## AMC 10 2018 Test A

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**Instructions**

- This is a 25-question, multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct.
- You will receive 6 points for each correct answer, 2.5 points for each problem left unanswered if the year is before 2006, 1.5 points for each problem left unanswered if the year is after 2006, and 0 points for each incorrect answer.
- No aids are permitted other than scratch paper, graph paper, ruler, compass, protractor and erasers (and calculators that are accepted for use on the test if before 2006. No problems on the test will require the use of a calculator).
- Figures are not necessarily drawn to scale.
- You will have
**75 minutes**working time to complete the test.

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

$\textbf{(A) } \dfrac58 \qquad \textbf{(B) }\dfrac{11}7 \qquad \textbf{(C) } \dfrac85 \qquad \textbf{(D) } \dfrac{18}{11} \qquad \textbf{(E) } \dfrac{15}8$

$\textbf{B}$

For all nonzero numbers $a,$ recall that $a^{-1}=\dfrac1a$ is the reciprocal of $a.$ The original expression becomes\begin{align*} \left(\left((2+1)^{-1}+1\right)^{-1}+1\right)^{-1}+1 &= \left(\left(3^{-1}+1\right)^{-1}+1\right)^{-1}+1 \\ &= \left(\left(\frac13+1\right)^{-1}+1\right)^{-1}+1 \\ &= \left(\left(\frac43\right)^{-1}+1\right)^{-1}+1 \\ &= \left(\frac34+1\right)^{-1}+1 \\ &= \frac47+1 \\ &= \frac{11}7 \end{align*}

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?

$\textbf{(A)}$ Liliane has $20\%$ more soda than Alice.$\\$

$\textbf{(B)}$ Liliane has $25\%$ more soda than Alice.$\\$

$\textbf{(C)}$ Liliane has $45\%$ more soda than Alice.$\\$

$\textbf{(D)}$ Liliane has $75\%$ more soda than Alice.$\\$

$\textbf{(E)}$ Liliane has $100\%$ more soda than Alice.

$\textbf{A}$

Let's assume that Jacqueline has $1$ gallon of soda. Then Alice has $1.25$ gallons and Liliane has $1.5$ gallons. Doing division, we find out that $\dfrac{1.5}{1.25}=1.2$, which means that Liliane has $20\%$ more soda. Therefore, the answer is A.

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?

$\textbf{(A) }\text{January 2}\qquad\textbf{(B) }\text{January 12}\qquad\textbf{(C) }\text{January 22}\qquad\textbf{(D) }\text{February 11}\qquad\textbf{(E) }\text{February 12}$

$\textbf{E}$

Converting $10!$ seconds to minutes by dividing by $60$, we get $9\cdot 8\cdot 7\cdot 5\cdot 4\cdot 3\cdot 2$ minutes. Converting minutes to hours by dividing by $60$ again, we get $9\cdot 8\cdot 7\cdot 2$ hours. Converting hours to days by dividing by $24$, we get $3\cdot 7\cdot 2 = 42$ days.

Now we need to count $42$ days after January 1. 30 days after January 1 is January 31. Then 12 days after January 31 is February 12. The answer is E.

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.)

$\textbf{(A) }3\qquad\textbf{(B) }6\qquad\textbf{(C) }12\qquad\textbf{(D) }18\qquad\textbf{(E) }24$

$\textbf{E}$

Ignoring distinguishability of classes, we can 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 $3$ non-distinguishable 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, we have $4 \cdot 6 = 24$ ways to choose the classes.

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 $d$?

$\textbf{(A) } (0,4) \qquad \textbf{(B) } (4,5) \qquad \textbf{(C) } (4,6) \qquad \textbf{(D) } (5,6) \qquad \textbf{(E) } (5,\infty)$

$\textbf{D}$

Think of the distances as if they are on a number line. Alice claims that $d > 6$, Bob says $d < 5$, while Charlie thinks $d < 4$. This means that all possible numbers less than $5$ and greater than $6$ are included. However, since the three statements are actually false, the distance to the nearest town is one of the numbers not covered. Therefore, the answer is $(5,6)$.

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?

$\textbf{(A) } 200 \qquad \textbf{(B) } 300 \qquad \textbf{(C) } 400 \qquad \textbf{(D) } 500 \qquad \textbf{(E) } 600$

$\textbf{B}$

Since $65\%$ of the votes were likes, $35\%$ of the votes were dislikes. $65\%-35\%=30\%$, so $90$ votes is $30\%$ of the total number of votes. Hence, we get the total number of votes $90/30\%=300$.

