Solution:

$$(1~\diamondsuit~(2~\diamondsuit~3))-((1~\diamondsuit~2)~\diamondsuit~3) =$$ $$|1-|2-3|| - ||1-2|-3| =$$ $$|1-1| - |1-3| = 0-2 = -2.$$

Thus, the answer is A.

Solution:

Since we have a rhombus, we know $$AB = AD = AP + PD = 5.$$ And by the Pythagorean Theorem, $$AP^2 + BP^2 = AB^2$$We know that $$9+ BP^2 = 25.$$ $$BP = 4.$$ The area of a rhombus is $bh,$ so the area is $5\cdot 4 = 20.$

Thus, the answer is D.

Solution:

First, we can choose any combination for the first two digits. This would have $9\cdot 10 = 90$ choices.

Then, if there are an odd number of even digits among them, I make the units digit odd, which can be done in $5$ ways. Otherwise, I make the units digit even, which can be done in $5$ ways. Regardless of my choice of the first two digits, I have $5$ ways to choose the units digit.

Therefore, there are $90$ ways to choose the first two digits, and $5$ ways to choose the last two, so the total number of ways is $90\cdot 5 = 450.$

Thus, the answer is D.

Solution:

Since we want to look at the $700$th hiccup, we need to look at time that is $699$ hiccups after the first one.

This would be $699\cdot 5 = 3495$ seconds. Note that $3495 = 60\cdot 58+15,$ so the time would be $58$ minutes and $15$ seconds after the first hiccup. This would therefore be $4:58$ and $15$ seconds.

Thus, the answer is A.

Solution:

Lets work with the denominator first. By using difference of two squares on each term, we get the denominator as $$\sqrt{(1-\frac{1}{3^2})(1-\frac{1}{5^2})(1-\frac{1}{7^2})}$$ $$=\sqrt{(1-\frac{1}{3})(1-\frac{1}{5})(1-\frac{1}{7})}$$ $$\cdot\sqrt{(1+\frac{1}{3})(1+\frac{1}{5})(1+\frac{1}{7})}$$ $$=\sqrt{ (\frac 23)(\frac 45)(\frac 67) } \sqrt{ (\frac 43)(\frac 65)(\frac 87) }.$$

Furthermore, we simplify the numerator as follows: $$(1+\frac{1}{3})(1+\frac{1}{5})(1+\frac{1}{7})$$$$= (\frac 43)(\frac 65)(\frac 87) .$$

Dividing the numerator and denominator yields $$\begin{align*}&\frac{(\frac 43)(\frac 65)(\frac 87)}{\sqrt{ (\frac 23)(\frac 45)(\frac 67) } \sqrt{ (\frac 43)(\frac 65)(\frac 87) }} \\&=\frac{\sqrt{ (\frac 43)(\frac 65)(\frac 87) }} {\sqrt{ (\frac 23)(\frac 45)(\frac 67) }} \\&= \frac{\sqrt 8 } { \sqrt 2} \\&= 2.\end{align*}$$

Thus, the answer is B.

Solution:

We claim that none of these numbers can ever be prime.

We prove this claim by noticing that the $n$th number is $$\sum_{k=0}^{2n} 10^k + 10^n = \sum_{k=0}^{n} 10^k +$$ $$\sum_{k=n}^{2n} 10^k =\sum_{k=0}^{n} 10^k + \sum_{k=0}^{n} 10^k \cdot 10^n$$ $$= (10^n+1)(\sum_{k=0}^{n} 10^k).$$ This shows that the number can be written as the product of two numbers greater than $1,$ so there are no primes.

Thus, the answer is A.

Solution:

Let the roots be $r,s.$ Then: $$\begin{align*}&x^2+kx+36 \\ &= (x-r)(x-s) \\&= x^2-(r+s)x+rs \\&=0 \end{align*}$$ And so, $rs = 36$ and $r + s = -k.$

Therefore, we need $r$ and $s$ distinct such that $rs = 36.$ All the possible factor pairs are $$\pm\{1,36\},\pm\{2,18\},\pm\{3,12\}$$ $$\text{and }\pm\{4,9\}.$$

Each of these unordered pairs produces a unique value for $k,$ so there are $8$ possible values for $k.$

Thus, B is the correct answer.

