Solution:

Each classroom has $18-2=16$ more students than rabbits.

Across all $4$ classrooms there are $4\cdot16=64$ more students than rabbits.

Thus, the correct answer is C.

Solution:

The circle is inscribed, so the width of the rectangle equals the diameter, $2\cdot5=10.$

The length is then $2\cdot10=20,$ so the area is $10\cdot20=200.$

Thus, the correct answer is E.

Solution:

Let $h$ be the number of holes the chipmunk dug. The chipmunk hid $3h$ acorns and the squirrel hid $4(h-4)$ acorns.

Since they hid the same number, $3h=4(h-4),$ which gives $h=16.$

The chipmunk hid $3\cdot16=48$ acorns.

Thus, the correct answer is D.

Solution:

Étienne's money is worth $400\cdot1.30=520$ dollars, while Diana has $500$ dollars.

The percent by which Étienne's value exceeds Diana's is $$\frac{520-500}{500}\cdot100\%=4\%.$$

Thus, the correct answer is B.

Solution:

The three successive pairs have sums $26,$ $41-26=15,$ and $57-41=16.$

A pair sums to an even number when its two integers share parity, and to an odd number when exactly one is even. Only the middle pair sums to an odd number, so it must contain at least one even integer.

The other two pairs can be all odd, so as few as $1$ even integer is possible, for example $1,25,1,14,1,15.$

Thus, the correct answer is A.

Solution:

Let $d \gt 0$ be the small amount. Xiaoli computes $(x+d)-(y-d)=(x-y)+2d.$

Since $2d \gt 0,$ her estimate exceeds the true value $x-y.$

Thus, the correct answer is A.

Solution:

The lights repeat in blocks of $5,$ so consecutive blocks start $5\cdot6=30$ inches, or $2.5$ feet, apart.

Each block has one odd-numbered red light beginning it. The $3$rd red light begins the $2$nd block and the $21$st red light begins the $11$th block.

The distance between them is $(11-2)\cdot2.5=22.5$ feet.

Thus, the correct answer is E.

Solution:

Friday is fixed as cake. Work outward from Friday.

Each of the other six days (Saturday, then Thursday, Wednesday, Tuesday, Monday, Sunday) can be any dessert except the one served on the neighboring already-chosen day, giving $3$ choices each.

The number of menus is $3^6=729.$

Thus, the correct answer is A.

Solution:

Let $x$ be Clea's walking rate and $r$ the escalator's rate, with the length equal to $60x.$ Walking on the moving escalator gives $24(x+r)=60x,$ so $r=\tfrac32x.$

Standing takes time $t$ with $rt=60x,$ so $\tfrac32xt=60x$ and $t=40$ seconds.

Thus, the correct answer is B.

Solution:

From $x^2+y^2=25$ we get $y^2=25-x^2.$ Substituting into $(x-4)^2+9y^2=81$ gives $x^2+x-20=0,$ so $x=4$ or $x=-5.$

The intersection points are $(-5,0),$ $(4,3),$ and $(4,-3).$

The vertical side from $(4,3)$ to $(4,-3)$ has length $6,$ and the horizontal distance to $(-5,0)$ is $9,$ so the area is $\tfrac12\cdot6\cdot9=27.$

Thus, the correct answer is B.

Solution:

Writing the numerals out, $132_A=A^2+3A+2,$ $43_B=4B+3,$ and $69_{A+B}=6(A+B)+9.$

With $B=A+1,$ the equation becomes $A^2+3A+2+4(A+1)+3=6(2A+1)+9,$ which simplifies to $(A-6)(A+1)=0.$ The positive solution is $A=6,$ so $B=7.$

(The case $B=A-1$ gives $A^2-5A-2=0,$ which has no integer solution.)

Therefore $A+B=13.$

Thus, the correct answer is C.

Solution:

Let $A$ be the sequences in which all zeros are consecutive and $B$ those in which all ones are consecutive.

For $A,$ there is one all-ones sequence, $20$ sequences with exactly one zero, and $\binom{20}{2}=190$ sequences with two or more zeros (choose the first and last zero position). So $|A|=1+20+190=211,$ and by symmetry $|B|=211.$

A sequence in $A\cap B$ is a block of zeros followed by a block of ones, or the reverse; there are $|A\cap B|=40$ of these.

Therefore $|A\cup B|=211+211-40=382.$

Thus, the correct answer is E.

Solution:

The parabolas meet where $x^2+ax+b=x^2+cx+d,$ i.e. $ax+b=cx+d.$ This has no solution exactly when the lines are parallel and distinct: $a=c$ and $b\neq d.$

The probability that $a=c$ is $\tfrac16,$ and the probability that $b\neq d$ is $\tfrac56,$ so the probability of no common point is $\tfrac16\cdot\tfrac56=\tfrac5{36}.$

The probability of at least one common point is $1-\tfrac5{36}=\tfrac{31}{36}.$

Thus, the correct answer is D.

