Answer: E

Solution:

We can add each individual digit, yielding $11,110.$

We can also get the sum by noticing that each digit has a sum of $10,$ so the sum is equal to $10\cdot 1111 = 11,110.$

Thus, the answer is E.

Answer: B

Solution:

The area is a triangle of area $$\dfrac {4\cdot 5}2 - \dfrac{2\cdot 4}2 = 10-4$$$$= 6$$ since we subtract the area of a smaller triangle from a larger triangle.

Thus, the answer is B.

Answer: E

Solution:

We can rewrite this as $$\dfrac{2021\cdot 2021}{2020\cdot 2021} -\dfrac{2020\cdot 2020}{2020\cdot 2021}$$ $$= \dfrac{2021^2-2020^2}{2020\cdot 2021}.$$

This can be simplified to $$\dfrac{(2021-2020)(2021+2020)}{2020\cdot 2021}$$ $$=\dfrac{4041}{2020\cdot 2021}.$$

Since $4041$ is coprime with both $2020$ and $2021,$ we know $4041$ is the numerator.

Thus, the answer is E.

Answer: C

Solution:

Let the temperature in Minneapolis be $m$ and let the temperature in St. Louis be $s.$ Then, we know that $$|(m-5)-(s+3)| = 2.$$ This means $$|((m-s) -8| = 2,$$ so $$m-s = 8 \pm 2,$$ thus making the difference $6$ or $10.$ Therefore, the product is $6\cdot 10 = 60.$

Thus, the answer is C.

Answer: E

Solution:

We know $$\begin{align*}n &= 8^{2022} \\&= 2^{6066} \\&= 4^{3033} .\end{align*}$$ Therefore, $$\frac n4 = n\cdot 4^{-1}=4^{3032} .$$

Thus, the answer is E.

Answer: B

Solution:

Before starting, note that if we can represent the prime factorization of an integer $z$ as $$z=p_1^{e_1} p_2^{e_2} \cdots,$$ then there are $(e_1+1)(e_2+1) \cdots$ distinct positive factors.

If the number in question has $2021$ factors, by the previous logic, $$2021=(e_1+1)(e_2+1) \cdots,$$ and as the prime factorization of $2021 = 43\cdot 47,$ then our number must be $p_1^{46}p_2^{42}$ or $p^{2020}.$

The smallest number we can make in either of these is making $p_1 = 2,p_2=3$ in the first configuration, yielding $$2^{46}3^{42} = 16\cdot 6^{42} .$$

Therefore, $$m=16$$$$k=42,$$ so $$m+k = 42+16 = 58.$$

Thus, the answer is B.

Answer: C

Solution:

Let the denominators of both fractions be $x,y.$ Therefore, their sums are $$\frac{15-x}x + \frac{15-y}y$$ $$= \frac{15}x + \frac{15}y -2.$$ Thus, we need to find the number of unique integers we can get from $$\frac{15}x + \frac{15}y .$$

If we have $x = 1,3,5,$ then our fraction is an integer, which would be $15,5,3.$ Adding each set of pairs yields $30,20,18,10,8,6.$

If we have $x = 2,6,10,$ then our fractional part is a half, which would be $7.5,2,5,1.5.$ Adding each set of pairs yields $15,10,9,5,4,3.$

If we have $x = 4,12,$ then our fractional part is a quarter or three quarters, which would be $3.75,1.25.$ Adding each set of pairs yields $7.5,5,2.5.$

The unique integers from this are $30,20,15,10,9,8,6,5,4,3,$ of which there are $11.$

Thus, the answer is C.

Answer: C

Solution:

We know $$\begin{align*}16,383 &= 16384-1 \\&= 2^{14}-1 \\&= (2^7-1)(2^7+1) \\&= 127\cdot 129.\end{align*}$$ Since $129=3\cdot 43,$ we get $$16383 = 3\cdot 43\cdot 127.$$ Therefore, $127$ is the largest prime factor, and the sum of its digits is $10.$

Thus, the answer is C.