Solution:

We can analyze the units digit of the first multiple of $7$ in each set.

If the last digit is $4,5,6,7,$ then adding $7$ yields numbers outside the set, so the sets with a multiple of $7$ such that its units digit is $4,5,6,7$ would have only one multiple.

If the last digit is $1,2,3$ then adding $7$ would yield a units digit of $8,9,0$ in the same set.

Out of the first $98$ sets, there are an equal number of occurrences of sets such that the first multiple of $7$ has each of the units digit of $1,2,3,4,5,6,7$ since they cycle every $7$ sets. These yield $42$ sets whose first multiple of $7$ contains two multiples of $7.$

The $99$th and $100$th set wouldn't work since they yield a set whose first multiples are $987$ and $994$ respectively, which can't have $2$ multiples of $7.$ This means we have $42$ sets that work.

Thus, the answer is B.

Solution:

We claim $$\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+\dots+\dfrac{n-1}{n!}$$$$= 1- \dfrac 1{n!}.$$

To prove this, we can use induction.

If $n=2,$ then the sum is $\dfrac 1{2!} = 1-\dfrac 1{2!} .$

If it works for $n-1,$ then $$\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+\dots+\dfrac{n-1}{n!}$$ $$= 1- \dfrac 1{(n-1)!}+\dfrac{n-1}{n!}$$$$= 1- \dfrac n{n!}+\dfrac{n-1}{n!} = 1 - \dfrac 1{n!}.$$

This means our formula is proven, so we can get our answer by plugging in $n=2022.$ Our answer is $1- \dfrac 1{2022!},$ yielding $a=1,b=2022,$ making our answer $2023.$

Thus, our answer is D.

Solution:

Let the integers in order be $a,b,c,d,e.$

The median of this list is $c.$ Since the mode is greater than the median, both $d$ and $e$ are equal to the mode, so we can write the list as $a,b,c,c+2,c+2.$

The mean is now $$\dfrac{a+b+c+c++c+2}5$$$$= c-2,$$ which yields $$a+b+3c+4 = 5c-10.$$ This means $$a+b=2c-14.$$ This means $a+b$ is even, and $a,b$ are different positive integers since we have a unique mode. This means $a+b \geq 4,$ since the uniqueness eliminates $a=1,b=1,$ so $a+b \neq 2.$ This means $a+b=4$ yielding a median of $c=9,$ making the mode $11.$

Thus, the answer is D.

Solution:

First, we have no information about schools that are smaller than Euclid HS, so we can eliminate all the choices that mention smaller schools. This leaves just B and C to look at.

Now, given our statement, we know that if a school is bigger than Euclid, then it couldn't have sold more than Euclid. This means if a school sold more, it couldn't have been bigger, which corresponds to choice B.

Thus, the answer is B.

Solution:

To compute this, we can also find the least $n$ such that the probability of not rolling a $7$ is less than $\dfrac 12.$ Each roll has an independent probability of $\dfrac 16$ of getting $7,$ so it has a $\dfrac 56$ probability of not landing on $7.$

Thus, the probability of none of the rolls being $7$ is $\left(\dfrac 56\right)^n.$ We must find the least $n$ such that $\left(\dfrac 56\right)^n < \dfrac 12.$

If $n=3,$ then the probability is $\dfrac{125}{216},$ which is greater than $\dfrac 12.$

If $n=4,$ then the probability is $\dfrac{625}{1296},$ which is less than $\dfrac 12.$ This makes the answer $4.$

Thus, the answer is C.

Solution:

Since the primes are $2$ away from each other, we can make them equal to $m-1,m+1,$ where $m$ is their average.

Then, $$(m+1)^3-(m-1)^3 = 31106 ,$$ making $$m^3+3m^2+3m+1$$$$-(m^3-3m^2+3m-1)$$$$= 6m^2+2 = 31106.$$

Therefore, $m^2= 5184,$ so $m=72.$

The primes are therefore $71,73.$ The least prime greater than both of those is $79,$ and its digit sum is $16.$

Thus, the answer is E.

