Solution:

Since $2 \diamond 3 = |2-3| = 1,$ we get $1 \diamond (2 \diamond 3) = 1 \diamond 1 = 0.$

Since $1 \diamond 2 = |1-2| = 1,$ we get $(1 \diamond 2) \diamond 3 = 1 \diamond 3 = |1-3| = 2.$

The value is $0 - 2 = -2.$

Thus, the correct answer is A.

Solution:

The side length is $AD = AP + PD = 5,$ so $AB = 5.$ In right triangle $APB,$ $$BP = \sqrt{AB^2 - AP^2} = \sqrt{25 - 9} = 4.$$

Taking $AD$ as the base and $BP$ as the height, the area is $AD \cdot BP = 5 \cdot 4 = 20.$

Thus, the correct answer is D.

Solution:

The $n$th term consists of $n$ ones, then a $2,$ then $n$ ones. It factors as a repunit times a number of the form $10^n + 1:$ $$121 = 11 \cdot 11, \quad 11211 = 111 \cdot 101, \quad 1112111 = 1111 \cdot 1001,$$ and in general the $n$th term equals $\underbrace{1\cdots1}_{n+1} \cdot (10^n + 1).$

For every $n \ge 1$ both factors exceed $1,$ so every term is composite. None of the ten numbers is prime.

Thus, the correct answer is A.

Solution:

If the roots are integers $p$ and $q,$ then $pq = 36$ and $k = -(p+q).$ Distinct roots must have the same sign, so we list factor pairs of $36$ with $p \ne q.$

The positive pairs are $(1,36), (2,18), (3,12), (4,9),$ and the negative pairs are $(-1,-36), (-2,-18), (-3,-12), (-4,-9).$ The pair $(6,6)$ is excluded since the roots must be distinct.

Each of these $8$ pairs gives a different value of $k.$

Thus, the correct answer is B.

Solution:

Relative to the center $(3, 1),$ the point is at $(-1 - 3,\, -2 - 1) = (-4, -3).$

A $270^\circ$ counterclockwise rotation sends $(x, y)$ to $(y, -x),$ so $(-4, -3)$ becomes $(-3, 4).$

Translating back gives $(3 - 3,\, 1 + 4) = (0, 5).$

Thus, the correct answer is B.

Solution:

Among $1$ to $1000$ there are $\left\lfloor \tfrac{1000}{7} \right\rfloor = 142$ multiples of $7.$ Because $10 \gt 7,$ each block of $10$ consecutive integers contains one or two multiples of $7.$

If $x$ blocks contain two and the remaining $100 - x$ contain one, then $2x + (100 - x) = 142,$ so $x = 42.$

Thus, the correct answer is B.

Solution:

List the numbers in increasing order with median $m.$ The mode is $m + 2 \gt m,$ so it can only occur among the two largest entries; for it to be the unique mode, both of them must equal $m + 2.$

The mean is $m - 2,$ so the total is $5(m-2).$ With the two largest equal to $m + 2$ and the median $m,$ the two smallest sum to $5(m-2) - m - 2(m+2) = 2m - 14.$

The two smallest are distinct positive integers, so $2m - 14 \ge 1 + 2 = 3,$ giving $m \ge 9.$ With $m = 9$ the list $1, 3, 9, 11, 11$ works, so the least mode is $m + 2 = 11.$

Thus, the correct answer is D.

Solution:

Rearranging, $y^4 - 2y^2 + 1 = x^4,$ so $(y^2 - 1)^2 = (x^2)^2.$ This factors as $$(y^2 - 1 - x^2)(y^2 - 1 + x^2) = 0.$$

Thus either $y^2 - x^2 = 1,$ which is a hyperbola, or $x^2 + y^2 = 1,$ which is a circle.

Thus, the correct answer is D.

Solution:

Dividing by $2^{27},$ we need $2^{a_7 - 27} = a_7.$ The only positive integer solution is $a_7 = 32,$ since $2^{32 - 27} = 2^5 = 32.$

With common difference $d \ge 1,$ we have $a_7 = a_0 + 7d = 32$ and $a_2 = a_0 + 2d = 32 - 5d.$ To minimize $a_2$ we maximize $d;$ since $a_0 = 32 - 7d \ge 1,$ the largest choice is $d = 4$ (giving $a_0 = 4$).