Answer: C

Solution:

Let $r,b$ be the number of red and blue knights, and let $r_m,b_m$ be the number of magical knights of each color. We know $$r_m + b_m = \frac 16.$$

We also know $$\frac{r_m}r = 2 \frac{b_m}b .$$ Note that $$\begin{align*}b &= 1-r \\&= 1-\frac 27 \\&= \frac 57.\end{align*}$$

Therefore, $$\dfrac{r_m}{\frac 27} = 2\dfrac{b_m}{\frac 57},$$ so $b_m = 1.25r_m.$

As such, $2.25r_m = \frac 16,$ so $r_m = \frac 2{27}.$ This makes the fraction that are magical equal to $$\dfrac{\frac 2{27} }{\frac 27} = \dfrac{7}{27} .$$

Thus, the answer is C.

Answer: A

Solution:

Alice saying that she doesn't know the number means that she doesn't have the largest or smallest possible numbers, which are $1,40.$ Bob, now knows who has the largest number. Thus, he may have $1$ or $40.$ He may also have $2$ or $39$ as he would know Alice doesn't have $1,40.$ This means $2$ would mean he has the lesser number, and $39$ means he has the greater number. Since his number is prime, his number must be $2.$

Since his number is $2,$ we know $200$ plus Alice's number is a sqaure. Since Alice's number is between $1$ and $40,$ we must find a square somewhere from $201$ to $240,$ which must be $225.$ This makes Alice's number $25.$ Therefore the sum is $$2+25 = 27.$$

Thus, the answer is A.

Answer: B

Solution:

The average of the area of the circle and the new figure is equal to the area of the hexagon. The hexagon's area can be taken as the area of $6$ equilateral triangles with length $1,$ making them each have area $\frac{\sqrt 3}4 .$ Thus, the total area of the hexagon is $$6\cdot \frac{\sqrt 3}4 = 3\frac{\sqrt 3} 2.$$ The radius of the circle is $1$ since thats the length of the equilateral triangle, so the area of the circle is $\pi\cdot 1^2 = \pi.$

Let the area of the shape be $a.$ Then, by the first statement, we know $$\frac{a + \pi} 2 = \frac{3\sqrt 3}2 ,$$ so $$a + \pi = 3 \sqrt 3,$$ making $$a = 3\sqrt 3 - \pi.$$

Thus, the answer is B.

Answer: D

Solution:

Notice $$x(x-y)+y(y-z)+z(z-x)$$ $$= x^2 +y^2 + z^2 -xy-xz-yz$$ $$=\frac 12 \left( (x-y)^2+(x-z)^2\right.$$$$\left.+(y-z)^2\right).$$

Thus, $$(x-y)^2+(x-z)^2+(y-z)^2$$$$= 2.$$ Since every term is a positive integer, we know that two of them are $1$ and the other is $0,$ or two of them are $0$ and the other is $2.$

Since they must be squares, it has to be the first condition. If we have a term with $(a-b)^2=0,$ then $a=b.$ If we have $(a-b)^2 = 1,$ then, $|a-b|=1.$

Thus, two of $x,y,z$ must be equal and the other number must be $1$ away from the equal numbers. This is guaranteed with the condition $x=z$ and $y-1=x.$

Thus, the answer is D.

Answer: B

Solution:

Firstly, note the area is equal to $$\frac{bh}2.$$

Now, if we take the smaller triangle and scale it up to the bigger triangle, we multiply by $\frac 32.$ The base of the smaller triangle is $3,$ so the base of the larger triangle is $\frac 32 \cdot 3 = \frac 92.$ This makes the area equal to $\frac 94h.$

Now, the heights of the squares go down in a scaling factor of $\frac 23.$ This means if we continually put smaller and smaller squares at the top, their lengths would be multiplied by $\frac 23,$ so the total height is $$\begin{align*}&3 + 3\cdot \frac 23 + 3\cdot \left(\frac 23\right)^2 \cdots \\&= 3(1+ \frac 23 + \left(\frac 23\right)^2 \cdots \\ &= 3 \cdot \frac{1}{1- \frac 23} \\&= 3 \cdot \frac{1}{ \frac 13} \\&= 9.\end{align*}$$

Therefore, the area is $$9\cdot \frac 94 = \frac{81}4 = 20\frac14 .$$

Thus, the answer is B.