Solution:

First, note that we can make a set with size $13$ using $S = \{13,14 \cdots ,25\}.$

Now, we prove no arbitrary set of size greater than $13$ work. Let $m$ be the maximum element of $S.$ Then, for all $i$ in $S,$ we know $m-i$ isn't in $S.$

This would eliminate $\lceil{\dfrac{m-1}{2}} \rceil$ of the numbers below $m.$ This means the maximum number of elements below $m$ is $\lfloor \dfrac {m-1}2 \rfloor ,$ making the maximum number of elements $\lfloor \dfrac{m-1}2 \rfloor +1.$

The maximum value of this has $m=25,$ yielding $13.$

Thus, the answer is B.

Solution:

Let the sequence be $a_1, a_2, \cdots.$ Create $a$ by making it the term before $a_1$ in the sequence. This would make $a_n = a+2n.$

This would make $$S_n = \sum_{i=1}^n (a+2i) =$$ $$a\cdot n + 2\sum_{i=1}^ni = an + n^2+n.$$

This makes $$\dfrac{S_{3n}}{S_n} = \dfrac{3an + 9n^2+3n}{an + n^2+n} =$$ $$\dfrac{3a + 9n+3}{a + n+1}.$$ Thus, we must find $a$ such that this value is constant.

If our given value is constant, than the given value minus $9$ is constant, so $$\dfrac{3a + 9n+3}{a + n+1}-9 = \dfrac{-6a-6}{a+n+1}$$ is constant. As $n$ increases, the numerator is constant and the denominator is increasing. Therefore, if the number is constant, the numerator must be $0.$

Since $-6a-6 = 0,$ we have $a=-1.$

With our formula from before, we have $$S_{20} = -1(20)+20^2+20 = 400.$$

Thus, the answer is D.

Solution:

Firstly, let's label the points as follows:

Since we have a rectangle, $AB = 4.$ By the Pythagorean Theorem, we have $AC = 3.$ Then, since $\angle ACB$ and $\angle DCE$ are complementary, $\angle BAC = \angle CDE = 90^\circ,$ and $BC = CE$ we know $ABC \cong CDE.$ Therefore, $ED = 3$ and $EF = 1.$

Since $\angle CED$ and $\angle GEF$ are complementary and $\angle GEF = \angle CDE = 90^\circ,$ we know $EFG$ and $CDE$ are similar. This means $\dfrac{EC}{CD} = \dfrac{EG}{EF},$ so $EG = 1.25.$ Since the shaded region is a trapezoid, we can get the area as $$\dfrac{(GE+BC)EC}2 = \dfrac{ 5(5+1.25)}2$$$$= \dfrac{5\cdot 6.25}{2} = 15.625.$$

This is equal to $15 \dfrac 58.$

Thus, the answer is D.

Solution:

Note that $a^n-b^n$ is divisible by $a-b.$

For A, let $a = 4,b=1,n=303.$ Then we get that $$4^{30}3-1 = 2^{606}-1$$ is divisible by $3.$

For B, let $a = 4,b=-1,n=303.$ Then we get that $$4^{303}-(-1)^{303} = 2^{606}-1$$ is divisible by $5.$

For D, since $2^{606}-1$ is divisible by $3,$ we have $2^{607} -2$ is divisible by $3,$ leaving $2^{607} +1$ being divisible by $3.$

For E, let $a = 3,b=-2,n=607.$ Then we get that $$3^{607}-(-2)^{607} = 3^{607}+2^{607}$$ is divisible by $5.$

For C, we know $2^{607} +1$ is divisible by $3,$ so our value isn't divisible by $3.$ We also know $2^{607} +1$ is divisible by $5$ from choice D so our value isn't divisible by $5.$

Also, $64^{101}-1 = 2^{606} -1$ is divisible by $7,$ so $2^{607} -2$ is divisible by $7.$ This means our value isn't divisible by $7.$ Since it isn't divisible by $2,$ it isn't divisible by a prime under $10.$

Thus, our answer is C.

