Answer: E

Solution:

The gap between them closes at $18+12=30$ miles per hour, so they meet after $\dfrac{45}{30}=1.5$ hours.

In that time Alicia has ridden $18\cdot 1.5=27$ miles from City $A.$

Thus, the correct answer is E.

Answer: A

Solution:

Let the pizza weigh $P$ pounds. Then $3\tfrac12$ cups weigh $\tfrac72\cdot\tfrac14=\tfrac78$ and $\tfrac12$ cup weighs $\tfrac12\cdot\tfrac14=\tfrac18.$

The equation is $$\tfrac13 P+\tfrac78=\tfrac34 P+\tfrac18.$$

Subtracting gives $\tfrac78-\tfrac18=\left(\tfrac34-\tfrac13\right)P,$ so $\tfrac34=\tfrac{5}{12}P$ and $P=\dfrac{9}{5}=1\tfrac45.$

Thus, the correct answer is A.

Answer: A

Solution:

A perfect square is divisible by $5$ only if its root is, so the squares are $(5k)^2=25k^2.$

Since $44^2=1936\lt 2023\lt 2025=45^2,$ the root can be $5,10,\ldots,40,$ which is $k=1$ through $8.$

Thus, the correct answer is A.

Answer: E

Solution:

Writing everything in primes, $$8^5\cdot 5^{10}\cdot 15^5=2^{15}\cdot 5^{10}\cdot 3^5\cdot 5^5=3^5\cdot 2^{15}\cdot 5^{15}.$$

This equals $3^5\cdot 10^{15}=243\cdot 10^{15},$ which is $243$ followed by $15$ zeros, for a total of $18$ digits.

Thus, the correct answer is E.

Answer: B

Solution:

The running total is increasing, so it hits $3$ exactly when one of these disjoint openings occurs: a first roll of $3;$ rolls $1,2;$ rolls $2,1;$ or rolls $1,1,1.$

Their probabilities are $$\dfrac16+\dfrac1{36}+\dfrac1{36}+\dfrac1{216} =\dfrac{36+6+6+1}{216}=\dfrac{49}{216}.$$

Thus, the correct answer is B.

Answer: D

Solution:

Let the $x$-coordinates be $x_1$ and $x_2.$ The midpoint gives $x_1+x_2=12,$ and the average of the $y$-values gives $\log_2 x_1+\log_2 x_2=4,$ so $x_1x_2=2^4=16.$

Then $$|x_1-x_2|=\sqrt{(x_1+x_2)^2-4x_1x_2}=\sqrt{144-64}=\sqrt{80}=4\sqrt5.$$

Thus, the correct answer is D.

Answer: E

Solution:

The year contributes the digits $2,0,2,3,$ so $2$ appears twice while $0$ and $3$ each appear once. For every digit to end up with an even count, the four digits of the month and day must supply an odd number of $0$'s, an odd number of $3$'s, and an even number of every other digit.

With only four digits available, the month-day string must use exactly one $0,$ one $3,$ and a repeated pair of some digit. Checking valid months and days leaves nine dates: $01\text{-}13,$ $01\text{-}31,$ $02\text{-}23,$ $03\text{-}11,$ $03\text{-}22,$ $10\text{-}13,$ $10\text{-}31,$ $11\text{-}03,$ and $11\text{-}30.$

Thus, the correct answer is E.

Answer: D

Solution:

Let the current mean be $m$ over $n$ quizzes, so the total is $S=mn.$ Adding one $11$ gives $$\dfrac{mn+11}{n+1}=m+1,$$ which simplifies to $m+n=10.$

Adding three $11$'s gives $$\dfrac{mn+33}{n+3}=m+2,$$ which simplifies to $3m+2n=27.$

Solving $m+n=10$ and $3m+2n=27$ gives $m=7.$

Thus, the correct answer is D.

Answer: C

Solution:

The outer square has side $\sqrt3$ and the inner square has side $\sqrt2.$ Each triangle is right, with legs $p$ and $q$ along an outer side, so $p+q=\sqrt3,$ and with hypotenuse an inner side, so $p^2+q^2=2.$

Then $(p+q)^2=3$ gives $2pq=1,$ so $p$ and $q$ are the roots of $t^2-\sqrt3\,t+\tfrac12=0,$ namely $\dfrac{\sqrt3\pm 1}{2}.$

The ratio of shorter to longer leg is $$\dfrac{\sqrt3-1}{\sqrt3+1} =\dfrac{(\sqrt3-1)^2}{2}=2-\sqrt3.$$

Thus, the correct answer is C.