Answer: C

Solution:

We will first count the number of ways to have the product be not divisible be $4.$ This can be done if the product is odd, where all numbers are odd, or the product is even but not a multiple of $4,$ in which $5$ die are odd and the other die is $2$ or $6.$

In the first case, we can do this with a probability of $\frac 12^6 = \frac 1{64} .$

In the second case, there is a $\frac 26$ probability that a chosen dice is $2$ or $6,$ a $\frac 12^5$ probability of the other die being even, and we have $6$ ways to choose the choosen dice. This makes the probability $6 \cdot \frac 26 \cdot \frac 12 ^5 = \frac 4{64} .$

Therefore, the total probability that the product isn't divisible by $4$ is $\frac{5}{64}$ making the probability that it is divisible equal to $\frac{59}{64} .$

Thus, the answer is C.

Answer: D

Solution:

We know $$\begin{align*}\angle QBR &= 90^\circ - \angle BQC \\&= 90^\circ - (90^\circ - \angle BCQ ) \\&= \angle BCQ.\end{align*}$$

Since $$\angle BCQ = \angle ABP,$$$$\angle PAB = \angle QBC,$$ and $AB = BC,$ we know $PAB \cong QBC.$

Therefore, $QC = PB = 13.$ As such, $QR = RC.$

Also, since $QRB \sim BRC,$ we know $$QR\cdot RC = RB^2 = 36.$$ Let $RC = x.$ Then, $$x+ \frac{36}x = 13,$$ so $$x^2-13x+36 = 0$$$$(x-4)(x-9)=0.$$ Thus, $x=4,9.$ We know $x=9$ since its the greater number. Then, $$\begin{align*}CB^2 &= BR^2 + CR^2 \\&= 9^2 + 6^2 \\&= 81+36 \\&= 117.\end{align*}$$ Therefore, the area is $$CB^2 = 117.$$

Thus, the answer is D.

Answer: D

Solution:

Let Chris choose his two balls.

There is a $\frac 15$ probability that Silva choose the same balls, which would make all the balls the same as the original, making it such that $5$ balls are the same.

There is a $\frac 25$ probability that Silva choose one of the same balls, which would make $2$ the balls the same as the original.

There is a $\frac 25$ probability that Silva choose none of the same balls as Chris, so $4$ balls are in different positions. This makes $1$ ball the same.

Therefore, the expected value is $$\begin{align*}5\cdot \dfrac 15 + 2\cdot \dfrac 25 + 1\cdot \dfrac 25 &= \dfrac{11}{5} \\&= 2.2 .\end{align*}$$

Thus, the answer is D.

Answer: D

Solution:

Let the line from the origin to the point $P$ be $p.$ Let the line from the origin to the point $P''$ be $r.$Let $\theta_\ell, \theta_m, \theta_r$ be the angle from each of the lines to the origin. If line $a$ is reflected across line $b,$ then the mean of the angles of $a$ and its reflection is equal to the angle of $b.$ If $a'$ is the reflection, we get $$\frac{\theta_a + \theta_{a'}}2 = \theta_b$$ which means $$\theta_{a'} = 2\theta_b-\theta_a.$$

Bringing this to our current problem, the angle after the first reflection is $2\theta_\ell-\theta_p.$ Then, the angle after the second reflection is $$2\theta_m-(2\theta_\ell-\theta_p)$$$$= 2(\theta_m-\theta_\ell) + \theta_p.$$ Notice that $P''$ is $P$ rotated $90^\circ$ clockwise, so it lowers the angle by $90^\circ.$ This means $$2(\theta_m-\theta_\ell) + \theta_p = \theta_p - 90^\circ.$$

