Solution:

We only need to consider the ones digits of each number (except for the first one so we avoid getting a negative answer): $$22 - 2 - 2 - 2 - 2 - 2 = 12$$ which has a ones digit of $2$.

Thus, B is the correct answer.

Solution:

We can simplify the fractions by taking out the common factor $11$: $\dfrac{44}{11}$ simplifies to $4$, $\dfrac{110}{44}$ simplifies to $\dfrac{10}{4} = \dfrac{5}{2} = 2.5$, and $\dfrac{44}{1100}$ simplifies to $\dfrac{4}{100} = 0.04$. Therefore, we have $4 + 2.5 + 0.04 = 6.54.$

Thus, C is the correct answer.

Solution:

The visible colored region includes the area of the square with side length $10$ subtracted by the area of the square with side length $9$ plus the area of the square with side length $7$ minus the area of the square with side length $4$. Hence, we get $100 - 81 + 49 - 16 = 52.$

Thus, E is the correct answer.

Solution:

To find the number that Yunji left out, we need to find the sum of the integers from $1$ to $9$ and find its difference with the largest perfect square below the sum. We can calculate the sum of the integers from $1$ to $9$ as follows: $$1 + \ldots + 9 = \dfrac{9(9+1)}{2} = 45$$ The largest perfect square less than $45$ would be $36$ and $45 - 36 = 9$.

Thus, E is the correct answer.

Solution:

can draw a 2x2 chart of possible rolls

We can simply list down all possible combinations of the two rolls: $$(1,6), (2,3), (2,6), (3,6),$$ $$(4,6), (5,6), (6,6)$$ Respectively, the sums would be $7, 5, 8, 9, 10, 11, 12$. Among the choices, only $6$ is not a possible sum.

Thus, B is the correct answer.

Solution:

From inspection, we can notice that Path R is the shortest path since it avoids going through the entire loop by taking straight-line shortcuts instead of going through the arc portions of the rink. This leaves us with two possible answer choices, D and E, which only differ in how they sort Path P and S.

To determine which path is longer between Path P and S, notice that the difference between the two paths can be related by a right triangle as shown here:

The hypotenuse of the triangle was traversed by Sergei in Path S while one of the legs is part of Path P. By the Pythagorean Theorem, we know that the hypotenuse will always be longer than the legs, and so Path S will be longer than Path P.

Thus, B is the correct answer.

Solution:

Notice that if we only fill the $3 \times 7$ rectangle using $2 \times 2$ and $1 \times 4$ tiles, then we will always be filling the rectangle in multiples of $4.$ From the answer choices, it will suffice to consider the cases where we are able to fill $16$ or $20$ tiles out of the $21$ tiles of the rectangle, since the former will leave us $5$ spaces for the $1 \times 1$ tiles while the latter will leave us with one. The other options are not possible since those numbers of tiles cannot be arrived by subtracting any multiple of $4$ from $21$.

By attempting to place the $2 \times 2$ and $1 \times 4$ tiles, we can immediately notice that it's not possible to fill $20$ tiles in the rectangle, and we can easily find cases where we are left with five tiles.

Thus, E is the correct answer.

Solution:

We can systematically exhaust the cases using a tree diagram where the upper branch covers the case where the amount gains $3 while the other branch represents the case where the amount doubles.

From the diagram, we can see that we end up with six unique dollar amounts after three days.

Thus, D is the correct answer.

Solution:

We can let $r$ be the number of red marbles that Maria has. Since Maria has half as many red marbles as green, then we know that she has $2r$ green marbles. Moreover, since she has twice as many blue marbles as green, then she will have $2(2r) = 4r$ blue marbles. Adding these together gives us $r + 2r + 4r = 7r,$ and so the answer must be a multiple of $7.$ Among the answer choices, only $28$ is a multiple of $7.$

Thus, E is the correct answer.

Solution:

There are $50$ years between $1980$ and $2030$, so we can expect the CO₂ reading to increase by $50 \times 1.515 = 75.75 \approx 76$ ppm by $2030$. Since the CO₂ reading in $1980$ was $338$ ppm, then we will have $338 + 76 = 414$ ppm by $2030$.