Solution:

Bernardo wins after a round when his doubled number $2n+50\ge1000$ but the previous numbers stayed below $1000.$ The smallest $n$ with $2n+50\ge1000$ is $475.$

Working backwards, the smallest starting values that lead to a win after two, three, and four rounds are the smallest integers with $2n+50\ge475,$ $\ge213,$ and $\ge82,$ namely $213,$ $82,$ and $16.$ No start wins after more than four rounds.

So $N=16,$ and the sum of its digits is $1+6=7.$

Thus, the correct answer is A.

Solution:

Each sector forms a cone with slant height $12.$ The smaller sector's arc length is $\tfrac{120}{360}\cdot2\pi\cdot12=8\pi,$ so its base radius is $4$ and its height is $\sqrt{12^2-4^2}=8\sqrt2.$

The larger sector (central angle $240^\circ$) has arc length $16\pi,$ base radius $8,$ and height $\sqrt{12^2-8^2}=4\sqrt5.$

The ratio of volumes is $$\frac{\tfrac13\pi\cdot4^2\cdot8\sqrt2}{\tfrac13\pi\cdot8^2\cdot4\sqrt5} =\frac{\sqrt{10}}{10}.$$

Thus, the correct answer is C.

Solution:

Each song is liked by exactly one of the three pairs, by a single girl, or by no one. Every pair must be represented.

Case 1: every song is liked by a pair. One pair gets two of the four songs ($\binom42=6$ ways, and $3$ choices for which pair), and the other two pairs get one song each ($2$ ways). This gives $3\cdot6\cdot2=36.$

Case 2: three songs go to the three pairs (one each) and the fourth song is liked by a single girl or no one. Assigning the four songs to these four roles gives $4!=24$ ways, and the leftover role has $4$ options (Amy, Beth, Jo, or no one): $24\cdot4=96.$

The total is $36+96=132.$

Thus, the correct answer is B.

Solution:

Let $\theta$ be the acute angle line $PQ$ makes with the $x$-axis. Sides $SR=PQ$ span the segment from $(3,0)$ to $(5,0)$ as $2\cos\theta,$ while $SP=QR$ span the segment from $(7,0)$ to $(13,0)$ as $6\sin\theta.$

Since the square has equal sides, $2\cos\theta=6\sin\theta,$ so $\tan\theta=\tfrac13.$ Thus lines $SP,RQ$ have slope $3$ and lines $SR,PQ$ have slope $-\tfrac13.$

The center lies on the line through $(4,0)$ with slope $3$ and the line through $(10,0)$ with slope $-\tfrac13:$ $$y=3(x-4),\qquad y=-\tfrac13(x-10).$$ These meet at $(4.6,1.8).$

The sum of the coordinates is $4.6+1.8=6.4.$

Thus, the correct answer is C.

Solution:

Once $a_1=k$ is fixed, the numbers $k,k+1,\ldots,10$ must appear left to right in increasing order, and the numbers $1,\ldots,k-1$ must appear from right to left in increasing order (so each new small number has its successor already placed).

For each $k,$ the list is determined by choosing which of the $9$ positions after the first hold the numbers below $k,$ giving $\binom{9}{k-1}$ lists.

Summing, $$\sum_{k=1}^{10}\binom{9}{k-1}=\sum_{j=0}^{9}\binom{9}{j}=2^9=512.$$

Thus, the correct answer is B.

Solution:

Place $P_1$ at the origin with edges along the axes, and let each of the three octahedron vertices near $P_1$ be a distance $t$ from $P_1;$ by symmetry the three near $P_1'$ are also a distance $t$ from $P_1'.$

Two vertices sharing $P_1,$ such as $(t,0,0)$ and $(0,t,0),$ are a distance $t\sqrt2$ apart. A vertex near $P_1,$ say $(t,0,0),$ and the appropriate vertex near $P_1',$ say $(1,1-t,1),$ must be the same distance apart.

Setting the two squared side lengths equal and using the cube's unit edges yields $t=\tfrac34,$ so the side length is $t\sqrt2=\dfrac{3\sqrt2}{4}.$

Thus, the correct answer is A.