Therefore, $\theta_m-\theta_\ell = -45^\circ.$ This further implies $\theta_m = \theta_\ell - 45^\circ.$

Since the slope of $\ell$ is $5,$ we get $\tan (\theta_\ell) = 5.$ The tangent subtraction formula yields $$\begin{align*}\tan( \theta_m) &= \frac{\tan(\theta_\ell) - \tan(45^\circ)}{1+\tan(\theta_\ell)\tan(45^\circ)} \\&= \frac{5-1}{1+5} \\&= \frac 46 \\&= \frac 23.\end{align*}$$

Therefore, our line is $$y = \frac 23 x,$$ or $$2x-3y=0.$$

Thus, the answer is D.

Answer: E

Solution:

This shape can be split into identical $24$ triangles, as shown in the diagram. These triangles have one angle of $45^\circ$ as it is an angle bisector of a right triangle.

Another angle is $15^ \circ$ as it is $\frac 1{24}$ of the full way around, so the angle is $\frac {360}{24} = 15.$ Thus, the last angle is $120^\circ.$

Now, we extend the shortest side to make a right triangle. This has angles $45,45,90.$ The altitude is $3,$ since its half of the length of the square. The base of the right triangle is $3,$ as its also half of the square. Now, we find the portion of the base in the original triangle by subtracting the portion outside.

The portion outside the original triangle creates a $30-60-90$ triangle when using the altitude, so its base is $$3 \tan(30^\circ) = \sqrt 3.$$ Thus, the base of the original triangle is $3 - \sqrt 3.$ Therefore, each triangle has an area of $$\dfrac{3(3-\sqrt 3)}2 = \dfrac{9-3\sqrt 3}2 .$$ Since there are $24$ such angles, the total area is $$24\left(\dfrac{9-3\sqrt 3}2\right)=108-36 \sqrt 3.$$

This makes $$a,b,c = 108,54,3$$ respectively, so $$a+b+c = 147.$$

Thus, the answer is E.

Answer: A

Solution:

We can take $$N = \dfrac{7\left(10^{313}-1\right)}{9} .$$ Notice that the first digit of a number doesn't change when divided by $10.$ Thus, multiplying by a power of $10$ would preserve the same leading digit.

Next, we can approximate $N$ to be $$N = \dfrac{7\left(10^{313}\right)}{9}$$ as this wouldn't be increase the leading digit. Now, we can case on each number functional value.

$f(2)$ means we have to find the first digit of $$\sqrt{\dfrac{7\left(10^{313}\right)}{9}},$$ which has the same units digit as $$\frac{\sqrt{\frac{7(10^{313})}{9}}}{10^{156} } = \sqrt{ 7.\overline 7}.$$ This has leading digit $2,$ so $f(2)=2.$

$f(3)$ means we have to find the first digit of $$\sqrt[3]{\dfrac{7\left(10^{313}\right)}{9}},$$ which has the same units digit as $$\frac{\sqrt[3]{\frac{7(10^{313})}{9}}}{10^{104} } = \sqrt[3]{ 7.\overline 7}.$$ This has leading digit $1,$ so $f(3)=1.$

$f(4)$ means we have to find the first digit of $$\sqrt[4]{\dfrac{7\left(10^{313}\right)}{9}},$$ which has the same units digit as $$\frac{\sqrt[4]{\frac{7(10^{313})}{9}}}{10^{78} } = \sqrt[4]{ 7.\overline 7}.$$ This has leading digit $1,$ so $f(4)=1.$

$f(3)$ means we have to find the first digit of $$\sqrt[5]{\dfrac{7\left(10^{313}\right)}{9}},$$ which has the same units digit as $$\frac{\sqrt[5]{\frac{7(10^{313})}{9}}}{10^{62} } = \sqrt[5]{ 777.\overline 7}.$$ Note that $3^5 =243,4^5 = 1024,$ so $\sqrt[5]{ 777.\overline 7}$ has leading digit $3.$ Thus, $f(5)=3.$

