Answer: D

Solution:

Simplify from the bottom. The innermost fraction is $3+\dfrac13=\dfrac{10}{3}.$

The next layer is $3+\dfrac{1}{10/3}=3+\dfrac{3}{10}=\dfrac{33}{10}.$

Finally, $3+\dfrac{1}{33/10}=3+\dfrac{10}{33}=\dfrac{109}{33}.$

Thus, the correct answer is D.

Answer: E

Solution:

Let the third number be $t.$ Then the first is $6t$ and the second is $t+40.$ Their sum is $$6t+(t+40)+t=8t+40=96,$$ so $t=7.$

The first number is $42$ and the second is $47,$ so the difference has absolute value $5.$

Thus, the correct answer is E.

Answer: B

Solution:

The five areas are $6,\ 8,\ 30,\ 14,$ and $6,$ which sum to $64.$ So the square is $8\times8.$

Placing $C$ ($5\times6$) across the top left, $D$ ($2\times7$) up the right side, $E$ ($2\times3$) in the lower left, and $A$ ($1\times6$) along the bottom leaves a central $2\times4$ gap, which is exactly rectangle $B.$

Thus, the correct answer is B.

Answer: B

Solution:

Since $180=2^2\cdot3^2\cdot5$ and $18=2\cdot3^2,$ the condition $\operatorname{lcm}(n,18)=180$ forces $n$ to contribute $2^2$ and $5,$ with its power of $3$ at most $2.$

From $\gcd(n,45)=\gcd(n,3^2\cdot5)=15=3\cdot5,$ the power of $3$ in $n$ is exactly $1$ and the power of $5$ is at least $1.$

Therefore $n=2^2\cdot3\cdot5=60,$ whose digits sum to $6.$

Thus, the correct answer is B.

Answer: C

Solution:

For each $k\ge1,$ the set $|x|+|y|=k$ contains exactly $4k$ lattice points, and $k=0$ gives the single origin.

The total is $$1+\sum_{k=1}^{20}4k=1+4\cdot\frac{20\cdot21}{2}=1+840=841.$$

Thus, the correct answer is C.

Answer: D

Solution:

The known numbers sum to $20,$ so the mean is $\dfrac{20+X}{6},$ which must equal an element of the set.

Setting it to $5$ gives $X=10;$ to $7$ gives $X=22;$ and to $X$ itself gives $20+X=6X,$ so $X=4.$ Values $1$ and $2$ give negative $X.$

The positive values are $10,22,4,$ summing to $36.$

Thus, the correct answer is D.

Answer: D

Solution:

The bottom-middle region shares a border with all four other regions. Color it first in $5$ ways.

The top-left region borders it, giving $4$ choices. Each of the three remaining regions borders exactly two already-colored regions, which have different colors, leaving $3$ choices apiece.

The total is $5\cdot4\cdot3\cdot3\cdot3=540.$

Thus, the correct answer is D.

Answer: A

Solution:

The $k$th factor is $10$ raised to the $k$-fold cube root, namely $10^{1/3^k}.$

The product is $10$ raised to $$\frac13+\frac19+\frac1{27}+\cdots=\frac{1/3}{1-1/3}=\frac12.$$

So the value is $10^{1/2}=\sqrt{10}.$

Thus, the correct answer is A.

Answer: A

Solution:

To "Are you a truth-teller?" the truth-tellers and liars both answer yes, and only alternaters who lie on this question answer yes. To "Are you an alternater?" the liars answer yes, and among alternaters only those telling the truth on this question answer yes. To "Are you a liar?" only alternaters lying on this question answer yes.

Split the alternaters by first response. Those starting with a lie answer (lie, truth, lie), so they say yes to all three questions; those starting truthful answer (truth, lie, truth) and say yes to none of the three. The $9$ yeses on the last question are exactly the lie-first alternaters, so there are $9$ of them.

The second question's $15$ yeses are the liars plus these $9,$ so there are $6$ liars. The first question's $22$ yeses are truth-tellers plus liars plus the $9,$ so the truth-tellers number $22-6-9=7.$

Truth-tellers answer yes only to the first question, receiving one candy each, for $7\cdot1=7$ pieces.

Thus, the correct answer is A.

Answer: E

Solution:

Any number $8$ or larger cannot be a smaller element (its double exceeds $14$), so $8$–$14$ are all larger elements and $1$–$7$ are all smaller elements.

Match each smaller $s$ to a larger $g\ge2s.$ Processing from the most restrictive: $s=7$ forces $g=14$ ($1$ way); then $s=6$ has $\{12,13\}$ left ($2$); $s=5$ has $3;$ $s=4$ has $4;$ $s=3$ has $3;$ $s=2$ has $2;$ $s=1$ has $1.$

The number of matchings is $1\cdot2\cdot3\cdot4\cdot3\cdot2\cdot1=144.$

Thus, the correct answer is E.