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

$\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }6 \qquad \textbf{(D) }8 \qquad \textbf{(E) }9 \qquad$

$\textbf{E}$

Note that\[4000\cdot \left(\frac{2}{5}\right)^n=\left(2^5\cdot5^3\right)\cdot \left(\frac{2}{5}\right)^n=2^{5+n}\cdot5^{3-n}\]Since this expression is an integer, we need:

$5+n\geq0,$ from which $n\geq-5.\newline$

$3-n\geq0,$ from which $n\leq3.$

Taking the intersection gives $-5\leq n\leq3.$ So there are $3-(-5)+1=9$ integer values of $n.$

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?

$\textbf{(A) } 0 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 4$

$\textbf{C}$

Let $x$ be the number of $5$-cent coins that Joe has. Therefore, he must have $(x+3) \ 10$-cent coins and $23-(x+3)-x=20-2x \ 25$-cent coins. Since the total value of his collection is $320$ cents, we have$$ 5x + 10(x+3) + 25(20-2x) = 320 $$ Hence, we get $x=6$. Joe has six $5$-cent coins, nine $10$-cent coins, and eight $25$-cent coins. Thus, our answer is $8-6 = 2.$

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 $DBCE$?

$\textbf{(A) } 16 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 22 \qquad \textbf{(E) } 24$

$\textbf{E}$

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 $DBCE=40-x=24$.

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

$\textbf{(A) }8 \qquad \textbf{(B) }\sqrt{33}+8\qquad \textbf{(C) }9 \qquad \textbf{(D) }2\sqrt{10}+4 \qquad \textbf{(E) }12 \qquad$

$\textbf{A}$

We notice that $$(\sqrt {49-x^2} + \sqrt {25-x^2})(\sqrt {49-x^2} - \sqrt {25-x^2}) = 49-x^2 - 25 +x^2 = 24$$ Given that $(\sqrt {49-x^2} - \sqrt {25-x^2}) = 3$, we get $(\sqrt {49-x^2} + \sqrt {25-x^2}) = \dfrac {24} {3} = 8$.

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$$\frac{n}{6^{7}},$$where $n$ is a positive integer. What is $n$?

$\textbf{(A) }42\qquad \textbf{(B) }49\qquad \textbf{(C) }56\qquad \textbf{(D) }63\qquad \textbf{(E) }84\qquad$

$\textbf{E}$

There are 3 ways to get a sum of 10:\[4,1,1,1,1,1,1\]\[3,2,1,1,1,1,1\]\[2,2,2,1,1,1,1\] The probability of the first case is $$\dfrac{\binom76\binom11}{6^7}=\dfrac{7}{6^7}$$ The probability of the second case is $$\dfrac{\binom75\binom21\binom11}{6^7}=\dfrac{42}{6^7}$$ The probability of the third case is $$\dfrac{\binom74\binom33}{6^7}=\dfrac{35}{6^7}$$ Hence, the total probability is $\dfrac{7}{6^7}+\dfrac{42}{6^7}+\dfrac{35}{6^7}=\dfrac{84}{6^7}$. The answer is $n=84$.

How many ordered pairs of real numbers $(x,y)$ satisfy the following system of equations?\begin{align*}

x+3y&=3 \\

\big||x|-|y|\big|&=1

\end{align*}

$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 8$

$\textbf{C}$

Note that $||x| - |y||$ can take on one of four values: $x + y$, $x - y$, $-x + y$, $-x -y$. So we have 4 cases:

$\textbf{Case 1:}$ $||x| - |y|| = x+y$\begin{align*}

x+3y&=3\\

x+y&=1

\end{align*} We get $(x,y)=(0,1)$

$\textbf{Case 2:}$ $||x| - |y|| = x-y$\begin{align*}

x+3y&=3\\

x-y&=1

\end{align*} We get $(x,y)=\left(\dfrac{3}{2},\dfrac{1}{2}\right)$

$\textbf{Case 3:}$ $||x| - |y|| = -x+y$\begin{align*}

x+3y&=3\\

-x+y&=1

\end{align*} We get $(x,y)=(0,1)$. This is the same solution as we got in Case 1.

$\textbf{Case 4:}$ $||x| - |y|| = -x-y$\begin{align*}

x+3y&=3\\

-x-y&=1

\end{align*} We get $(x,y)=(-3,2)$.