Solution:

There are $2^9 =512$ total configurations. Now, we can use complementary counting to determine how many have more than one solution.

If a configuration has $3$ equations which don't contain redundant information, then it has only one solution.

This means every equation has to be different. Also, if any equation has $a,b,c =0,$ then it doesn't provide any information, making it redundant. This means we have $7$ choices for the first equation, $6$ choices for the second, and $5$ choices for the third.

This yields $210$ configurations. However, some configurations may still yield redundant information. If two equations add to the other equation, then there is a redundancy.

There are two cases for this to happen.

Case $1:$ $1$ of the equations has $a,b,c=1,$ another equation has $1$ of the variables being $1$ and the other equation has $2$ variables being $1.$ There are $3$ ways to choose which equation has every variable as $1.$ Then, there are $2$ ways to choose which variables have one variable being $1,$ and this equation has $3$ ways to choose which variable is $1.$ This case has $3\cdot 2\cdot 3=18$ configurations to exclude.

Case $2:$ $1$ of the equations has $2$ variables being $1,$ and the other two equations have only one variable being $1,$ with those variables being different from each other, but one of the variables chosen in the first equation. There are $3$ ways to choose the equation with $2$ variables being $1,$ there are $3$ ways to choose which variables are $1,$ and $2$ ways to choose the order of the other equations. This case has $3\cdot 2\cdot 3=18$ configurations to exclude.

There are a total of $210-18-18=174$ cases which have only one solution. This means $512-174=338$ configurations have multiple solutions, making at least one nonzero.

Thus, the answer is B.

Solution:

Suppose the center is initially filled. Then, there are either either $2$ or $3$ other filled squares, each of which can't have $2$ or $3$ filled neighbors.

This means that there are at most $4$ filled squares, so each square has at most $3$ neighbors. Since they don't have $2$ or $3$ neighbors, they must have at most $1$ neighbor. The center square is a neighbor, so they can't have any other neighbor.

Suppose I have a filled square on an edge. Since there is some filled square that isn't a neighbor of the square, we can examine the two edges which are neighbors of the filled edge. If I have a filled edge on the corner, the edge on the same side as the corner would have three neighbors. If I choose the opposite edge, the adjacent edges would have three neighbors.

Suppose I choose a corner. Then, I need to choose another corner. If I choose the adjacent corner, then the edge between would have three neighbors, making it filled. Therefore, it must be the adjacent corner. This has $2$ configurations.

Suppose the center is initially empty. Then, there are $3$ filled neighbors of the center, each with at most $1$ neighbors. This means no square has two filled neighbors. This makes it only possible to do in the following ways:

The first three can rotated making $4$ configurations, and the last one can be rotated and reflected making $8$ configurations. There are $20$ configurations with the center being empty. This means there are $20+2=22$ different configurations.

Thus, the answer is C.

Solution:

First, we extend $BE$ and $AD$ such that they meet at $G.$ Since $$\angle GDE = \angle ECB,$$ $$\angle GED = \angle BEC \text{ and } DE = EC,$$ we know $GDE \cong BCE.$ Therefore, $DG =BC = AD.$ This means that if we construct a circle with center $D$ that includes $A,$ $C,G$ are also on it.

Also, since $AFG$ is a right triangle, the drawn circle would be its circumcircle, placing $F$ on the circle.

Since $\angle GDC = 134^\circ,$ we can get $$\angle CFE = \angle CFG = \dfrac {\overset{\Large\frown}{CG}} 2$$$$= \dfrac{134^\circ}{2} = 67^\circ.$$ Therefore, $$\angle BFC = 180^\circ - 67^\circ = 113^\circ.$$

Thus, the answer is D.

Solution:

Since $P(x)$ has a remainder of $x+2$ when divided by $x^2+x+1,$ it must be able to be written as $$P(x) =$$$$(x^2+x+1)Q(x)+x+2$$ for some polynomial $Q(x).$ Note that the remainder of $P(x)$ when divided by $x^2+1$ is equal to the remainder when $xQ(x) + x+2.$

If $Q(x)=c$ for some constant, then the remainder when $P(x)$ is $(c+1)x+2,$ which can't happen since we need a $1$ for our constant.