Answer: E

Solution:

The right-hand distance is $|\log_6 10-1|=\log_6\dfrac53,$ so twice it is $2\log_6\dfrac53=\log_6\dfrac{25}{9}.$

Thus $\left|\log_6\dfrac{x}{9}\right|=\log_6\dfrac{25}{9},$ giving $\dfrac{x}{9}=\dfrac{25}{9}$ or $\dfrac{x}{9}=\dfrac{9}{25},$ so $x=25$ or $x=\dfrac{81}{25}.$

Their product is $25\cdot\dfrac{81}{25}=81.$

Thus, the correct answer is E.

Answer: B

Solution:

Take edge length $1.$ Since $M$ is the midpoint of $\overline{AB},$ segments $CM$ and $DM$ are altitudes of the equilateral faces, each of length $\dfrac{\sqrt3}{2}.$ Also $CD=1.$

By the Law of Cosines in $\triangle CMD,$ $$\cos(\angle CMD)=\frac{\frac34+\frac34-1}{2\cdot\frac34}=\frac{1/2}{3/2}=\frac13.$$

Thus, the correct answer is B.

Answer: A

Solution:

Adding a disk of radius $1$ to every point of the segment sweeps out all points within distance $1$ of it. The segment from $3$ to $4i$ has length $\sqrt{3^2+4^2}=5.$

This "stadium" is a $5\times2$ rectangle plus two half-disks of radius $1,$ with area $$5\cdot2+\pi(1)^2=10+\pi\approx13.14.$$

The closest integer is $13.$

Thus, the correct answer is A.

Answer: C

Solution:

Let $u=\log 2.$ Then $\log 5=1-u,$ $\log 20=1+u,$ $\log 8=3u,$ and $\log 0.25=-2u.$

With $a=1-u,\ b=1+u,$ we have $a+b=2$ and $ab=1-u^2,$ so $$a^3+b^3=(a+b)\big((a+b)^2-3ab\big)=2\big(4-3(1-u^2)\big)=2+6u^2.$$

The last term is $(3u)(-2u)=-6u^2,$ so the total is $2+6u^2-6u^2=2.$

Thus, the correct answer is C.

Answer: D

Solution:

Let the roots be $r,s,t.$ By Vieta's formulas, $r+s+t=\dfrac{39}{10},$ $rs+rt+st=\dfrac{29}{10},$ and $rst=\dfrac{6}{10}=\dfrac35.$

The new volume is $$(r+2)(s+2)(t+2)=rst+2(rs+rt+st)+4(r+s+t)+8=\frac35+\frac{58}{10}+\frac{156}{10}+8=30.$$

Thus, the correct answer is D.

Answer: D

Solution:

Square triangular numbers satisfy the recurrence $N_k=34N_{k-1}-N_{k-2}+2.$ Starting from $N_2=36$ and $N_3=1225,$ the next term is $N_4=34\cdot1225-36+2=41616.$

Indeed $41616=204^2=\dfrac{288\cdot289}{2},$ so it is both a perfect square and $t_{288}.$

The sum of its digits is $4+1+6+1+6=18.$

Thus, the correct answer is D.

Answer: A

Solution:

Since $\sin x\ne0$ on $(0,\pi),$ divide by $\sin x:$ $$a(1+2\cos x)=4\cos^2 x-1=(2\cos x-1)(2\cos x+1).$$

When $\cos x=-\tfrac12$ (that is, $x=\tfrac{2\pi}{3}$) both sides vanish, so this is a solution for every $a.$ Otherwise we may cancel $1+2\cos x$ to get $a=2\cos x-1,$ i.e. $\cos x=\dfrac{a+1}{2}.$

This yields a second solution in $(0,\pi)$ exactly when $-1\lt\dfrac{a+1}{2}\lt1,$ that is $a\in(-3,1),$ and it is distinct from $x=\tfrac{2\pi}{3}$ unless $a=-2.$

So more than one solution occurs for $a\in(-3,-2)\cup(-2,1),$ giving $p+q+r=-3-2+1=-4.$

Thus, the correct answer is A.

Answer: A

Solution:

Rotating a point at angle $\theta$ by $k^\circ$ gives $\theta+k,$ and reflecting across the $y$-axis sends angle $\phi$ to $180-\phi.$ So $T_k$ sends $\theta$ to $(180-k)-\theta.$

Starting from $(1,0)$ at angle $0,$ applying $T_1,T_2,\ldots$ gives angles $179,-1,178,-2,177,\ldots.$ After an even number $2m$ of steps the angle is $-m,$ and after an odd number $2m+1$ it is $179-m.$

For the point to return, the angle must be a multiple of $360^\circ.$ The even case needs $m=360,$ i.e. $n=720.$ The odd case needs $179-m=0,$ i.e. $m=179$ and $n=359,$ where the net reflection fixes $(1,0).$

The least such $n$ is $359.$

Thus, the correct answer is A.