$f(3)$ means we have to find the first digit of $$\sqrt[6]{\dfrac{7\left(10^{313}\right)}{9}},$$ which has the same units digit as $$\frac{\sqrt[6]{\frac{7(10^{313})}{9}}}{10^{52} } = \sqrt[6]{ 7.\overline 7}.$$ This has leading digit $1,$ so $f(6)=1.$

We now know $$f(2)+f(3)+f(4)+f(5)+f(6)$$$$= 2+1+1+3+1$$$$=8.$$

Thus, the answer is A.

Answer: C

Solution:

The probability that Hugo rolled a $5$ given that he won is equal to the probability that Hugo rolled a $5$ and won divided by the probability that he won. The probability that he wins is $\frac 14,$ so dividing by this is equal to multiplying by $4.$ Thus, we need to just find 4 times the probability that Hugo rolled a $5$ and won.

Now, to find the probability that he won given that he rolled a $5,$ we need to case on the number of people who tied with him.

Case 1: No one ties

This has a probability of $$\frac 16 \cdot \frac 46 ^3 = \frac{64}{6^4}$$ as there are $3$ players who choose something from $1$ to $4.$

Case 2: One person ties

This has a probability of $$\frac 16^2 \cdot \frac 46 ^2 \cdot \binom{3}{1} = \frac{48}{6^4}$$ as there are $3$ players who choose something from $1$ to $4.$ Now, the probability that Hugo wins here is $\frac 12$ given this event occured, so the total probability is $$\frac{48}{6^4} \cdot \frac 12 = \frac{24}{6^4}$$

Case 3: Two people tie

This has a probability of $$\frac 16^3 \cdot \frac 46 ^1 \cdot \binom{3}{2}= \frac{12}{6^4}$$ as there are $3$ players who choose something from $1$ to $4.$ Now, the probability that Hugo wins here is $\frac 13$ given this event occured, so the total probability is $$\frac{12}{6^4} \cdot \frac 13 = \frac{4}{6^4}$$

Case 4: Three people tie

This has a probability of $$\frac 16^4= \frac{64}{6^4}$$ as there are $3$ players who choose something from $1$ to $4.$ Now, the probability that Hugo wins here is $\frac 14$ given this event occured, so the total probability is $$\frac{1}{6^4} \cdot \frac 14 = \frac{0.25}{6^4}$$

The combined probability is $$\frac{64 + 24+4+0.25}{6^4} = \frac{92.25}{1296}.$$ This now has to be multiplyied by $4,$ yielding $$\frac{369}{1296} = \frac{41}{144} .$$

Thus, the answer is C.

Answer: E

Solution:

Suppose we have two regular polygons with $a$ and $b$ sides such that $a \geq b.$ On, the circle, there area $b$ arcs from one point of the polygon to the other. If a line goes from $1$ arc to another arc, then it intersects the polygon twice, as there are two lines. The polygon with $a$ sides goes across each of the $b$ arcs, so it crosses $2b$ lines.

To get the intersections of the points in our given setup, we can take the sum of the intersections with each pair of polygons. In each of the pairs, we take $2$ times the lower number, and take the sum of this. This would be $$3\cdot 2\cdot 5+2\cdot 2\cdot 6+1\cdot 2\cdot 7$$$$= 68.$$

Thus, the answer is E.