Finally, we need to test all the results $(0, 1)$, $(-3,2)$, and $\left(\dfrac{3}{2}, \dfrac{1}{2}\right)$ in the original system of equations \begin{align*}

x+3y&=3 \\

\big||x|-|y|\big|&=1

\end{align*}

All of them satisfy the original equations. So the answer is $3$.

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?

$\textbf{(A) } 1+\dfrac12 \sqrt2 \qquad \textbf{(B) } \sqrt3 \qquad \textbf{(C) } \dfrac74 \qquad \textbf{(D) } \dfrac{15}{8} \qquad \textbf{(E) } 2$

$\textbf{D}$

The crease line is the perpendicular bisector of side $AB$. Call the midpoint of $AB$ point $D$. Draw this line and call the intersection point with $AC$ as $E$. Now, $\triangle ACB$ 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, the answer is $\dfrac{15}{8}$.

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

$\textbf{(A) }80\qquad \textbf{(B) }81 \qquad \textbf{(C) }96 \qquad \textbf{(D) }97 \qquad \textbf{(E) }625\qquad$

$\textbf{A}$

Notice that\[\frac{3^{100}+2^{100}}{3^{96}+2^{96}}=\frac{3^{96}}{3^{96}+2^{96}}\cdot\frac{3^{100}}{3^{96}}+\frac{2^{96}}{3^{96}+2^{96}}\cdot\frac{2^{100}}{2^{96}}=\frac{3^{96}}{3^{96}+2^{96}}\cdot 81+\frac{2^{96}}{3^{96}+2^{96}}\cdot 16\]We see that our number is a weighted average of 81 and 16, extremely heavily weighted toward 81. Hence, the number is ever so slightly less than 81. The answer is $80$.

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 $\dfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?

$\textbf{(A) } 21 \qquad \textbf{(B) } 29 \qquad \textbf{(C) } 58 \qquad \textbf{(D) } 69 \qquad \textbf{(E) } 93$

$\textbf{D}$

Let the center of each circle be $X$, $Y$, $Z$, and the tangent points be $A$, $B$, as shown below.

We have $XY=XZ=13-5=8$, $YZ=2\cdot5=10$, $XA=XB=13$. Notice that $\triangle XAB\sim\triangle XYZ$, we have $$\dfrac{AB}{YZ}=\dfrac{XA}{XY}\rightarrow AB=\dfrac{XA\cdot YZ}{XY}=\dfrac{13\cdot10}{8}=\dfrac{65}{4}$$ So the answer is $65+4=69$.

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 $\overline{AC}$?

$\textbf{(A) }5 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }13 \qquad \textbf{(E) }15 \qquad$

$\textbf{D}$

The hypotenuse has length $29$. The altitude from $B$ to $AC$ is $BP=\dfrac{20\cdot 21}{29}$, which is between $14$ and $15$.

Note that if a circle with an integer radius $r$ centered at vertex $B$ intersects hypotenuse $\overline{AB}$, the lines drawn from $B$ to the points of intersection are integer lengths. The shortest distance $14<\overline{BP}<15$. As a result, a circle of $14$ will not reach the hypotenuse and thus does not intersect it. We also know that a circle of radius $21$ intersects the hypotenuse once and a circle of radius $\{15, 16, 17, 18, 19, 20 \}$ intersects the hypotenuse twice. So the answer is $6\cdot2+1=13$.

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 $S$?

$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 7$

$\textbf{C}$

We start with $2$ because every number is a multiple of $1$. We can 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.

Now we try with $3$. Our set becomes $\{3,4,5,7,11\}$ but we can add no more valid numbers.

Starting with $4,$ we find that the sequence $4,5,6,7,9,11$ works. So the answer is 4.

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 $0\le i \le 7$?

$\textbf{(A) } 512 \qquad \textbf{(B) } 729 \qquad \textbf{(C) } 1094 \qquad \textbf{(D) } 3281 \qquad \textbf{(E) } 59,048$

$\textbf{D}$

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 $i=0, 1, 2, 3, ... 7$ is $3^8=6561$. $\dfrac{6561-1}{2}=3280$ gives the number of possible negative integers. The question asks for the number of non-negative integers, so subtracting from the total gives $6561-3280= 3281$.

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 $1$?