Answer: D

Solution:

From $(y-x)^2=4y^2$ we get $y-x=\pm 2y.$ The choice $y-x=2y$ gives $x=-y\lt 0,$ impossible, so $y-x=-2y,$ meaning $x=3y.$

Substituting into $y^3=x^2=9y^2$ gives $y=9,$ hence $x=27$ and $x+y=36.$

Thus, the correct answer is D.

Answer: C

Solution:

The tangent of the angle between the lines is $$\left|\dfrac{2-\tfrac13}{1+2\cdot\tfrac13}\right| =\left|\dfrac{\tfrac53}{\tfrac53}\right|=1.$$

The acute angle with tangent $1$ is $45^\circ.$

Thus, the correct answer is C.

Answer: D

Solution:

Group into pairs $(2k)^3-(2k-1)^3$ for $k=1,\ldots,9.$ Expanding, $$(2k)^3-(2k-1)^3=12k^2-6k+1.$$

Summing for $k=1$ to $9,$ with $\sum k^2=285$ and $\sum k=45,$ gives $$12\cdot 285-6\cdot 45+9=3420-270+9=3159.$$

Thus, the correct answer is D.

Answer: B

Solution:

Let there be $L$ left-handed and $2L$ right-handed players, for $3L$ players and $\binom{3L}{2}$ games total.

If right-handers win $R$ games, left-handers win $1.4R,$ so the total is $2.4R=\tfrac{12}{5}R.$ For this to be an integer count, the total number of games must be a multiple of $12.$

Testing $L=1,2,3$ gives totals $3,15,36;$ only $36$ is a multiple of $12,$ and it is achievable (the $3$ left-handers can take all $18$ mixed games plus their $3$ internal games for $21=1.4\cdot 15$ wins).

Thus, the correct answer is B.

Answer: E

Solution:

Taking magnitudes gives $|z|^5=|z|,$ so $|z|=0$ or $|z|=1.$ The value $z=0$ works, giving one solution.

If $|z|=1,$ multiply the equation by $z$ to get $z^6=z\overline{z}=|z|^2=1.$ This has $6$ distinct roots, all of modulus $1.$

Altogether there are $1+6=7$ solutions.

Thus, the correct answer is E.

Answer: A

Solution:

Every segment of the zigzag spans the full width $30,$ so it has length $\dfrac{30}{\sin\theta}$ and advances $\dfrac{30}{\tan\theta}$ horizontally.

Summing over all segments, the total length is $120$ and the total horizontal advance is $100.$ Their ratio is $$\dfrac{\text{length}}{\text{horizontal}} =\dfrac{1/\sin\theta}{1/\tan\theta}=\dfrac{1}{\cos\theta}=\dfrac{120}{100}.$$

Therefore $\cos\theta=\dfrac56,$ so $\theta=\arccos\dfrac56.$

Thus, the correct answer is A.

Answer: B

Solution:

Write $z=x+yi.$ Then $1+z+z^2=(1+x+x^2-y^2)+y(1+2x)i,$ and the constraint is $$(1+x+x^2-y^2)^2+y^2(1+2x)^2=16.$$

Setting the derivative of $y$ with respect to $x$ to zero factors as $(1+2x)\bigl(P+2y^2\bigr)=0,$ where $P=1+x+x^2-y^2.$ The factor $P+2y^2=0$ is impossible for real $x,y,$ so $x=-\tfrac12.$

Then $1+2x=0,$ so the constraint reduces to $\left(\tfrac34-y^2\right)^2=16.$ Taking $\tfrac34-y^2=-4$ gives $y^2=\tfrac{19}{4},$ so the maximum is $y=\dfrac{\sqrt{19}}{2}.$

Here $m=19$ and $n=2,$ so $m+n=21.$

Thus, the correct answer is B.

Answer: E

Solution:

Let $a_n$ be the probability that Flora ever lands exactly on $n,$ with $a_0=1.$ Conditioning on the first jump, $$a_n=\sum_{k=1}^{n}\dfrac{1}{2^k}\,a_{n-k}.$$

Then $a_1=\tfrac12,$ $a_2=\tfrac12,$ and by induction $a_n=\tfrac12$ for all $n\ge 1:$ each new term averages the previous values, all equal to $\tfrac12.$

Hence the probability of landing on $10$ is $\dfrac12.$

Thus, the correct answer is E.

Answer: D

Solution:

Put the centers at $O_1=\left(-\tfrac14,0\right)$ and $O_2=\left(\tfrac14,0\right).$ By symmetry $C_3$ is centered at the origin, and internal tangency to $C_1$ gives radius $1-\tfrac14=\tfrac34.$

Let $C_4$ have radius $r,$ centered at $(0,k)$ on the axis of symmetry. External tangency to $C_3$ gives $k=\tfrac34+r,$ and internal tangency to $C_1$ gives $\sqrt{\tfrac{1}{16}+k^2}=1-r.$

Substituting, $\tfrac{1}{16}+\left(\tfrac34+r\right)^2=(1-r)^2,$ which simplifies to $\tfrac72 r=\tfrac38,$ so $r=\dfrac{3}{28}.$

Thus, the correct answer is D.