Thus, B is the correct answer.

Solution:

Consider the base of the triangle to be $\overline{\rm AB}$ which has length $11-5=6$. Given that the area of the triangle is $12$, its height must be of length $\dfrac{2(12)}{6}=4$. Since $y>7$, then $y$ must be $7+4=11$.

Thus, D is the correct answer.

Solution:

Let $x$ be the number of guppies in the 1st tank. Hence, there are $x+1$ guppies in the 2nd tank, $x+3$ guppies in the 3rd tank, and $x+6$ guppies in the 4th tank. We then use the fact that there are a total of 90 guppies in the 4 tanks to find $x$: $$x + x + 1 + x + 3 + x + 6 = 90$$ $$4x+10=90$$ $$x=20.$$

Note that we are not yet done since we are asked for the number of guppies in the 4th tank and not the 1st. There are $x+6=20+6=26$ guppies in the 4th tank.

Thus, E is the correct answer.

Solution:

We can deduce from the choices that it is possible to exhaust all possible cases for this problem. Note that all sequences must start with up $(U)$ and end with down $(D)$, and that it should not be possible to go down more times than Buzz has gone up so far. Keeping this in mind, we can arrive at the following possible cases: $$UUUDDD$$ $$UUDUDD$$ $$UUDDUD$$ $$UDUDUD$$ $$UDUUDD$$ which is a total of five possible sequences.

Thus, B is the correct answer.

Solution:

A systematic way of tracking the shortest overall distance to $Z$ is to consider the shortest distance to get to each town from $A$. For instance, the shortest distance to get to town $X$ from $A$ is $5$ km, trivially.

Then, for town $M$, going to town $X$ first will be shorter compared to going directly from $A$, so the shortest path to town $M$ has a length of $7$ km.

For town $Y$, it will take us $15$ km if we come from town $X$ and only $13$ km coming from $M,$ so $13$ km is the length of shortest path to $Y$ from $A$.

Doing the same for town $C$ will give us $18$ km as the shortest distance by coming from town $Y$.

Finally, for town $Z$, we can either come from town $Y,$ $C,$ or $M$. The total distance if we come from each three towns respectively would be $30,$ $28,$ and $32$. Hence, $28$ km is the shortest distance from $A$ to $Z$.

The diagram below summarizes our process. Here, the town name is replaced by the shortest distance to get to that town from town $A$.

Thus, A is the correct answer.

Solution:

Firstly, note that $$\text{FLYFLY} = 1001(\text{FLY})$$ and, similarly, $$\text{BUGBUG}=1001(\text{BUG})$$ so the equation can be simplified to $8 \cdot \text{FLY} = \text{BUG}.$

For $\text{BUG}$ to remain three digits, $F$ must be $1.$ Moreover, $L$ must also be less than $3$ to avoid carrying over $2$ to the hundreds digit and making the product $4$ digits. Since we need $\text{FLY}$ to be the greatest number, $L$ must be $2.$

To identify the possible values for $Y,$ we note that so far we have $8(12)=96$, so we must avoid carrying $4$ to the tens digit to keep the resulting product three digits. Hence, $Y\le5$. We can try $4$ and verify that the resulting product has unique digits that haven't been used yet: $8(124) = 992$, which does not have unique digits. Trying $3,$ we get $8(123) = 984$, which satisfies our criteria.

Hence, $$\text{FLY} + \text{BUG} = 123 + 984 = 1107.$$

Thus, C is the correct answer.

Solution:

We first need to note that there are $27$ multiples of $3$ from $1$ through $81$. If any row or column has a multiple of $3$, then the product of the numbers in the row or column will be divisible by $3$. From this, it is evident that we must try to place all $27$ multiples of $3$ together in some corner of the grid so the least number of products will be divisible by $3$.

We can place $25$ multiples of $3$ in a $5 \times 5$ grid so only $5$ rows and $5$ columns will have products that are divisible by $3$. The remaining $2$ multiples of $3$ can then be placed in the 6th column along the 1st and 2nd row of the $5 \times 5$ grid so in total we will only have $6$ columns and $5$ rows that have products that are divisible by $3$, for a total of $6+5=11$ products divisible by $3$.