Answer: B

Solution:

The products that are added from $S_{n-1}$ to $S_n$ are those where $k=n$ and $j < n.$ The sum of these are the sums of all positive integers less than $n-1$ times $n,$ making it $$n\cdot \frac{n(n-1)}2 = \frac{n^2(n-1)}2.$$ This makes $$S_n = S_{n-1}+\frac{n^2(n-1)}2.$$

If $n \equiv 0,1 \mod 3,$ then $$\dfrac{n^2(n-1)}2$$ is a multiple of $3,$ so $$S_n \equiv S_{n-1} \mod 3.$$

If $n \equiv 2 \mod 3,$ then $$\dfrac{n^2(n-1)}2 \equiv \frac{2^2\cdot 1}2$$$$\equiv 2 \mod 3,$$ so $S_n \equiv S_{n-1} + 2 \mod 3.$

This means that the remainder of $S_n$ when divided by $3$ is increased by $2$ when $n \equiv 2 \mod 3,$ and its remainder is the same otherwise.

$$S_2=1\cdot 2$$$$=2$$ as the only pair $(j,k)$ is $(1,2).$

The it must increase by $2 \mod 3$ twice to be a multiple of $3,$ so the first $n$ that works is $n=8.$ Then, $n=8,9,10$ have $S_n$ being multiples of $3.$ After $8,$ the next occurence is after $3$ more changes, making $n=17,18,19$ work, and then $n=26,27,28$ with $3$ changes after that. Finally, the $10$th number that works is $n=35.$

We can now find the sum to be $$3\cdot 9+3\cdot 18+3\cdot 27+35$$$$=197.$$ (Note that I did 3 times some number as it was the average of the tuple of 3 numbers around it.)

Thus, the answer is B.

Answer: D

Solution:

We will do this with complementary counting, meaning we will find the probability that no triangle has three sides of the same stroke first.

Now we will case on the stroke configurations of the outside edges. I will use the notation $a-b-c-d$ or some shortened version of this to specify the stroke, with the numbers being how many of one stroke there is, and a dash means that it switches strokes afterwards. There must be an even number of numbers in the configurations, except for the case where there is just one number. Therefore, the configurations that are possible can be found by adding an even number of whole numbers that add to $5.$ Therefore, these are the possible configurations: $$5,$$$$4-1,$$$$3-2,$$$$2-1-1-1.$$ Notice how all the configurations require at least one adjacent pair of edges to be the same stroke type. If a pairing of adjacent sides exists like this, then the connection must be the opposite stroke as shown below

Case 1: The configuration $5$

This can occur with probability of $\frac{1}{16}$ as there are two stroke types, each with a $\frac{1}{32}$ chance of happening. Due to the pairing of every adjacent side, we know the diagonals must all be the same stroke, which has a probability of $\frac 1{32} .$ Thus, the probability of this case is $\frac 1{16\cdot 32} = \frac 1{512} .$

Case 2: The configuration $4-1$

This can occur with probability of $\frac{5}{16}$ as there are two strokes that can be the $1$ and $5$ positions for the edge, each with a $\frac{1}{32}$ chance of happening. After filling the required diagonals, we have the configuration as shown.

This has the long triangle with a base at the bottom having all of one stroke, so there are no possible configurations.

Case 3: The configuration $3-2$

This can occur with probability of $\frac{5}{16}$ as there are two stroke types that can be the $1$ and $5$ positions for the edge, each with a $\frac{1}{32}$ chance of happening. After filling the required diagonals, we have the configuration as shown.

The other two diagonals have to be solid to have the longer triangles with the upper right and left edge have at least one distinct side. This makes the long triangle with the bottom side all solid, so there are no possible configurations.

Case 3: The configuration $2-1-1-1$

This can occur with probability of $\frac{5}{16}$ as there are two strokes that can be the $2$ and $5$ positions for the adjacent pair, each with a $\frac{1}{32}$ chance of happening. After filling the required diagonals, we have the configuration as shown.

The other two diagonals have to be solid to have the longer triangles with the upper right and left edge have at least one distinct side. This makes the long triangle with the bottom side all solid, so there are no possible configurations. Then we fill the diagonals in the following way.

This has only one configuration, with probability $\frac{1}{32},$ so the total probability is $\frac{5}{16\cdot 32} = \frac{5}{512} .$