Small Hint:

A $1:20$ scale means the replica height is the real height divided by $20$.

Big Hint:

After dividing $289$ by $20$, round to the nearest whole foot.

Video solution:

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Written solution:

The replica is $\dfrac{289}{20}=14.45$ feet tall, which rounds to $14$ feet.

Thus, the correct answer is A.

Small Hint:

Rewrite each factor $1+\frac1n$ as $\frac{n+1}{n}$.

Big Hint:

The product telescopes after the fractions are rewritten.

Video solution:

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Written solution:

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.

Small Hint:

Simulate only the numbers that cause someone to leave.

Big Hint:

Keep the counting order after each person leaves; the next number goes to the next person still in the circle.

Video solution:

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Written solution:

The first five removal numbers are $7,14,17,21,27.$ Tracking only those turns gives:

Arn says $7$, Cyd says $14$, Fon says $17$, Bob says $21$, and Eve says $27$.

Dan is the only student remaining, so D is the correct answer.

Small Hint:

Enclose the figure in an easy rectangle or split it into a central square and small triangles.

Big Hint:

The slanted parts form four congruent right triangles of area $1$.

Video solution:

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Written solution:

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.

Small Hint:

Pair each positive odd number after $1$ with the even number just before it.

Big Hint:

Count how many $1$’s remain after pairing.

Video solution:

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Written solution:

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.

Small Hint:

Use the coastal-road information to find Anh’s coastal-road speed.

Big Hint:

The highway speed is three times the coastal-road speed, so the highway time follows from the $50$ highway miles.

Video solution:

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Written solution:

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.

Small Hint:

Use the divisibility-by-$9$ rule to determine $U$.

Big Hint:

Once $U$ is known, only the last three digits matter for the remainder modulo $8$.

Video solution:

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Written solution:

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.

Small Hint:

Compute the weighted average from the bar heights.

Big Hint:

First find the total number of students, then find the total number of reported exercise days.

Video solution:

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Written solution:

The bar heights for $1,2,3,4,5,6,7$ days are $1,3,2,6,8,3,2$, for a total of $25$ students.

The total number of reported exercise days is $$\begin{aligned} &1\cdot1+2\cdot3+3\cdot2+4\cdot6\\ &\qquad+5\cdot8+6\cdot3+7\cdot2=109. \end{aligned}$$

The mean is $\frac{109}{25}=4.36$ days.

Thus, C is the correct answer.

Small Hint:

Count the $1$-foot border tiles first, being careful not to double-count the corners.

Big Hint:

After removing a $1$-foot border, the remaining rectangle is $10$ feet by $14$ feet.

Video solution:

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Written solution:

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.

Small Hint:

Average the reciprocals of $1,2,$ and $4$.

Big Hint:

The harmonic mean is the reciprocal of that average.

Video solution:

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Written solution:

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.

Small Hint:

First choose Abby’s seat, then count how many seats are adjacent to it.

Big Hint:

Middle seats and corner seats have different numbers of adjacent seats.

Video solution:

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Written solution:

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 beside Abby in the same row, 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.

Small Hint:

Compare how much time the car clock gains to how much real time passes.

Big Hint:

The car clock’s $35$ minutes correspond to $30$ actual minutes.

Video solution:

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Written solution:

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.

Small Hint:

Let the repeated score be $f$ and the last test score be $l$.

Big Hint:

Use $4f+l=410$ and the fact that $l$ is larger than the average.

Video solution:

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Written solution:

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.

Small Hint:

To maximize the five-digit number, maximize the digits from left to right.

Big Hint:

Each chosen digit must divide the remaining required product.

Video solution:

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Written solution:

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 $4$th and $5$th 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.

Small Hint:

Compare the radius of a smaller circle to the radius of the larger circle.

Big Hint:

Area scales by the square of the scale factor.

Video solution:

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Written solution:

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

Small Hint:

Treat the two Arabic books as one block and the four Spanish books as one block.

Big Hint:

Arrange the five objects, then arrange the books within the Arabic and Spanish blocks.

Video solution:

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Written solution:

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.

Small Hint:

Since Ella is $5$ times as fast as Bella, split the distance in the ratio $1:5$.

Big Hint:

After finding Bella’s walking distance in feet, divide by $2\frac12$ feet per step.

Video solution:

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Written solution:

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.

Small Hint:

Prime-factorize $23,232$.

Big Hint:

If $n=p^a q^b r^c$, then its number of positive factors is $(a+1)(b+1)(c+1)$.

Video solution:

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Written solution:

The prime factorization is $23{,}232=2^6\cdot3\cdot11^2.$ A divisor may use any exponent from $0$ through $6$ on $2$, from $0$ through $1$ on $3$, and from $0$ through $2$ on $11$. Therefore, the number of positive divisors is $(6+1)(1+1)(2+1)=42.$

Thus, E is the correct answer.

Small Hint:

If you know one lower cell and the cell above a pair, the other lower cell is forced.

Big Hint:

Work downward from the top, counting how many choices appear at each new row.

Video solution:

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Written solution:

Suppose we have two cells and the cell above them. If we are given the bottom left cell and the top cell, we can always find the bottom right cell as follows:

If the top cell is $+,$ then the bottom right cell must be the same as the bottom left cell, and if the top cell is $-,$ the bottom right cell must be the opposite of the bottom left cell.

Now, suppose we are given a row. Then, suppose we choose a value for the cell below and to the left of the leftmost cell in our given row. We then can inductively determine the entire row below our given by first finding the bottom-right cell of the leftmost cell in our row, and using that newly found cell as the bottom-left reference for the second to the left cell in the given row to find its bottom-right counterpart. The process continues on until the row below the given row is fully solved.

Therefore, since we know that the top row has a cell labelled $+,$ we have $2$ choices for the row below, depending on our choice of the bottom-left cell. Similarly, we have $2$ choices for the third row, and thus $2$ choices for the fourth row. This makes $2\cdot2\cdot2=8$ total choices for the bottom row of the sign pyramid.

Thus, the correct answer is C.