The diagram below illustrates our process. The highlighted cells contain the multiples of $3$ while the dotted lines mark the rows or columns that have products which are divisible by $3.$

Thus, D is the correct answer.

Solution:

Firstly, we note that a king cannot be on the center square as it will attack any other piece on the $3 \times 3$ grid. To solve this problem, we can simply consider some possible cases: one where a king is in the corner and another when one king is on an edge (but not a corner).

When one king is placed in any corner, then the other king can be placed in $5$ other squares without them attacking each other.

Since there are four corners, we have $5 \times 4 =20$ possibilities for this case.

Another possible case is when one king is on an edge that is not a corner.

In this scenario, the other king can be in $3$ other squares without the two kings attacking each other. There are $4$ edges in the grid so we have $4(3)=12$ possibilities for this case.

Adding the number of possibilities, we get $20+12=32$ total number of ways.

Thus, E is the correct answer.

Solution:

Let $\theta$ be the measure of $\angle BOC.$

One component of the shaded region is the area of the circle with radius $2$ minus the area of the circle with radius $1.$ This part has area $4\pi-\pi=3\pi.$ The remaining area is a sector of the biggest circle minus the area of the circle with radius $2$. This has area $\dfrac{\theta}{360}(9\pi-4\pi) = \dfrac{\theta}{360}(5\pi).$ Hence, the total area of the shaded region is $3\pi + \dfrac{\theta}{360}(5\pi).$

Next, we note that the unshaded region is composed of the smallest circle and the unshaded portion of the outer ring. This will have a total area of $\pi + \dfrac{360-\theta}{360}(5\pi)$

Lastly, we equate the area of both regions and solve for $\theta:$ $$3\pi + \dfrac{\theta}{360}(5\pi) = \pi + \dfrac{360-\theta}{360}(5\pi)$$ $$2\pi = \dfrac{360-\theta-\theta}{360}(5\pi)$$ $$\dfrac{2}{5} = 1 - \dfrac{2\theta}{360}$$ $$2\theta = \dfrac{3}{5}(360)$$ $$\theta = 108.$$

Thus, A is the correct answer.

Solution:

Jordan has $\dfrac{3}{5} \times 15 = 9$ pairs of red sneakers and $6$ pairs of white sneakers. Moreover, $\dfrac{2}{3} \times 15 = 10$ are high-top and $5$ are low-top. If we want to minimize the number of red high-top sneakers, then we can set all $6$ white sneakers to be high-top, leaving $10-6=4$ red sneakers as high-top. Hence, the fraction of red high-top sneakers would be $\dfrac{4}{15}$.

Thus, C is the correct answer.

Solution:

We first note that we can only form equilateral triangles if we go through the diagonals of the square faces, otherwise at least one angle of the triangle will be different. Afterwards, it is easy to exhaust all possible equilateral triangles that can be formed: $\triangle PVT,$ $\triangle PRT,$ and $\triangle PRV.$

Thus, D is the correct answer.

Solution:

We can let $g$ be the number of green frogs and $y$ be the number of yellow frogs. Initially, we have $g=3y$. Then, after some frogs moved, we have the following proportion: $$\dfrac{g+5-3}{y+3-5} = \dfrac{4}{1}.$$

Substituting $3y$ for $g$ will allow us to determine the number of yellow frogs originally ($y$): $$\dfrac{3y+2}{y-2} = 4$$ $$3y+2 = 4(y-2) = 4y-8$$ $$y = 10$$

Hence, there were $10$ yellow frogs and $3 \times 10 = 30$ green frogs initially. After some frogs moved, we now have $10+3-5 = 8$ yellow frogs and $30+5-3=32$ green frogs, giving us a difference of $32-8=24$ between the number of green and yellow frogs.

Thus, E is the correct answer.

