Answer: D

Solution:

Let's first notice that when we subtract a negative number, it is the same as adding it's positive counterpart. In other words: $$a-(-b) = a+b$$ With that in mind, we can rewrite the above expression as: $$1+2-3+4-5+6$$ Which can be solved to yield $5$ as our final answer.

Thus, D is the correct answer.

Answer: E

Solution:

Recall that a cube with a side length of $a$ has a volume of $a^3.$ With this in mind, Carl's cubes (each with side length $1$) have a volume of $1^3=1$ each. Therefore, all of Carl's $5$ cubes will have a total volume of $5\cdot 1=5.$

Similarly, Kate's cubes (each with side length $2$) have a volume of $2^3=8$ each. Therefore, all of Kates's $5$ cubes will have a total volume of $5\cdot 8=40.$

Therefore, the total volume of these 10 cubes is $5+40=45.$

Thus, the correct answer is E.

Answer: E

Solution:

Let's begin by restating the following: $$w:x = \dfrac{w}{x}=\dfrac43=4:3$$ $$y:z = \dfrac{y}{z}=\dfrac32=3:2$$ $$z:x = \dfrac{z}{x}=\dfrac16=1:6$$ Armed with these three equations, we want to find $w:y= \dfrac wy.$ To do this, we must try and represent $\dfrac wy$ using $\dfrac wx, \dfrac yz,$ and $\dfrac zy.$

Notice that we can represent $\dfrac wy$ as $\dfrac wz \cdot \dfrac zy.$

Further notice that $\dfrac wz = \dfrac wx \cdot \dfrac xz.$ Therefore, combining these two facts shows us that: $$\begin{align*}\dfrac wy &= \dfrac wx \cdot \dfrac xz \cdot \dfrac zy\\ &= \dfrac wx \cdot \dfrac {1}{(\dfrac zx)} \cdot \dfrac {1}{(\dfrac yz)} \\ &= \dfrac 43 \cdot \dfrac{1}{(\dfrac 16)} \cdot \dfrac{1}{(\dfrac 32)} \\ &= \dfrac 43 \cdot \dfrac 61 \cdot \dfrac 23\\ &= \dfrac{48}{9}\\ &= \dfrac{16}{3} \end{align*}$$ As such: $$w:y = \dfrac wy = \dfrac {16}{3} = 16:3$$ Thus, E is the correct answer.

Answer: D

Solution:

We know that the interior angles of a triangle add up to $180^{\circ},$ and since the triangle in question is a right triangle, by definition one of the interior angles must measure $90^{\circ}.$ The remaining two acute angles, $a^{\circ}$ and $b^{\circ},$ must therefore have a sum of $180^{\circ}-90^{\circ}=90^{\circ}.$

Let's begin by exploring the largest values of $a^{\circ}$ and going from there, as those will naturally yield the smallest values of $b^{\circ}.$

The greatest possible value of $a^{\circ}$ is $89^{\circ}.$ This makes $b^{\circ}=90^{\circ}-89^{\circ}=1^{\circ},$ which is not prime, so this is not a valid possible case.

Moving on, the next largest possible value of $a^{\circ}$ is $83^{\circ}.$ This makes $b^{\circ}=90^{\circ}-83^{\circ}=7^{\circ},$ which is prime! Therefore, this is the smallest possible value of $b.$

Thus, D is the correct answer.

Answer: B

Solution:

We have $1+1+2+3=7$ total tiles, and as such, there are $7!$ total ways to order them. However, notice that this is overcounting, as it assumes that all tiles are indistinguishable, when in reality, the yellow and green tiles can be reordered amongst themselves without any change to the overall tiling pattern.

Let's remove this overcounting as follows. There are $3$ yellow tiles, and as such, there are $3!=6$ ways to switch them around amongst themselves such that the overall tiling pattern is unchanged, as they are indistinguishable.

Similarly, there are $2$ green tiles, and as such, there are $2!=2$ ways to switch them around amongst themselves such that the overall tiling pattern is unchanged, as they are indistinguishable.