$\textbf{(A) } \dfrac{1}{5} \qquad \textbf{(B) } \dfrac{1}{4} \qquad \textbf{(C) } \dfrac{3}{10} \qquad \textbf{(D) } \dfrac{7}{20} \qquad \textbf{(E) } \dfrac{2}{5}$

$\textbf{E}$

$\textbf{Case 1:}\ m=11$. The units digit of $11^n$ must be 1. So the probability in this case is $\dfrac15$.

$\textbf{Case 2:}\ m=13$. When the units digit of $13^n$ is 1, $n$ must be a multiple of 4. We have five multiples of 4 in the set $\{1999,2000,2001,\ldots,2018\}$. So the probability in this case is $\dfrac15\cdot\dfrac{5}{20}=\dfrac1{20}$.

$\textbf{Case 3:}\ m=15$. The units digit of $15^n$ can't be 1 when $n$ is chosen from $\{1999,2000,2001,\ldots,2018\}$. So the probability is 0.

$\textbf{Case 4:}\ m=17$. When the units digit of $17^n$ is 1, $n$ must be a multiple of 4 (just like case 2). So the probability is $\dfrac15\cdot\dfrac{5}{20}=\dfrac1{20}$.

$\textbf{Case 5:}\ m=19$. When the units digit of $19^n$ is 1, $n$ must be a even number. We have ten even numbers in the set $\{1999,2000,2001,\ldots,2018\}$. So the probability in this case is $\dfrac15\cdot\dfrac{10}{20}=\dfrac1{10}$.

Therefore, the answer is $\dfrac15+\dfrac1{20}+0+\dfrac1{20}+\dfrac1{10}=\dfrac25$.

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 $\textit{symmetric}$ 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?

$\textbf{(A)} \text{ 510} \qquad \textbf{(B)} \text{ 1022} \qquad \textbf{(C)} \text{ 8190} \qquad \textbf{(D)} \text{ 8192} \qquad \textbf{(E)} \text{ 65,534}$

$\textbf{B}$

Start from the center and label all protruding cells symmetrically.

In the grid, 10 letters are used: $A$, $B$, $C$, $D$, $E$, $F$, $G$, $H$, $J$, and $K$. Each of the letters must have its own color, either white or black. This means, for example, all $K$'s must have the same color for the grid to be symmetrical.

So there are $2^{10}$ ways to color the grid, including a completely black grid and a completely white grid. Since the grid must contain at least one square with each color, the number of ways is $2^{10}-2=1024-2=$ $1022$.

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?

$\textbf{(A) }a=\dfrac14 \qquad \textbf{(B) }\dfrac14 < a < \dfrac12 \qquad \textbf{(C) }a>\dfrac14 \qquad \textbf{(D) }a=\dfrac12 \qquad \textbf{(E) }a>\dfrac12 \qquad$

$\textbf{E}$

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 has at least one intersection at $(0,-a)$ (and is in fact a double root).

The other two intersection points have $x$ coordinates $\pm\sqrt{2a-1}$. We must have $2a-1> 0$. Thus, the answer is $a>\dfrac12$.

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 $a$?

$\textbf{(A)} \text{ 5} \qquad \textbf{(B)} \text{ 7} \qquad \textbf{(C)} \text{ 11} \qquad \textbf{(D)} \text{ 13} \qquad \textbf{(E)} \text{ 17}$

$\textbf{D}$

The GCD information tells us that $24$ divides $a$, both $24$ and $36$ divide $b$, both $36$ and $54$ divide $c$, and $54$ divides $d$.

Note that we have the prime factorizations:\begin{align*} 24 &= 2^3\cdot 3,\\ 36 &= 2^2\cdot 3^2\\ 54 &= 2\cdot 3^3 \end{align*} Hence we get\begin{align*} a &= 2^3\cdot 3\cdot w\\ b &= 2^3\cdot 3^2\cdot x\\ c &= 2^2\cdot 3^3\cdot y\\ d &= 2\cdot 3^3\cdot z \end{align*}for some positive integers $w,x,y,z$. Now if $3$ divides $w$, then $\gcd(a,b)$ would be at least $2^3\cdot 3^2$, which is too large, so $3$ does not divide $w$. Similarly, if $2$ divides $z$, then $\gcd(c,d)$ would be at least $2^2\cdot 3^3$, which is too large, so $2$ does not divide $z$. Therefore,\[\gcd(a,d)=2\cdot 3\cdot \gcd(w,z)\]where neither $2$ nor $3$ divide $\gcd(w,z)$. In other words, $\gcd(w,z)$ is divisible only by primes that are at least $5$. The only possible value of $\gcd(a,d)$ between $70$ and $100$ and which fits this criterion is $78=2\cdot3\cdot13$. So the answer is $13$.

