They ride toward each other, so their speeds add. That closes the $45$-mile gap at $18 + 12 = 30$ mph, and they meet after $45/30 = 1.5$ hours. Alice starts at $A,$ so by then she's gone $18 \cdot 1.5 = 27$ miles. Thus, E is the correct answer.

Let $p$ be the pizza's weight. A cup of orange slices is $\frac14$ pound, so the two sides balance as $\frac13 p + \frac72 \cdot \frac14 = \frac34 p + \frac12 \cdot \frac14,$ that is $\frac13 p + \frac78 = \frac34 p + \frac18.$ Collect the pizza terms: $\frac68 = \left(\frac34 - \frac13\right)p = \frac{5}{12}p.$ So $p = \frac34 \cdot \frac{12}{5} = \frac95 = 1\frac45.$ Therefore, the answer is A.

If a perfect square is divisible by $5,$ it's divisible by $25,$ so it looks like $(5k)^2 = 25k^2.$ We need $25k^2 \lt 2023,$ i.e. $k^2 \lt 80.9.$ That allows $k = 1, 2, \ldots, 8,$ which is $8$ squares. Thus, A is the correct answer.

In any quadrilateral each side is shorter than the sum of the other three. Call the longest side $s.$ The rest sum to $26 - s,$ so $s \lt 26 - s,$ which gives $s \lt 13$ and hence $s \le 12.$ Can we hit $12?$ The sides $4, 12, 9, 1$ work, since $12 \lt 4 + 9 + 1.$ So the greatest length is $12.$ Therefore, the answer is D.

Factor everything into primes. $8^5 \cdot 5^{10} \cdot 15^5$ $= 2^{15} \cdot 5^{10} \cdot 3^5 \cdot 5^5$ $= 2^{15} \cdot 5^{15} \cdot 3^5$ $= 10^{15} \cdot 243.$ That's $243$ followed by $15$ zeros, so it has $3 + 15 = 18$ digits. Thus, E is the correct answer.

Count by incidences. Each edge lies on $2$ faces, so the six face values together are $2$ times the total of all edge values. Each vertex lies on $3$ edges, so the total edge value is $3$ times the vertex sum. Chaining these, the cube's value is $2 \cdot 3 \cdot 21 = 126.$ Therefore, the answer is D.

The total can only reach exactly $3$ through the opening rolls, and these ways are disjoint: $3$ alone (probability $\frac16$), then $1,2$ and $2,1$ (each $\frac1{36}$), and $1,1,1$ (probability $\frac1{216}$). Add them up: $\frac{36}{216} + \frac{6}{216} + \frac{6}{216} + \frac{1}{216} = \frac{49}{216}.$ Thus, B is the correct answer.

The Breadus reading is linear in Fahrenheit through $(110, 0)$ and $(350, 100),$ so $B = \frac{100}{350 - 110}(F - 110)$ $= \frac{5}{12}(F - 110).$ Plug in $F = 200:$ $B = \frac{5}{12} \cdot 90 = 37.5.$ Therefore, the answer is D.

The year $2023$ already gives two $2$s (even), one $0$, and one $3$. So to make every digit occur an even number of times, the four digits of $MM$ and $DD$ must supply one more $0$, one more $3$, and two equal digits, while keeping the number of $2$s even. Checking the legal dates gives $01/13,01/31,02/23$, $03/11,03/22,10/13$, and $10/31,11/03,11/30$, exactly $9$ dates. Thus, E is the correct answer.

Let $m$ be the current mean over $n$ quizzes. One more $11$ makes the mean $m + 1:$ $\frac{mn + 11}{n + 1} = m + 1,$ which tidies up to $m + n = 10.$ Three more $11$s make it $m + 2:$ $\frac{mn + 33}{n + 3} = m + 2,$ i.e. $3m + 2n = 27.$ Solve the pair and $m = 7.$ Therefore, the answer is D.

Each corner right triangle has legs $a$ and $b.$ A side of the outer square gives $a+b=\sqrt3,$ and a side of the inscribed square gives $a^2+b^2=2.$ Subtract to find the product: $2ab=(a+b)^2-(a^2+b^2)$ $=3-2$ $=1,$ so $ab=\tfrac12.$ Then $a$ and $b$ are the roots of $t^2-\sqrt3\,t+\tfrac12=0,$ namely $\tfrac{\sqrt3\pm1}{2}.$ The ratio of the smaller leg to the larger is $\dfrac{\sqrt3-1}{\sqrt3+1}$ $=\dfrac{(\sqrt3-1)^2}{2}$ $=2-\sqrt3.$ Thus, C is the correct answer.