Then $a_2 = 32 - 20 = 12.$

Thus, the correct answer is B.

Solution:

Place the hexagon with center at the origin: $A = (-1, \sqrt3),$ $B = (1, \sqrt3),$ $C = (2, 0),$ $D = (1, -\sqrt3),$ $E = (-1, -\sqrt3),$ $F = (-2, 0).$

Then $G = (0, \sqrt3)$ and $H = (0, -\sqrt3).$ By symmetry all four sides of $GCHF$ are equal, and $$GC = \sqrt{2^2 + (\sqrt3)^2} = \sqrt7.$$

The perimeter is $4\sqrt7.$

Thus, the correct answer is D.

Solution:

The two bases are the primitive cube roots of unity, $\omega = e^{2\pi i/3}$ and its conjugate $\omega^2 = e^{-2\pi i/3}.$ So $f(n) = \omega^n + \omega^{-n} = 2\cos\dfrac{2\pi n}{3}.$

Since $2022$ is a multiple of $3,$ $\omega^{2022} = 1,$ so $f(2022) = 1 + 1 = 2.$

Thus, the correct answer is E.

Solution:

Sort each die into low $\{1,2\},$ mid $\{3,4\},$ or high $\{5,6\};$ each has probability $\tfrac13,$ so the $3^4 = 81$ category patterns are equally likely.

We need at least one high die (a number greater than $4$) and at least two dice that are greater than $2$ (mid or high). Let $A$ be the event of at least one high and $B$ the event of at most one low die.

There are $2^4 = 16$ patterns with no high die, $9$ patterns with at most one non-low die, and $5$ patterns with neither a high die nor two non-low dice. By inclusion-exclusion the count of good patterns is $81 - 16 - 9 + 5 = 61.$

The probability is $\dfrac{61}{81}.$

Thus, the correct answer is D.

Solution:

Place the rectangle as $[0,8] \times [0,4].$ The tilted square, using the $3$-$4$-$5$ right triangles, has vertices $(4, 0),$ $(0, 3),$ $(3, 7),$ and $(7, 4).$

The entire square lies inside the rectangle except for the triangle poking above the top edge $y = 4.$ That triangle has vertices $(0.75, 4),$ $(3, 7),$ and $(7, 4),$ with area $$\dfrac12 \cdot (7 - 0.75) \cdot (7 - 4) = \dfrac{75}{8}.$$

The region inside both is $25 - \dfrac{75}{8} = \dfrac{125}{8} = 15\dfrac58.$

Thus, the correct answer is D.

Solution:

Factoring, $x^2 + 2x - 15 = (x+5)(x-3),$ so $A = (-5, 0)$ and $C = (3, 0),$ and the $y$-intercept is $B = (0, -15).$

Then $\vec{BA} = (-5, 15)$ and $\vec{BC} = (3, 15).$ Using the cross and dot products, $$\tan(\angle ABC) = \dfrac{|(-5)(15) - (15)(3)|}{(-5)(3) + (15)(15)} = \dfrac{120}{210} = \dfrac47.$$

Thus, the correct answer is E.

Solution:

Every option is odd, so only the primes $3, 5, 7$ need checking.

Option A: $2^{606} \equiv 1 \pmod 3,$ so $2^{606} - 1$ is divisible by $3.$ Option B: $2^{606} \equiv 4 \pmod 5,$ so $2^{606} + 1$ is divisible by $5.$ Option D: $2^{607} \equiv 2 \pmod 3,$ so $2^{607} + 1$ is divisible by $3.$ Option E: modulo $5,$ $2^{607} + 3^{607} \equiv 3 + 2 = 5 \equiv 0.$

For $2^{607} - 1:$ it is $\equiv 1 \pmod 3,$ $\equiv 2 \pmod 5,$ and (since $2^3 \equiv 1 \pmod 7$ and $607 \equiv 1 \pmod 3$) $\equiv 1 \pmod 7.$ So it is not divisible by any prime below $10.$

Thus, the correct answer is C.