Therefore, there are $$\begin{align*}\dfrac{7!}{6\cdot 2}&=7\cdot 5\cdot 4\cdot 3 \\ &= 7\cdot 60 \\ &= 420\end{align*}$$ possible tilings.

Thus, B is the correct answer.

Answer: B

Solution:

We want to find the smallest palindrome larger than $15951,$ and to do so, we want to increase the value of the middle digit. This is because if we increased the value of any other digit with less place value (say, the ones place), we would also need to increase the value of the digit with greater place value (i.e. ten-thousands place).

Therefore, we must replace the number $9$ to the next greatest integer: $10.$ As all digits must be between $0$ and $9$ (inclusive), we write a zero in place of the nine, and increase the thousands digit (and the tens — to preserve the palindrome).

As such, the next smallest palindrome larger than $15951$ is $16061.$

As Megan drives $16061-15951$$=110$ miles in $2$ hours, she drives at an average of $55$ mph.

Thus, B is the correct answer.

Answer: A

Solution:

If a number is even, it must be a multiple of $2,$ and as such, if a number is a positive even multiple of $3,$ it is a multiple of $3$ and $2,$ meaning that it is a multiple of $6.$

As such, we can write arbitrary positive multiples of six as $6k,$ for $k ≥ 0.$

We want to find the number of positive even multiples of $3$ — or more simply, positive integers of the form $6k$ — that are also perfect squares. Therefore, we are looking for integers less than $2020$ of the form: $\left(6k\right)^2.$

Notice that for $$k=8, (6k)^2 = 2304 ≥ 2020,$$ and for $$k=7, (6k)^2 = 1764 \le 2020.$$ This means that the maximum $k$ we can have is $7,$ and as $k > 0$ all $k: 1\le k\le 7$ are valid solutions.

As such, there are $7$ solutions.

Thus, A is the correct answer.

Answer: D

Solution:

We know that the area of $\triangle PQR,$ denoted $A(\triangle PQR),$ is equal to $\dfrac 12 \cdot PQ \cdot h_R,$ where $h_R$ is equal to the height of $\triangle PQR.$

As $A(\triangle PQR) = 12$ and $PQ=8,$ we can see that $h_R = 3.$

WLOG, allow $P=(-4,0)$ and $Q=(4,0).$ With this in mind, let's case on the position of the right angle:

$\angle P = 90^{\circ}.$ This implies that $R=(-4,\pm 3).$

$\angle Q = 90^{\circ}.$ This implies that $R=(4,\pm 3).$

$\angle R = 90^{\circ}.$ This implies that $$\begin{align*} 8^2 &= PR^2+QR^2\\ 8^2 &=(-4-x)^2 + y^2 \\&+ (4-x)^2 + y^2\\ 8^2 &=(-4-x)^2 + (\pm 3)^2 \\&+ (4-x)^2 + (\pm 3)^2\\ 8^2 &=(4+x)^2 + (4-x)^2 + 18 \\ 46 &= (4+x)^2+ (4-x)^2\\ 46 &=16+8x+x^2\\&+16-8x+x^2\\ 14&= 2x^2\\ x&= \pm \sqrt{7} \end{align*}$$ Thus, we have $R = (\pm \sqrt{7},\pm 3).$

As such, we have $2+2+4=8$ possible locations for $R.$

Thus, the correct answer is D.

Answer: D

Solution:

We can rewrite the equation as $$y^2-2y+x^{2020}=0.$$ Using the quadratic formula, we know that: $$y = \dfrac{2 \pm \sqrt{4-4(1)(x^{2020}) }}{2}$$

As $y$ must be an integer, and therefore a real number, we must have $4-4x^{2020} \ge 0$ to prevent negative square roots. This suggests: $$\begin{align*}4-4x^{2020} &\ge 0 \\ x^{2020} &\le 1\\ -1 \le x \le 1. \end{align*}$$ Since $x$ must be an integer for $y$ to be an integer, this gives us three $x$ values, each with either one or two corresponding $y$ values. With this, we have the following four solutions: $$(-1,1)$$ $$(0,0)$$ $$(0,2)$$ $$(1,1)$$

Thus, D is the correct answer.