Answer: D

Solution:

Let $\text{pos}(k)$ be the position of card $k.$ A fresh pass is needed exactly when $\text{pos}(k+1)\lt\text{pos}(k),$ so the number of passes is one more than the number of descents in the sequence $\text{pos}(1),\text{pos}(2),\ldots,\text{pos}(13).$

Exactly two passes means exactly one descent. The number of permutations of $13$ elements with exactly one descent is the Eulerian number $$\left\langle{13\atop1}\right\rangle=2^{13}-13-1=8178.$$

Thus, the correct answer is D.

Answer: B

Solution:

Place the trapezoid symmetric about the $y$-axis: $A=(-p,0),$ $D=(p,0),$ $B=(-q,h),$ $C=(q,h),$ with $P=(x,y).$

Then $PA^2-PD^2=4px=1-16=-15$ and $PB^2-PC^2=4qx=4-9=-5.$ Dividing gives $\dfrac{p}{q}=3.$

Since $AD=2p$ and $BC=2q,$ we get $\dfrac{BC}{AD}=\dfrac{q}{p}=\dfrac13.$

Thus, the correct answer is B.

Answer: E

Solution:

If $\zeta$ is a root of a divisor, then $P(\zeta)=\zeta^{2022}+\zeta^{1011}+1=0$ is required.

The roots of $x^6+x^3+1$ are the primitive $9$th roots of unity, so $\zeta^9=1.$ Reducing exponents modulo $9,$ $2022\equiv6$ and $1011\equiv3,$ giving $\zeta^6+\zeta^3+1.$ Here $\omega=\zeta^3$ is a primitive cube root of unity, so this equals $\omega^2+\omega+1=0.$

The other four options fail: substituting their roots yields nonzero values (for instance, the primitive cube roots of unity give $P=3$).

Thus, the correct answer is E.

Answer: A

Solution:

If the roots are non-real, then $z_1=\sqrt{10}\,e^{i\theta}$ and $z_2=\overline{z_1},$ since $z_1z_2=10.$ Then $\dfrac{1}{z_1}=\dfrac{1}{\sqrt{10}}e^{-i\theta}$ and $\dfrac{1}{z_2}=\dfrac{1}{\sqrt{10}}e^{i\theta}.$

These four points form a trapezoid with two vertical sides symmetric about the real axis. Its area works out to $$\frac{99}{20}\sin 2\theta,$$ which is maximized at $\theta=45^\circ.$

Then $c=z_1+z_2=2\sqrt{10}\cos45^\circ=2\sqrt5\approx4.47,$ closest to $4.5.$

Thus, the correct answer is A.

Answer: D

Solution:

Always $k_n\mid L_n,$ so $k_n\lt L_n$ exactly when some prime $p$ divides both $L_n$ and the numerator $N=\sum_{k=1}^n \tfrac{L_n}{k}$ (i.e. a prime cancels).

For a prime $p$ with maximal power $p^a\le n,$ only the terms with $v_p(k)=a$ keep $p$ out of $L_n/k;$ all others are divisible by $p.$ So $p$ cancels iff $\sum_{v_p(k)=a}\tfrac{L_n}{k}\equiv0\pmod p.$

Checking each $n,$ cancellation occurs precisely for $n=6,7,8,18,19,20,21,22,$ which is $8$ values.

Thus, the correct answer is D.

Answer: E

Solution:

Sort the five digits as $d_{(1)}\le d_{(2)}\le\cdots\le d_{(5)}.$ The requirement "at least $j$ digits less than $j$" is equivalent to $d_{(j)}\le j-1$ for $j=1,2,3,4,$ i.e. $d_{(1)}=0,\ d_{(2)}\le1,\ d_{(3)}\le2,\ d_{(4)}\le3$ (with $d_{(5)}\le4$ automatic).

Counting the ordered strings of digits from $\{0,1,2,3,4\}$ whose sorted values obey these bounds gives $1296.$

Thus, the correct answer is E.

Answer: E

Solution:

The circle centered $(r,r)$ with radius $r$ is tangent to both axes. A segment from $(0,a)$ to $(b,0)$ with $a^2+b^2=c^2$ is tangent to it when $r$ equals either the inradius $\tfrac{a+b-c}{2}$ or the semiperimeter $\tfrac{a+b+c}{2}$ of the right triangle with legs $a,b.$

The inradius case rearranges to $(a-2r)(b-2r)=2r^2,$ so the number of such segments grows with the number of divisors of $2r^2.$ The least $r$ that admits $14$ distinct segments is $r=6:$ the semiperimeter case gives the $3$-$4$-$5$ triangle (both orientations, $c=5$), and the inradius case gives twelve more, up to $13$-$84$-$85$ with $c=85.$

Then $c_1=5$ and $c_{14}=85,$ so $\dfrac{c_{14}}{c_1}=\dfrac{85}{5}=17.$

Thus, the correct answer is E.