Solution:

For a trapezoid with parallel sides $a\lt c$ and legs $b,d,$ translating a leg forms a triangle with sides $b,$ $d,$ and $c-a.$ The triangle inequality forces the longer parallel side to be $c=11.$

If $a=3,$ the triangle has sides $5,7,8$ with area $10\sqrt3,$ and the trapezoid has area $\tfrac{35}{2}\sqrt3.$ If $a=5,$ the triangle has sides $3,6,7$ with area $4\sqrt5,$ giving trapezoid area $\tfrac{32}{3}\sqrt5.$ If $a=7,$ the triangle has sides $3,4,5,$ a right triangle, giving trapezoid area $27.$

The total is $\tfrac{35}{2}\sqrt3+\tfrac{32}{3}\sqrt5+27,$ so $r_1+r_2+r_3+n_1+n_2=\tfrac{35}{2}+\tfrac{32}{3}+27+3+5=63+\tfrac16.$

The greatest integer at most this value is $63.$

Thus, the correct answer is D.

Solution:

Extend $EF$ and $CB$ to a line through $A$ perpendicular to both, meeting them at $H$ and $J.$ Since $\angle ABJ=60^\circ,$ we have $BJ=20$ and $AJ=20\sqrt3.$ With $u=BX,$ the Pythagorean theorem gives $s^2=(20+u)^2+(20\sqrt3)^2.$

The equiangular angles make the four corner triangles congruent, and chasing the equal segments along $EF$ yields $$u+20\sqrt3=41(\sqrt3-1)+\frac{20+u}{\sqrt3},$$ so $u=21\sqrt3-20.$

Since $20+u=21\sqrt3,$ we get $s^2=(21\sqrt3)^2+(20\sqrt3)^2=3(441+400)=3\cdot29^2,$ giving $s=29\sqrt3.$

Thus, the correct answer is A.

Solution:

Label the seven columns of forward (rightward) segments; a path with no back segment simply chooses one forward segment in each column. The numbers of choices are $2,2,4,4,4,2,2,$ giving $2^{10}$ paths.

Let $s_1,s_2,s_3$ be the three left-pointing back segments (in columns $2,4,6$). Analyzing which columns become forced once a back segment is traversed gives $2^8$ paths for each of $\{s_1\}$ and $\{s_3\},$ $2^6$ for $\{s_1,s_3\},$ $2^9$ for $\{s_2\},$ $2^7$ for each of $\{s_1,s_2\}$ and $\{s_2,s_3\},$ and $2^5$ for $\{s_1,s_2,s_3\}.$

Adding, $$2^{10}+2\cdot2^8+2^6+2^9+2\cdot2^7+2^5=2400.$$

Thus, the correct answer is E.

Solution:

Because $|z_0|=1,$ applying the triangle inequality to the identity $4z_0^5-(z_0-1)P(z_0)=z_0^4(4-a)+z_0^3(a-b)+z_0^2(b-c)+z_0(c-d)+d$ forces equality throughout, so all but one of the coefficient differences vanish.

Working through the cases (including $z_0=-1$ and $z_0=\gamma$ a primitive cube root of unity), the polynomials are exactly $4z^4+4z^3+4z^2+4z+4,$ $4z^4+4z^3+4z^2,$ and $4z^4+4z^3+bz^2+bz$ for $0\le b\le4.$

Their values at $1$ are $20,$ $12,$ and $8+2b;$ summing gives $20+12+\sum_{b=0}^{4}(8+2b)=32+40+20=92.$

Thus, the correct answer is B.

Solution:

If $N_1\mid N_2$ then $f_1(N_1)\mid f_1(N_2),$ so if $S_{N_1}$ is unbounded so is $S_{N_2}.$ Call $N$ essential if it is unbounded but no proper divisor is. An essential $N$ must have all exponents at least $2,$ and $(p_1\cdots p_k)^2\le400$ forces at most two primes.

Checking prime powers and prime pairs, the essential values are $2^5=32,$ $3^4=81,$ $7^3=343,$ and $2^4\cdot5^2=400.$

Their multiples up to $400$ number $\lfloor400/32\rfloor=12,$ $\lfloor400/81\rfloor=4,$ $\lfloor400/343\rfloor=1,$ and $\lfloor400/400\rfloor=1,$ with no overlaps, for a total of $12+4+1+1=18.$

Thus, the correct answer is D.

Solution:

Isosceles right triangles contribute $f(t)=1.$ For a scalene right triangle, reflecting across a suitable line pairs it with a triangle $t_1$ so that $f(t)f(t_1)=\tan(\angle CBA)\tan(\angle ACB)=1.$

Successive reflections (across $x=2,$ then $x=y,$ then $y=\tfrac52$) reduce the product to the reciprocal of the product over just six triangles of the form $OYZ$ with $Y$ on the top row.

Those six give $$\frac15\cdot\frac25\cdot\frac35\cdot\frac45\cdot\frac{3\sqrt2}{\sqrt2}\cdot \frac{4\sqrt2}{\sqrt2}\cdot\frac{1}{2}=\frac{144}{625},$$ so the required product is its reciprocal, $\dfrac{625}{144}.$

Thus, the correct answer is B.