Answer: C

Solution:

Remember that the volume of a cone is equal to $\dfrac 13 \pi r^2 h.$ Further notice that the circumference of the base of the cone is equal to the remaining circumference of the circle: $\dfrac 34$ the circumference of the complete circle.

Therefore, the circumference of the base of the cone is equal to: $$\begin{align*} C &= \dfrac 34 \cdot 2(\pi) (4)\\ &= 6\pi \end{align*}$$ This suggests that the radius of the cone's base ($r'$) is equal to: $$\begin{align*} C &= 2\pi r' \\ 6\pi &= 2\pi r' \\ r' &= 3. \end{align*}$$ Also, when we tape together the marked radii to form the cone, the radii in question become the slanted height of the cone. This can be used to find the actual height of the cone, which by the Pythagorean Theorem, is equal to: $$\begin{align*}SL^2 &= h^2 + (r')^2\\ 4^2 &= h^2 +3^2 \\ 16 &= h^2 + 9\\ 7 &= h^2 \\ h &= \sqrt{7}. \end{align*}$$ Thus, the volume of the cone is equal to: $$\begin{align*}V &= \dfrac 13 \pi r^2 h \\ &= \dfrac 13 \pi (r')^2 h\\ &= \dfrac 13 \pi (3)^2 \cdot \sqrt{7} \\ &= \dfrac 13 \pi 9\sqrt{7} \\ &= 3\pi\sqrt{7} \end{align*}$$

Thus, the correct answer is C.

Answer: D

Solution:

Assume that Harold has already picked his $5$ books. Of these five books, there are $\binom{5}{2}$ ways that Betty can have picked exactly two of the same books as Harold, and $\binom{5}{3}$ ways that Betty can choose her other three books from the $5$ books not on Harold's list.

As such, there are $\binom{5}{2}\binom{5}{3}=100$ ways for Betty to choose her books such that she chooses exactly two books on Harold's list and three books not on Harold's list.

Therefore, as there are $\binom{10}{5}=252$ ways that Betty can choose her books arbitrarily, and $100$ of those choices satisfy the above conditions, the probability that they have exactly two books in common is: $$\dfrac{100}{252} = \dfrac{25}{63}$$

Thus, D is the correct answer.

Answer: D

Solution:

Before tackling the problem itself, let's notice that a pattern emerges when dividing 1 by different integers, where $a$ is one digit: $$\begin{align*} \dfrac{1}{a} &= 0.\underline{k} \cdots \\ \dfrac{1}{10a} &= 0.0\underline{k} \cdots\\ \dfrac{1}{10^2a} &= 0.00\underline{k} \cdots\\ \vdots \end{align*}$$ for some digit $k$ (the dots represent the fact that there may be digits after $k$).

It seems that $1\div (10^n a)$ will have $n$ zeros after the decimal point.

Returning to the original problem, notice that $20^{20} = 10^{20} \cdot 2^{20}.$ As $2^{10}=1024 \approx 1000 = 10^3,$ it follows that $2^{20} \approx 10^6.$ Thus: $$20^{20} \approx 10^6\cdot 10^{20} = 10^{26}.$$ Therefore: $$\dfrac{1}{20^{20}} \approx \dfrac{1}{10^{26}},$$ which, by the pattern established above, implies that there are $26$ zeroes after the decimal point.

Thus, the correct answer is D.

Answer: B

Solution:

Let's begin by seeing how each step changes the Andy's coordinate:

Step 1: Andy moves one unit east and then turns left. Therefore, his new position is $(-19,20),$ facing north.

Step 2: Andy moves two units north and then turns left. Therefore, his new position is $(-19,22),$ facing west.

Step 3: Andy moves three units west and then turns left. Therefore, his new position is $(-22,22),$ facing south.

Step 4: Andy moves four units south and then turns left. Therefore, his new position is $(-22,18),$ facing east.

