Let the height of the replica be $h.$ Since the ratio of the scale model to the real world is $1:20,$ we know that $$h:289 = 1:20$$ Therefore: $$\begin{align*} \dfrac{h}{289}&= \dfrac{1}{20} \\ h&=\dfrac{289}{20} \\ h&=14.45 \approx 14\end{align*}$$Thus, the correct answer is A.

Let's first note that if we are given an expression of the form $1 + \frac{1}{n},$ we can rewrite this as $$\frac{n}{n} + \frac{1}{n} = \frac{n+1}{n}.$$ With that in mind, we can rewrite the expression given to us in the problem, as shown below: $$\begin{align*} &\left(1+\frac{1}{1}\right)\cdot\left(1+\frac{1}{2}\right)\\ &\quad{}\cdot\left(1+\frac{1}{3}\right)\cdot\left(1+\frac{1}{4}\right)\\ &\quad{}\cdot\left(1+\frac{1}{5}\right)\cdot\left(1+\frac{1}{6}\right)\\ &=\dfrac{2}{1} \cdot \dfrac{3}{2} \cdot \dfrac{4}{3} \cdot\dfrac{5}{4} \cdot\dfrac{6}{5} \cdot\dfrac{7}{6}\\ &=7 \end{align*}$$ Thus, the correct answer is D.

Notice that the first 5 numbers that contains a $7$ as its digit or are a multiple of $7$ are $7,14,17,21,27.$ Any player who lands on one of these numbers must leave the circle.

With this in mind, let's start counting. Initially, we have all $6$ people, starting with $A$(Arn). After everyone says a number, $A$ must say $7,$ so he leaves the circle.

The circle now has $5$ members: $B$ (Bob), $C$ (Cyd), $D$ (Dan), $E$ (Eve), and $F$ (Fon) -- with $B$ restarting his counting at $8.$ Everyone in the circle counts without incident, and it loops around such that Bob says $13.$ However, this leaves Cyd to say $14,$ and he leaves the circle.

The circle now has $4$ members: $B,D,E,F$ -- with $D$ continuing the counting at $15.$ $E$ says $16,$ and $F$ says $17,$ and therefore leaves the circle.

The circle now has $3$ members: $B,D,E$ -- with $B$ continuing the counting at $18.$ The counting loops around, and $B$ says $21,$ and therefore leaves the circle.

The circle now has $2$ members: $D,E$ -- with $D$ starting at $22.$ They go back and forth, with $D$ saying even numbers and $E$ saying odd numbers. As such, eventually, $E$ must say $27,$ and as such, leaves the circle. This makes $D$ -- Dan -- the last one left in the circle.

Thus, D is the correct answer.

To solve for the area of the figure, we separate the compound shape into parts that are easier to work with, as such:

As is now clear, there is the center $3 \times 3$ square, with $4$ smaller shaded triangles surrounding it.

The area of the square is $3 \cdot 3 = 9.$ The other triangles each have a base of $2$ and a height of $1,$ so their area is equal to $$\dfrac{bh}{2} = \dfrac{2\cdot 1}{2} =1 .$$ There are $4$ of these triangles, so their total area is $1\cdot 4 = 4.$

Therefore, the total area is $9+4 = 13.$

Thus, the correct answer is C.

Rearranging the terms, notice that the expression in the question is equal to: $$\begin{align*}&1 + (3-2) + (5-4) + \cdots +\\ &(2017-2016) + (2019-2018). \end{align*}$$ Each term is equal to $1,$ and there are $\frac{2019-1}2+1 = 1010$ terms, so the total sum is $1010\cdot1 = 1010.$

Thus, E is the correct answer.

Anh drove $10$ miles on the coastal road in $30$ minutes. Therefore, his speed on the coastal road (notated as $v_c$) is $$\begin{align*}v_c&=\dfrac{10}{30}\\ &=\dfrac13.\end{align*}$$ This is $\dfrac13$ mile per minute. Since he drives $3$ times as fast on the highway (i.e. $v_h=3v_c$), his highway speed is $$3\cdot \dfrac13 = 1$$ mile per minute. Armed with these two facts, we know that Anh drove for $30$ minutes on the coastal road, and he drove $50$ miles at $1$ mile per minute. This means it takes $50$ minutes to drive the $50$ miles on the highway.

As such, the total travel time is $50+30=80$ minutes.

Thus, the correct answer is C.

