - b = a^2 \cdot \frac{10^n - 1}{9}.$ This is a linear equation in $10^n$ . Thus, if two distinct values of $n$ satisfy it, then all values of $n$ will. Matching coefficients, we need $c = \frac{a^2}{9} \quad \text{and} \quad c - b = -\frac{a^2}{9} \implies b = \frac{2a^2}{9}.$ To maximize $a + b + c = a + \tfrac{a^2}{3}$ , we need to maximize $a$ . Since $b$ and $c$ must be integers, $a$ must be a multiple of 3. If $a = 9$ then $b$ exceeds 9. However, if $a = 6$ then $b = 8$ and $c = 4$ for an answer of $\boxed{\textbf{(D)} \text{ 18}}$ . (CantonMathGuy)

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