A cyclic pattern seems to emerge, as every four steps Andy takes, it seems that if his original position is $(a,b)$ facing east, is new position is $(a-2,b-2)$ facing east.

As there are $505$ such $4$-step cycles in the $2020$ steps Andy takes, it follows that Andy's final position is: $$\begin{align*}&(-20-505(2),20-505(2))\\ =&(-20-1010,20-1010)\\ =& (-1030,-990)\end{align*}$$

Thus, B is the correct answer.

Answer: D

Solution:

Firstly construct the following lines between points on the hexagon:

Observe that the area of the shaded region is simply $6$ times the area of the following shaded region:

As the entire figure is simply two equilateral triangles with side length $1,$ the total area is: $$2\cdot \dfrac{\sqrt{3}}{4}=\dfrac{\sqrt{3}}{2}$$ The unshaded region has an area equal to a circular sector with radius $1$ and an angle $\dfrac{\pi}{3}$ (as the triangle is equilateral). As such, the area of the unshaded region is: $$\dfrac{\frac{\pi}{3}}{2\pi} \cdot \pi = \dfrac{1}{6} \cdot \pi = \dfrac{\pi}{6}$$ Meaning the area of the shaded region is: $$\dfrac{\sqrt{3}}{2}-\dfrac{\pi}{6}$$ And as such, the area of the larger, complete shaded area in the hexagon is: $$6\cdot \left(\dfrac{\sqrt{3}}{2}-\dfrac{\pi}{6}\right) = 3\sqrt{3}-\pi$$ Thus, D is the correct answer.

Answer: D

Solution:

Let's begin by analyzing the list of digits after the removal of the third digits.

Although the list initially cycles with a period of $5$ (as in, the list repeats itself every $5$ digits), as lcm($3,5)=15,$ the relative positions from which we delete digits from repeats every $15$ digits. Therefore: $$123451234512345$$ $$12\cancel{3}45\cancel{1}23\cancel{4}51\cancel{2}34\cancel{5}$$ Therefore, the new list is: $$1245235134$$ and repeats every $10$ digits.

Similarly, we consider the same procedure for removing the fourth digits.

As lcm$(4,10)=20,$ let's consider the first $20$ digits of this list when making our deletions, as after this point, the relative positions of our deletions will repeat. $$12452351341245235134$$ $$124\cancel{5}235\cancel{1}341\cancel{2}452\cancel{3}513\cancel{4}$$ Therefore, the new list is: $$124235341452513$$ and repeats every $15$ digits.

Lastly, we consider the same procedure for removing the fifth digits.

As lcm$(5,15)=15,$ let's consider the first $15$ digits of this list when making our deletions, as after this point, the relative positions of our deletions will repeat. $$124235341452513$$ $$1242\cancel{3}5341\cancel{4}5251\cancel{3}$$ Therefore, the new list is: $$124253415251$$ and repeats every $12$ digits.

As $2019 \equiv 3 \bmod{12},$ we know that the $2019$th digit is equal to the $3$rd digit in the list above. As such, it follows that: $$\mathrm{position } ~2019 = 4$$ $$\mathrm{position } ~2020 = 2$$ $$\mathrm{position } ~2021 = 5$$ Therefore, the sum of these values is $4+2+5=11.$

Thus, the correct answer is D.

Answer: A

Solution:

Consider the following strategy:

Bela begins by choosing the middle value $\dfrac{n}{2}$ in $[0,n].$ From then on, whatever value say — $k$ — Jenn chooses, Bela will always choose $n-k.$ As Bela goes first, and chooses the middle value, she will always be able to choose the mirrored value of Jenn's move and Jenn will run out of number to pick first.

Therefore, following this strategy, Bela will always win.

Thus, A is the correct answer.

Answer: C

Solution:

We case upon the number of opposite pairs. As in, we count the number of possible pairs when we have $0,1,2,\cdots,5$ pairs whose members are across the circle from one another. Then, we will add up all these values to find the total number of ways to split up the $10$ people.