Solution:

Let $a = \log_2 x$ and $b = \log_2 y.$ Taking $\log_2$ of $x^y = 2^{64}$ gives $y \log_2 x = 64,$ i.e. $a \cdot 2^b = 2^6.$

Taking $\log_2$ of the second equation gives $b \log_2 a = 7,$ so $a = 2^{7/b}.$ Substituting, $2^{7/b} \cdot 2^b = 2^6,$ so $b + \dfrac7b = 6,$ i.e. $b^2 - 6b + 7 = 0.$

Thus $b = 3 \pm \sqrt2,$ and the greatest value of $\log_2 y$ is $3 + \sqrt2.$

Thus, the correct answer is C.

Solution:

The row with sum $4$ is all $1$s and the column with sum $4$ is all $1$s. There are $4!$ ways to assign the row sums $1, 2, 3, 4$ to the four rows, and $4$ choices for which column has sum $4.$

Delete that column. The remaining $4 \times 3$ array has row sums $0, 1, 2, 3$ and must have column sums $1, 2, 3.$ The all-zero and all-one rows are forced; the rows of reduced sum $1$ and $2$ can be placed in $6$ ways to produce column sums $1, 2, 3$ in some order.

The total is $24 \cdot 4 \cdot 6 = 576.$

Thus, the correct answer is D.

Solution:

Only the inner $3 \times 3$ squares can start filled. For the center to be filled afterward, if it began empty it needs exactly $3$ filled neighbors, and if it began filled it needs $2$ or $3.$

Every other square must end empty. The key restriction is that no border square may acquire exactly three filled neighbors, which rules out filling all three squares along an outer edge of the $3 \times 3.$

Enumerating the arrangements subject to these conditions, one finds every valid configuration has exactly three filled cells: there are $20$ with the center initially empty and $2$ with the center initially filled, for $22$ in total.

Thus, the correct answer is C.

Solution:

Let $a = BC,$ $b = CA,$ $c = AB.$ Since $E$ is the midpoint of $AC,$ $AE = \tfrac{b}{2}.$ The centroid gives $AG = \tfrac23 m_a$ and $GE = \tfrac13 m_b,$ where $m_a, m_b$ are the medians from $A$ and $B.$

Equilateral $\triangle AGE$ means $AG = GE = AE.$ From $\tfrac23 m_a = \tfrac{b}{2}$ we get $m_a = \tfrac34 b,$ which with $m_a^2 = \tfrac{2b^2 + 2c^2 - a^2}{4}$ gives $2c^2 - a^2 = \tfrac{b^2}{4}.$ From $\tfrac13 m_b = \tfrac{b}{2}$ we get $m_b = \tfrac32 b,$ giving $a^2 + c^2 = 5b^2.$

Solving, $c^2 = \tfrac{7b^2}{4}$ and $a^2 = \tfrac{13b^2}{4}.$ Taking $b = 2$ gives $a^2 = 13,$ $c^2 = 7,$ so $$\cos C = \dfrac{a^2 + b^2 - c^2}{2ab} = \dfrac{13 + 4 - 7}{2 \cdot \sqrt{13} \cdot 2} = \dfrac{5\sqrt{13}}{26}.$$

Then $m + n + p = 5 + 26 + 13 = 44.$

Thus, the correct answer is A.

Solution:

The least-degree solution is a cubic. Write $P(x) = (x + 2) + (x^2 + x + 1)(px + q),$ which has remainder $x + 2$ upon division by $x^2 + x + 1.$

Reducing modulo $x^2 + 1$ (so $x^2 \equiv -1$) gives remainder $(q + 1)x + (2 - p).$ Setting this equal to $2x + 1$ gives $q = 1$ and $p = 1.$

Then $P(x) = x^3 + 2x^2 + 3x + 3,$ and the sum of the squares of the coefficients is $1 + 4 + 9 + 9 = 23.$

Thus, the correct answer is E.