Notice that a number is divisible by $9$ if and only if the sum of its digits is also divisible by $9.$

The sum of the digits of the 5-digit number in the problem is: $$2+0+1+8+U= 11+U.$$ As $\underline{2}$ $\underline{0}$ $\underline{1}$ $\underline{8}$ $\underline{U}$ is divisible by $9,$ $11+U$ must also be divisible by $9.$ Also, as $U$ is a digit, we know that $0\le U\le 9.$ This means that $U$ can only be $7.$

Now we know that the 5-digit number in question is 20187, and we want to find the remainder when we divide 20187 by 8. To solve this, simply use long division to see that $20187=2523\cdot 8 + 3.$ Therefore, the remainder is $3.$

Thus, the correct answer is B.

Counting the number of each occurrence, we can see that there are 1 $1$s, 3 $2$s, 2 $3$s, 6 $4$s, 8 $5$s, 3 $6$s, and 2 $7$s.

Therefore, there are $1+3+2+6+8+3+2 = 25$ students in total.

Therefore, the total number of days of exercise is $$\begin{align*}&1\cdot1+3\cdot2 + 2\cdot 3 + 6\cdot 4 \\ &+ 8\cdot 5 + 3\cdot 6 + 2\cdot 7 \\ &= 109\end{align*}$$

The mean number of days of exercise is the total number of days divided by the number of students: $$\dfrac{109}{25} \approx 4.36$$

Thus, C is the correct answer.

Note that each square foot of the border would require one tile, meaning that the border will take $16+12+16+12=56$ tiles. However, notice that this will cause overlapping tiles in each of the four corners, so to fix this, we subtract $4.$ Therefore, the border will take $56-4=52$ $1\times 1$ square tiles to completely tile.

Since we have removed one foot from each side due to the border, the remaining rectangle is $10$ feet by $14$ feet. This must be tiled completely by $2 \times 2$ tiles, so it will take $$\dfrac{10\cdot14}{2\cdot2} = 35$$ tiles in total to tile this area.

As it takes $52$ $1\times 1$ square tiles to tile the border, and $35$ $2\times 2$ square tiles to tile the remaining area, it will take $52+35=87$ tiles in total to fill in Tyler's entire living room floor.

Thus, the correct answer is B.

The reciprocals of $1$, $2$, and $4$ are $\dfrac11$, $\dfrac12$, and $\dfrac14$, respectively. The average of these reciprocals is $$\begin{align*}\dfrac{\left(1 + \dfrac12 + \dfrac14\right)}{3} &= \dfrac{\left(\dfrac{7}{4}\right)}{3} \\&= \dfrac{7}{12}. \end{align*}$$

As the harmonic mean is the reciprocal of the average of the reciprocals of the numbers (which we just calculated to be $\dfrac{7}{12}$), we conclude that the harmonic mean is $\dfrac{12}{7}.$

Thus, the correct answer is C.

We can split the problem into two cases. In case 1, Abby is in one of the middle two seats, and in case 2, she is in one of the outer 4 seats.

Firstly notice that there is a $\dfrac26 = \dfrac13$ probability of case 1 being true (i.e. Abby is in the middle two seats). For Bridget to be adjacent to Abby in this case, she must be in either of the two seats on the left or the two seats on the right of Abby, or she is in the same column as her. There are $3$ ways to make this happen out of a possible $5$ open seats, so there is a $\frac35$ chance of this happening. Therefore, the total probability of this case is $$\dfrac13 \cdot \dfrac35 = \dfrac{3}{15} .$$

Next, notice that there is a $\dfrac46 = \dfrac23$ probability of case 2 being true (i.e. Abby is in the outer four seats). For Bridget to be adjacent to Abby in this case, she must either be in the single seat next to Abby in the same row, or she is in the same column as Abby. There are $2$ ways to make this happen out of a possible $5$ open seats, so there is a $\frac25$ chance of this happening. Therefore, the total probability of this case is $$\frac23 \cdot \frac25 = \frac{4}{15} .$$

Therefore, the final probability of either of these cases happening is $\dfrac{3}{15} + \dfrac{4}{15} = \dfrac{7}{15} .$

Thus, C is the correct answer.

Starting from $12:00$ noon, after $30$ minutes of time elapsed, the car clock went $35$ minutes ahead.

Therefore, for every minute the car clock goes ahead, $\frac {30}{35} = \frac 67$ minutes of actual time pass by. From the time $12:00$ to $7:00,$ the car clock goes ahead $7\cdot 60 = 420$ minutes, and therefore, $420 \cdot \frac67 = 360$ minutes, or $6$ hours, of actual time have passed by. If we start at $12:00$ and $6$ hours pass by, the time is $6:00 .$

Thus, B is the correct answer.

Since the average score on the five tests is $82,$ the total score of those five tests must be $5\cdot82 = 410 .$

Now, let $f$ be the score on the first 4 tests and let $l$ be the score for the last test.