$0$ opposite pairs

In this case, we select pairs by either matching person $1$ with person $2,$ and person $3$ with person $4,$ and so on. Or, we can match person $1$ with person $10,$ and person $9$ with person $8,$ and so on. As such, there are $2$ ways to pair up individuals in this case.

$1$ opposite pair

In this case, we have $5$ possible pairs where a person is matched with the person across from them, and everyone else pairs with the person next to them.

$2$ opposite pairs

We cannot have $2$ opposite pairs, as there would be one person on either side that will not have a pair adjacent to them.

$3$ opposite pairs

In this case, we have $\binom{5}{3} = 10$ possible ways to match a person with the person across from them, where everyone else pairs with the person next to them. However, notice that there are duplicates, so we divide the $10$ possibilities by $2$ to get $5.$

$4$ opposite pairs

We cannot have $4$ opposite pairs, as there would be one person on either side that will not have a pair adjacent to them.

$5$ opposite pairs

In this case, there is only one way to do this.

As such, there are $2+5+5+1=13$ ways to pair up individuals such that the members of each pair know each other.

Thus, the correct answer is C.

Answer: B

Solution:

Begin by noting that there are $\binom{4}{2}=6$ ways to arrange two red and two blue balls in different orders.

Now, given an arbitrary arrangement of these four balls, we want to find the probability of said arrangement happening. We will show that the probability of each arrangement is the same.

We first want to find the total number of choices we have to choose from, and as there are $2$ choices for the first step, $3$ choices for the second step, $4$ choices for the third step, and $5$ choices for the fourth step, it follows that there are $$2\cdot 3\cdot 4\cdot 5 = 120$$ different ways to choose an arrangement.

Regardless of the details of the specific arrangement we aim to get, the first time a red ball is added will come from $1$ choice, and the second time will come from $2$ choices, as that is the initial number of red balls in the urn. This also holds for the greens. As such, the number of successful choices is: $$(1\cdot 2)^2 = 4$$

As such, for an arbitrary ordering of two red and two blue balls, we have a $\dfrac{4}{120} = \dfrac{1}{30}$ probability. As there are $6$ possible arrangements, the total probability is $\dfrac 15.$

Thus, the correct answer is B.

Answer: A

Video solution:

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Written solution:

The number of distinct hands that can be dealt to the player is equal to: $$\begin{align*} &\binom{52}{10} \\ &= \dfrac{52!}{42!\cdot 10!}\\ &=10\cdot 17\cdot 13\cdot 7\cdot 47\cdot 46\cdot 11\cdot 43\\ &= 158A00A4AA0 \end{align*}$$

Therefore: $$\begin{align*} &17\cdot 13\cdot 7\cdot 47\cdot 46\cdot 11\cdot 43\\ &= 158A00A4AA \end{align*}$$

To find the units digit, we can find the value of this expression mod $10$: $$\begin{align*} A &\equiv 17\cdot 13\cdot 7\cdot 47\cdot \\&46\cdot 11\cdot 43 \bmod{10}\\ &\equiv 7\cdot 3\cdot 7\cdot 7\cdot 6\cdot 1\cdot 3 \bmod{10}\\ &\equiv 1\cdot 9\cdot 6\cdot 3 \bmod{10}\\ &\equiv 9\cdot 8 \bmod{10}\\ &\equiv 2 \bmod{10}\\ \end{align*}$$

Therefore, as $0\le A \le 9,$ and $A\equiv 2\bmod{10},$ it follows that $A=2.$

Thus, A is the correct answer.