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?

$\textbf{(A) } \dfrac{25}{27} \qquad \textbf{(B) } \dfrac{26}{27} \qquad \textbf{(C) } \dfrac{73}{75} \qquad \textbf{(D) } \dfrac{145}{147} \qquad \textbf{(E) } \dfrac{74}{75}$

$\textbf{D}$

Note that the hypotenuse of the field is $5,$ and the area of the field is $6.$ Let $x$ be the side-length of square $S.$ We partition the field into a red triangle, a yellow triangle, and a green triangle, as shown below:

Let the brackets denote areas. By area addition, we set up an equation for $x:$\begin{align*} [\text{Red Triangle}]+[\text{Yellow Triangle}]+[\text{Green Triangle}]&=[\text{Field}] \\ \frac{3x}{2}+\frac{4x}{2}+\frac{5\cdot2}{2}&=6, \end{align*}from which $x=\dfrac27.$ Therefore, the answer is\[\frac{[\text{Field}]-[S]}{[\text{Field}]}=\frac{6-x^2}{6}=\frac{145}{147}\]

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 $FDBG$?

$\textbf{(A) }60 \qquad \textbf{(B) }65 \qquad \textbf{(C) }70 \qquad \textbf{(D) }75 \qquad \textbf{(E) }80 \qquad$

$\textbf{D}$

Let the area of $\triangle ABG$ be $x$, and the area of $\triangle ACG$ be $y$. Then we have $x+y=120$. Since $AG$ bisects $\angle BAC$, the altitude from $G$ to $AB$ in $\triangle ABG$ is the same as the altitude from $G$ to $AC$ in $\triangle ACG$. Hence, we get $\dfrac{x}{y}=\dfrac{AB}{AC}=5$. Combined with $x+y=120$, we get $x=100$.

We see that $DE\parallel BC$ and $D$ is the middle point of $AB$. So the area of quadrilateral $FDBG$ is $\dfrac34x=75$.

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 $C_n - B_n = A_n^2$?

$\textbf{(A) } 12 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 20$

$\textbf{D}$

By geometric series, we have\begin{alignat*}{8} A_n&=a\bigl(\phantom{ }\underbrace{111\cdots1}_{n\text{ digits}}\phantom{ }\bigr)&&=a\left(1+10+10^2+\cdots+10^{n-1}\right)&&=a\cdot\frac{10^n-1}{9}, \\ B_n&=b\bigl(\phantom{ }\underbrace{111\cdots1}_{n\text{ digits}}\phantom{ }\bigr)&&=b\left(1+10+10^2+\cdots+10^{n-1}\right)&&=b\cdot\frac{10^n-1}{9}, \\ C_n&=c\bigl(\phantom{ }\underbrace{111\cdots1}_{2n\text{ digits}}\phantom{ }\bigr)&&=c\left(1+10+10^2+\cdots+10^{2n-1}\right)&&=c\cdot\frac{10^{2n}-1}{9}. \end{alignat*}By substitution, we rewrite the given equation $C_n - B_n = A_n^2$ 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 $n > 0,$ it follows that $10^n > 1.$ We divide both sides by $\dfrac{10^n-1}{9}$ and then rearrange:\begin{align*} c\left(10^n+1\right) - b &= a^2\cdot\frac{10^n-1}{9} \\ 9c\left(10^n+1\right) - 9b &= a^2\left(10^n-1\right) \\ \left(9c-a^2\right)10^n &= 9b-9c-a^2 \end{align*} Given that there are at least two values of $n$, we get \begin{align*} 9c-a^2&=0 \\ 9b-9c-a^2&=0 \end{align*}The first equation implies that $c=\dfrac{a^2}{9}.$ Substituting this into the second equation gives $b=\dfrac{2a^2}{9}.$

To maximize $a + b + c = a + \dfrac{a^2}{3},$ we need to maximize $a.$ Clearly, $a$ must be divisible by $3.$ The possibilities for $(a,b,c)$ are $(9,18,9),(6,8,4),$ or $(3,2,1),$ but $(9,18,9)$ is invalid. Therefore, the greatest possible value of $a + b + c$ is $6+8+4=18.$