When we reverse $N,$ its last digit is the first digit of $N.$ For the reversal to be divisible by $5,$ that digit is $0$ or $5.$ A three-digit number can't start with $0,$ so $N$ starts with $5,$ meaning $500 \le N \le 599$ (and the reversal ends in $5,$ always fine). Now just count multiples of $7$ here: from $7 \cdot 72 = 504$ to $7 \cdot 85 = 595,$ that's $14$ numbers. Therefore, the answer is B.

Let $A$ be Abdul, $C$ be Chiang with $AC = 48,$ and $B$ be Bharat with $\angle B = 60^\circ.$ Every point seeing $AC$ at $60^\circ$ lies on one circular arc, so all valid $B$ sit on a circle where chord $AC$ subtends $60^\circ.$ The law of sines gives its diameter, $\frac{AC}{\sin 60^\circ} = \frac{48}{\sqrt3/2} = 32\sqrt3.$ Now $AB$ is a chord, and a chord is longest when it's a diameter. So $AB = 32\sqrt3$ and $AB^2 = 1024 \cdot 3 = 3072.$ Thus, C is the correct answer.

A number $n \le 100$ can only have a divisor divisible by $11$ when $11 \mid n,$ so $n \in \{11, 22, \ldots, 99\}.$ Write $n = 11m$ with $m \le 9.$ Here $11 \nmid m,$ so $d(11m) = 2\,d(m),$ and the divisors that are multiples of $11$ are exactly the $d(m)$ numbers $11d.$ That makes the chance $\frac{d(m)}{2\,d(m)} = \frac12$ for each such $n.$ Averaging over all $100$ starting numbers, the probability is $\frac{1}{100}\sum_{m=1}^{9}\frac12 = \frac{9}{200}.$ Therefore, the answer is B.

A circle of radius $r$ has area $\pi r^2.$ So the shaded ring between radius $2k$ and $2k-1$ has area $\pi\big((2k)^2 - (2k-1)^2\big)$ $= (4k-1)\pi.$ With $2n$ circles the shaded total is $\pi\sum_{k=1}^{n}(4k-1) = \pi(2n^2 + n).$ We want $2n^2 + n \ge 2023.$ At $n = 31$ it's $1953,$ at $n = 32$ it's $2080.$ So $n = 32,$ which means $2n = 64$ circles. Thus, E is the correct answer.

Say there are $L$ left-handers and $2L$ right-handers, so $3L$ players and $\binom{3L}{2}$ games. Every game has one winner, and left wins are $1.4$ times right wins, so the wins split $7 : 5$ and the total must be a multiple of $12$. Left-handers can win at most all games involving at least one left-hander, namely $\binom{L}{2} + 2L^2$. Hence $$\frac{7}{12}\binom{3L}{2} \leq \binom{L}{2} + 2L^2,$$ which gives $L \leq 3$. For $L = 1$ and $L = 2$, the total is not divisible by $12$. For $L = 3$, there are $\binom{9}{2} = 36$ games. This is attainable if the left-handers win all $18$ cross-group games and all $3$ games among themselves, while the right-handers win their $15$ internal games. The win totals are $21$ and $15$, so the answer is $36$. Therefore, the answer is B.

Set $A = (0,0)$, $B = (30,0)$, $C = (30,28)$, $D = (0,28)$, with $P = (30, p)$ on $BC$ and $Q = (30 - q, 28)$ on $CD$. The three right triangles give $AP = \sqrt{30^2 + p^2}$, $QA = \sqrt{28^2 + (30 - q)^2}$, and $PQ = \sqrt{(28 - p)^2 + q^2}$. For $0 \leq p \leq 28$, the equation $(AP-p)(AP+p)=900$ gives only $p=0$ and $p=16$, with $AP=30$ and $AP=34$. Similarly, setting $x=30-q$, the equation $(QA-x)(QA+x)=784$ with $0 \leq x \leq 30$ gives $x=0$ or $x=21$. Testing these four combinations in the formula for $PQ$, only $p=16$, $x=21$ works. Thus $q=9$, $QA=35$, and $PQ=\sqrt{12^2+9^2}=15$. The perimeter of $\triangle APQ$ is $34+15+35=84$. Thus, A is the correct answer.

Each rhombus has $4$ edges, and every edge is shared by $2$ faces, so $E = \frac{12 \cdot 4}{2} = 24.$ With $F = 12,$ Euler's formula gives $V = 2 - F + E = 14.$ Suppose $x$ vertices have $3$ edges and the other $14 - x$ have $4.$ The degrees sum to twice the edge count: $3x + 4(14 - x) = 2E = 48,$ so $x = 8.$ Therefore, the answer is D.