Answer: B

Video solution:

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Written solution:

Consider $S(r)$ at the following regions:

The rectangular prism itself $(B):$

In this region, $r=0,$ and therefore the volume of $S(0) = d = 1\cdot 3\cdot 4 = 12.$

The extensions of the faces of $B:$

The volumes of this region are equal to the sums of the volumes of the boxes produced by extending every face on $B$ to a distance $r.$ Therefore, the total volume of this region is: $$2(1\cdot 3+1\cdot 4+3\cdot 4)\cdot r = 38r$$

The quarter-cylinders at each of the edges of $B:$

There are $12$ edges on $B,$ meaning that there are $12$ quarter-cylinders. $4$ of which have length $4,$$4$ of which have length $3,$ and $4$ of which have length $2.$ All of these cylinders have radius $r.$ Therefore, the volumes is: $$\dfrac 14 \cdot 4\pi r^2(4+3+1)=8\pi r^2$$

The eighth-spheres at every vertex of $B:$

There are $8$ vertices on $B,$ and the eighth-sphere corresponding to each one has a radius $r.$ As such, the total volume is: $$8\cdot \dfrac 18\cdot\left(\dfrac 43 \pi r^3\right) =\dfrac 43 \pi r^3$$

Therefore, the total area enclosed by $S(r) = \dfrac 43 \pi r^3 + 8\pi r^2 + 38r + 12$

As such: $$a=\dfrac 43 \pi$$ $$b= 8\pi$$ $$c= 38$$ $$d=12$$

Therefore:$$\dfrac{bc}{ad} = \dfrac{8\cdot 38\pi}{16\pi} =19$$

Thus, the correct answer is B.

Answer: B

Video solution:

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Written solution:

As we know that the sum of the areas of $$\begin{align*}&AEH+BFIE+ \\&DHJG+FCGJI \\&= 1+1+1+1 \\&= 4,\end{align*}$$ we know that the side length of the square is $2.$

Similarly, as $AEH$ has an area of $1,$ we know that $AH=AE=\sqrt{2}.$

Now, let us extend the figure as follows:

With the construction of $K$ as the eventual intersection of $FB$ and $IE,$ we can notice that $BEK$ is a right isosceles triangle, with side length $2-\sqrt{2}.$ Therefore, its area is $3-2\sqrt{2}.$

From this, we can further deduce that $FIK$ is also a right isosceles triangle, with area $$\begin{align*}\dfrac 12 FI\cdot IK &= \dfrac 12 FI^2 \\&= 1+3-2\sqrt{2} \\&= 4-2\sqrt{2},\end{align*}$$ implying that $$FI^2 = 8-4\sqrt{2}.$$

Thus, the correct answer is B.

Answer: D

Video solution:

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Written solution:

Observe that: $$\begin{align*} 2^{202}+202 &= \left(2^{101}\right)^2+2\cdot 2^{101} +1 \\&-2\cdot 2^{101} + 201 \\ &= \left(2^{101}+1\right)^2 \\&-2\cdot 2^{101} + 201 \\ &= a^2-b^2+201, \end{align*}$$ where $a=2^{101}+1$ and $b=2^{51}.$ Thus, by difference of squares: $$\begin{gather*}a^2-b^2+201\\=(a+b)(a-b)+201\end{gather*}$$ Which means that: $$\begin{gather*}2^{202}+202 =\\(2^{101}+2^{51}+1)(2^{101}-2^{51}+1)\\+201.\end{gather*}$$ Therefore, $$\begin{gather*} 2^{202} + 202 \equiv \\ (2^{101}+2^{51}+1)(2^{101}-2^{51}+1)+ \\ 201 \equiv 201 \bmod{(2^{101}+2^{51}+1).} \end{gather*}$$

Thus, the correct answer is D.

Answer: C

Video solution:

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Written solution:

Notice that all of these transformations result in each vertex being moved to some adjacent vertex. For example, after one transformation, vertex $A$ will always be at either $(-1,1)$ or $(1,-1).$

It follows that after any odd number of transformations, vertex $A$ will always be either in quadrant II or IV, and after any even number of transformations, $A$ will always be in either quadrant I or III. This rule holds for all other vertices, and after $19$ moves, we are left with the following four possible states:

• $\begin{pmatrix}A&B\\D&C\end{pmatrix}$

• $\begin{pmatrix}C&B\\D&A\end{pmatrix}$

• $\begin{pmatrix}A&D\\B&C\end{pmatrix}$

• $\begin{pmatrix}C&D\\B&A\end{pmatrix}$