Solution:

We have a huge margin of error for this problem so we can freely estimate values. Firstly, we note that the entire roll of tape is 1-inch thick, and given that a single tape is $0.015$ inches thick, then there are $\dfrac{1}{0.015}= \dfrac{2}{0.03} = \dfrac{200}{3}$ layers of tape in the entire roll.

Then, we must identify an estimate for the circumference of one layer of tape. Near the center, one layer of tape will have a circumference of $2\pi$ while the layers near the outer section will have a circumference of about $4\pi$. A good estimate would be to take the average circumference which is $3\pi$. With this, we can estimate the total length for the entire roll: $$\dfrac{200}{3}(3\pi) = 200\pi \approx 200(3) \approx 600.$$

Thus, B is the correct answer.

Solution:

From the given example, observe that if we consider a different line segment with the same slope, for instance the line connecting points $(0,2)$ and $(1,0)$, then the number of colored cells will be halved.

In general, it is possible to scale down the problem as long as we still have the same slope for the line.

Next, note that the line segment passing through points $(2000, 3000)$ and $(5000, 8000)$ has a slope of $\dfrac{5}{3}$. Hence, a scaled down version that we can consider is a segment connecting the points $(0,0)$ and $(3,5).$

From the diagram, it is evident that the line passes through $7$ cells. We know that this will happen $1000$ times as the segment passes through points $(2000, 3000)$ and $(5000, 8000)$. Hence, Rodrigo will need to color $7(1000) = 7000$ cells.

Thus, C is the correct answer.

Solution:

Observe that the we can extend the two mountains at the intersection point to make three triangles as shown in the diagram below.

All triangles formed are $45^\circ\text{-}45^\circ\text{-}90^\circ$ triangles and so based on the given height information, we can determine that the side lengths of the triangles, from left to right, are $8\sqrt{2}, h\sqrt{2},$ and $12\sqrt{2}$. Hence, we have areas of $64, h^2,$ and $144$, respectively.

To find $h$, we use the total area of the diagram which we can obtain by adding the area of the two large triangles and subtracting the area of the triangle with height $h$ since this area was counted twice. Doing this, we get $$64+144-h^2=183$$ $$h^2 = 144+64-183 = 25$$ $$h=5$$

Thus, B is the correct answer.

Solution:

For simplicity, we can disregard the order in which passengers are seated so we only consider the number of ways that the $12$ seats can be filled by $8$ passengers for the total number of possibilities, which is given by ${12 \choose 8} = 495.$

Out of these $495$ possibilities, we consider the number of ways where no adjacent seats are available, then subtract this from $495$. This scenario can happen when two passengers are occupying the edge seats of one row or one passenger is seated in the middle seat of a row. Hence, we consider the following cases:

• No passenger is in the middle seat (all $8$ passengers are on the edge seats): ${8 \choose 8} = 1$ way.

• Exactly $1$ passenger is in the middle seat ($7$ are seated on the edge seats): There are ${4 \choose 1} = 4$ ways where one row can contain a passenger in the middle seat and ${2 \choose 1} = 2$ ways for the eighth passenger to be seated on the row where the one passenger is sat on the middle seat. Hence, the total for this case is $4(2) = 8$ ways.

• Exactly $2$ rows have a passenger in the middle seat: ${4 \choose 2} = 6$ ways to select rows with occupied middle seat and another ${4 \choose 2} = 6$ ways for the remaining $2$ passengers to be seated in the rows with occupied middle seats. Thus, this case has a total of $6(6) = 36$ possibilities.

• Exactly $3$ rows have occupied middle seats: ${4 \choose 3} = 4$ ways to select rows with occupied middle seats and ${6 \choose 3} = 20$ ways for the remaining $3$ passengers to be seated on the rows with occupied middle seats, for a total of $4(20) = 80$ ways.

• All $4$ rows have occupied middle seats: ${8 \choose 4} = 70$ ways for the remaining $4$ passengers to be seated.

Adding all the possibilities in each case, we get $1+8+36+80+70 = 195$ ways for there to be no two adjacent seats available. Hence, the probability that the couple will be seated together would be $\dfrac{495-195}{495} = \dfrac{20}{33} .$

Thus, C is the correct answer.