Answer: C

Solution:

Let $a=\log_{2023}7$ and $b=\log_{2023}289.$ Since $2023=7\cdot 289,$ we have $a+b=1.$ Writing $t=\log_{2023}x,$ each logarithm becomes a reciprocal, and the equation turns into $$(1+t)=(a+t)(b+t).$$

Expanding and using $a+b=1,$ the linear terms cancel, leaving $t^2+(ab-1)=0.$ Its two roots satisfy $t_1+t_2=0.$

The corresponding solutions multiply to $x_1x_2=2023^{t_1}\cdot 2023^{t_2}=2023^{\,t_1+t_2} =2023^0=1.$

Thus, the correct answer is C.

Answer: C

Solution:

Let $S_n$ be the sum of row $n.$ Each interior entry is $1$ more than the sum of the two entries above it, and summing over the row gives the recurrence $$S_n=2S_{n-1}+(n-2).$$

With $S_1=1,$ this solves to $S_n=2^n-n$ (check: $S_5=32-5=27=1+7+11+7+1$).

So $S_{2023}=2^{2023}-2023.$ Since powers of $2$ cycle with units digits $2,4,8,6$ and $2023\equiv 3\pmod 4,$ $2^{2023}$ ends in $8.$ Then $8-3=5$ gives units digit $5.$

Thus, the correct answer is C.

Answer: A

Solution:

Fix $R.$ Among the other $11$ vertices of the icosahedron, $5$ are at distance $1,$ $5$ are at distance $2,$ and $1$ (the antipode) is at distance $3.$

Choosing ordered distinct $Q,S,$ the probability that $d(Q,R)=d(R,S)$ is $$\dfrac{5\cdot 4+5\cdot 4}{11\cdot 10}=\dfrac{40}{110}=\dfrac{4}{11}.$$

By the symmetry between $Q$ and $S,$ $$P(d(Q,R)\gt d(R,S))=\dfrac{1-\tfrac{4}{11}}{2} =\dfrac{7}{22}.$$

Thus, the correct answer is A.

Answer: B

Solution:

Setting $n=1$ gives $f(1)=1.$ For a prime $p,$ $n=p$ gives $f(p)+p\cdot f(1)=1,$ so $f(p)=1-p.$ For $n=p^2,$ $f(p^2)+p\,f(p)+p^2 f(1)=1$ gives $f(p^2)=1-p.$

Since the defining relation is a Dirichlet convolution of multiplicative functions, $f$ is multiplicative. With $2023=7\cdot 17^2,$ $$f(2023)=f(7)\cdot f(17^2)=(1-7)(1-17)=(-6)(-16)=96.$$

Thus, the correct answer is B.

Answer: B

Solution:

By AM-GM, $1+2a\ge 2\sqrt{2a},$ $2+2b\ge 4\sqrt{b},$ and $2a+b\ge 2\sqrt{2ab}.$ Multiplying, $$(1+2a)(2+2b)(2a+b)\ge 16\sqrt{2a}\cdot\sqrt{b}\cdot\sqrt{2ab}=32ab.$$

Equality requires $1=2a,$ $2=2b,$ and $2a=b$ simultaneously. These give $a=\tfrac12,$ $b=1,$ which are consistent, so there is exactly one solution.

Thus, the correct answer is B.

Answer: C

Solution:

For a fixed length $n,$ each element of $\{1,\ldots,10\}$ independently either never appears or first appears in one of $A_1,\ldots,A_n,$ giving $n+1$ choices. Hence there are $(n+1)^{10}$ chains of length $n.$

Summing, $$K=\sum_{n=1}^{10}(n+1)^{10}=\sum_{k=2}^{11}k^{10}.$$ Modulo $10,$ the terms $k=2,\ldots,11$ reduce to $4,9,6,5,6,9,4,1,0,1,$ which sum to $45\equiv 5.$

Thus, the correct answer is C.

Answer: C

Solution:

By De Moivre, $(\cos x+i\sin x)^{2023}=\cos 2023x+i\sin 2023x.$ Expanding the left side and taking the ratio of imaginary to real parts gives $\tan 2023x$ as the stated rational function of $\tan x$ after dividing numerator and denominator by $\cos^{2023}x.$

The coefficient $a_{2023}$ is the coefficient of $\tan^{2023}x$ in the numerator, which comes from the $k=2023$ term: $$a_{2023}=(-1)^{(2023-1)/2}\binom{2023}{2023}=(-1)^{1011}=-1.$$

Thus, the correct answer is C.