Solution:

The first two circles are concentric with radii $2$ and $8.$ A circle tangent to both either has radius $3$ with center at distance $5$ from the origin, or radius $5$ with center at distance $3$ from the origin.

The third circle has center $(5, 0)$ and radius $\sqrt3.$ Imposing tangency with it, exactly four of the radius-$3$ circles and four of the radius-$5$ circles work (two tangency types, each giving a symmetric pair).

The sum of the areas is $4 \cdot \pi(3)^2 + 4 \cdot \pi(5)^2 = 36\pi + 100\pi = 136\pi.$

Thus, the correct answer is E.

Solution:

Because each $t_n \lt 1,$ Amelia always completes at least two steps. She stops after exactly two steps when $t_1 + t_2 \gt 1,$ which happens with probability $\tfrac12;$ otherwise she takes all three steps.

The increments are independent of the times. If she takes two steps, her position is $x_1 + x_2,$ and $P(x_1 + x_2 \gt 1) = \tfrac12.$ If she takes three, her position is $x_1 + x_2 + x_3,$ and $P(x_1 + x_2 + x_3 \gt 1) = 1 - \tfrac16 = \tfrac56.$

The answer is $\tfrac12 \cdot \tfrac12 + \tfrac12 \cdot \tfrac56 = \tfrac14 + \tfrac{5}{12} = \tfrac23.$

Thus, the correct answer is C.

Solution:

Since $S_n$ is the integer formed by the low $n$ bits, the condition $7S_n \equiv 1$ means $S_n \equiv 7^{-1} \pmod{2^n}$ for every $n.$ Thus the digits $x_k$ are the base-$2$ digits of $\tfrac17$ as a $2$-adic number.

Long division in base $2$ gives digits $x_0 = x_1 = x_2 = 1,$ and thereafter the block repeats with period $3:$ for $k \ge 1,$ $x_k = 0$ exactly when $3 \mid k,$ and $x_k = 1$ otherwise.

Since $3 \mid 2019$ and $3 \mid 2022,$ while $2020 \equiv 1$ and $2021 \equiv 2 \pmod 3,$ we get $x_{2019} = 0,$ $x_{2020} = 1,$ $x_{2021} = 1,$ $x_{2022} = 0.$ The sum is $0 + 2 + 4 + 0 = 6.$

Thus, the correct answer is A.

Solution:

A chord joining two vertices $d$ steps apart has squared length $2 - 2\cos\dfrac{2\pi d}{7},$ and there are $7$ chords for each of $d = 1, 2, 3.$ The required sum is $$7 \sum_{d=1}^{3} \left(2 - 2\cos\tfrac{2\pi d}{7}\right)^2.$$

Using $\displaystyle\sum_{d=1}^{3} \cos\tfrac{2\pi d}{7} = -\tfrac12$ and $\displaystyle\sum_{d=1}^{3} \cos^2\tfrac{2\pi d}{7} = \tfrac54,$ the inner sum expands to $4\left(3 + 1 + \tfrac54\right) = 21.$

Therefore the total is $7 \cdot 21 = 147.$

Thus, the correct answer is C.

Solution:

Center the square at the origin with vertices $\left(\pm\tfrac12, \pm\tfrac12\right).$ Each hexagon shares one edge with the square and extends across to the opposite side; the hexagon on the bottom edge, for instance, has its far (top) edge from $\left(-\tfrac12, \sqrt3 - \tfrac12\right)$ to $\left(\tfrac12, \sqrt3 - \tfrac12\right).$

The outer boundary is a $12$-gon with flat edges at distance $\sqrt3 - \tfrac12$ from the center, convex vertices such as $\left(\sqrt3 - \tfrac12, \tfrac12\right),$ and four reflex notches where adjacent hexagons' slanted edges meet, at $\left(\tfrac52 - \sqrt3, \tfrac52 - \sqrt3\right)$ and its symmetric images.

Applying the shoelace formula to these $12$ vertices gives area $16\sqrt3 - 23,$ so $m = 16,$ $n = 3,$ $p = -23,$ and $m + n + p = -4.$

Thus, the correct answer is B.