If $P(x) = ax+b,$ then the remainder when divided by $x^1$ is equal to the remainder when $$(ax+b)x+x+2 =$$$$ax^2 + (b+1)x+2$$ is divided by $x^2+1.$ We can get the remainder by subtracting $a(x^2+1),$ yielding $(b+1)x +(2-a).$ This means $b+1=2$ and $2-a=1,$ so $a=b=1.$

This means $$P(x) =$$$$(x+1)(x^2+x+1)+x+2 =$$$$x^3+2x^2+3x+3.$$ The sum of the squares of the coefficients is $23.$

Thus, the answer is E.

Solution:

Let $x^2 + y^2 = 64$ be circle $O,$ $x^2 + y^2 = 5$ be circle $P,$ and $(x - 5)^2 + y^2 = 3$ be circle $Q.$

First note that every circle, $R,$ in $S$ is internally tangent to $O.$ Then we case on the tangency of $R$ with $P$ and $Q.$

Case $1:$ This corresponds to the pink circle. This is where $P$ and $Q$ are internally tangent to $R.$

Case $2:$ This corresponds to the bluish circle. This is where $P$ and $Q$ are externally tangent to $R.$

Case $3:$ This corresponds to the green circle. This is where $P$ is externally and $Q$ is internally tangent to $R.$

Case $4:$ This corresponds to the red circle. This is where $P$ is internally and $Q$ is externally tangent to $R.$

We can consider cases $1$ and $4$ together. Note that $O$ and $P$ have the same center. This means that the line connecting the center of $R$ and $O$ passes through the tangency point of both $S$ and $O$ and $S$ and $P.$

This line is the diameter of $R,$ and it has length $$r_P + r_O = 2 + 8 = 10.$$ Therefore, the radius of $R$ is $5.$

Consider cases $2$ and $3$ together. Similarly to above, the line connecting the center of $R$ and $O$ will pass through the tangency points.

This time, however, the diameter of $R$ is $$r_P - r_O = 8 - 2 = 6.$$ This makes the radius of $R$ $3.$

$S$ contains $8$ circles: $4$ of which have radius $5$ and $4$ of which have radius $3$ (this is because we can flip all the circles in the diagram over the x-axis to get $4$ more circles).

The total area of the circles in $S$ is therefore $$4 (5^2 \pi + 3^2 \pi) = 136 \pi.$$ Thus, E is the correct answer.

Solution:

We begin by breaking the question into two smaller problems:

• The probability such that if $x,y$ were random numbers in the interval $(0,1),$ that $x+y < 1.$

• The probability such that if $x,y,z$ were random numbers in the interval $(0,1),$ that $x+y+z < 1.$$$$$

To solve both of these, we use geometric probability. That means, for a 2-D or 3-D coordinate system, that $x+y < 1$ and $x+y+z < 1.$

To solve the first one, we take the area of the region where $x+ y < 1,$ and $x,y > 0.$ This would be $\frac 12$ since it is a triangle with base and height of 1. Since the total area is just $1^2 = 1,$ the probability is also $\frac{1}{2}.$

To solve the second one, we take the area of the region where $x+ y + z < 1,$ and $x,y > 0.$ This would be $\frac 16$ since it is a volume of a pyramid with base of area $\frac 12$ and height of $1.$ Again, the volume is $1^3 = 1,$ so the probability is also $\frac{1}{2}.$

Now, we can look at the cases of the problem. Annie can either go until $n=2$ if $t_1 + t_2 > 1$ or until $n=3$ otherwise.

Case $1:$ The probability that $t_1+t_2 > 1$ is $\frac 12$ since the probability that $t_1+t_2 < 1$ is $\frac 12.$ Similarly, the probability that $x_1+x_2 > 1$ is $\frac 12.$ The total probability with this case is $\frac 12 \cdot \frac 12 = \frac 14.$