We know that $$f < l \leq 100$$ and $$4f + l = 410.$$ And as $410 = 4f + l < 5l ,$ we know $\frac{410}{5} = 82 < l .$

Also, since $4f + l = 410 ,$ and dividing $410$ by $4$ gives us a remainder of 2, we know that dividing $l$ by $4$ must leave a remainder of $2$ as $4f$ will leave no remainder when divided by $4.$ Equivalently: $$l \equiv 2 \mod 4 .$$ Since $82 < l \leq 100$ and $l \equiv 2 \mod 4,$ the only options for $l$ are $86,90,94,98.$ This yields four distinct solutions as follows: $$\begin{align*} (f,l) =& (81,86);\\ &(80,90);\\ &(79,94);\\ &(78,98) \end{align*}$$ Therefore, there are $4$ solutions, and A is the correct answer.

To make the largest possible $5$ digit number, we must maximize the first digit (the digit in the ten-thousands place).

The largest number that is strictly less than $10$ and divides $120$ is $8,$ so the first digit must be $8.$ Therefore, the product of the remaining number is $15.$

Similarly, we must now maximize the second digit.

The largest number that is less than $10$ and divides $15$ is $5,$ so the second digit is $5.$ Therefore, the product of the remaining number is $3.$

We must then maximize the third digit.

The largest number that is less than $10$ and divides $3$ is $3,$ so the third digit is $3.$ Therefore, the product of the remaining number is $1.$ This means the 4th and 5th digits are $1.$

This makes $N = 85311,$ so the sum of the digits is $$8+5+3+1+1=18$$

Thus, D is the correct answer.

Let $A$ be the area of the large circle.

Since the diameter of each of the two smaller circles is itself the radius of the larger circle, the radius of each smaller circle is half that of the larger circle.

Symbolically, if we allow $r_l$ to be the radius of the large circle and $r_s$ to be the radius of each of the smaller circles: $$r_s = \dfrac12 r_l$$ As the area of the larger circle is equal to $A=\pi r_l^2,$ the area of the smaller circles are equal to $$\begin{align*}\pi r_s^2 &= \pi \left(\dfrac12 r_l\right)^2 \\&= \dfrac14 (\pi r_l^2)\\&=\dfrac14 A.\end{align*}$$ As the area of two of these smaller circles combined is equal to 1 square unit, then it follows that $2\cdot \dfrac14 A=1$ square unit, implying that $A=2$ square units.

As the area of the shaded region is equal to the area of the larger circle $(A)$ minus the combined area of the two smaller circles $(1),$ the area of the shaded region is $A - 1=2-1=1$ square unit.

Thus, the correct answer is D

Since we are keeping the Arabic books together and the Spanish books together, we can look at each group as a single block.

As such, there are 5 objects on the bookshelf: three German books, one collection of Arabic books, and one collection of Spanish books. There are $5!$ ways to order the $5$ objects. As we already have the books together, there are $2!$ ways of ordering the Arabic books and $4!$ ways of ordering the Spanish books. Therefore, the total ways to order the books is $$\begin{align*}5! \cdot 4! \cdot 2! &= 120 \cdot 24 \cdot 2 \\&= 5760 \end{align*}$$

Thus, the correct answer is C.

Since for every foot Bella walks, Ella rides 5 feet, we know that Bella will walk $\frac16$ of the distance between the two houses, and so she walks $$\dfrac16 \cdot 10560=1760$$ feet. Since she walks $2.5$ feet per step, she takes $$\dfrac{1760}{2.5} = 704$$ steps by the time she meets Ella.

Thus, A is the correct answer.

Begin by finding the prime factorization of $23232.$ To do this, we repeatedly factor out the smallest prime factor from the number, a process that terminates when the number is a prime number. This process is outlined below: $$\begin{align*}23232 &= 2\cdot 11616\\ &= 2^2\cdot 5808\\ &= 2^3\cdot 2904\\ &= 2^4\cdot 1452\\ &= 2^5\cdot 726\\ &= 2^6\cdot 363\\ &= 2^6\cdot 3 \cdot 121\\ &=2^6\cdot 3\cdot 11^2 \end{align*}$$ An arbitrary factor of $23232$ can be created by taking the product of any number of prime factors. More explicitly, as $23232$ can be represented $p_1^{e_1} p_2^{e_2} \cdots p_m^{e_m}$ where $p$ is a prime number, each factor has $(e_1+1) \cdots (e_m+1)$ options of prime factorizations to choose from, and thus, there are $(e_1+1) \cdots (e_m+1)$ factors. Plugging in values, we can see that there are $$\begin{align*}(6+1)(1+1)(2+1) &= 7\cdot2\cdot3\\ &= 42 \end{align*}$$ factors of $23232.$