Solution:

The cheapest popsicles come from the $5$-popsicle box, at $\dfrac{\$3}{5}=\$0.60$ each. Even at that rate, $14$ popsicles would cost $14\cdot\$0.60=\$8.40,$ more than $\$8.$

So Pablo can buy at most $13,$ and he achieves this with two $5$-boxes for $\$6$ and one $3$-box for $\$2,$ giving $2\cdot5+3=13$ popsicles.

Thus, the correct answer is D.

Solution:

Let the numbers be $x$ and $y,$ so $x+y=4xy.$

Dividing both sides by $xy$ gives $$\dfrac{x+y}{xy}=4,$$ and the left side is exactly $\dfrac{1}{y}+\dfrac{1}{x}.$ So the sum of the reciprocals is $4.$

Thus, the correct answer is C.

Solution:

The promise is "all right $\Rightarrow$ A." An implication is equivalent only to its contrapositive: "not A $\Rightarrow$ not all right."

"Not all right" means at least one question was wrong, which is exactly statement B. The converse and inverse do not follow, and getting "all wrong" is a much stronger claim than the negation.

Thus, the correct answer is B.

Solution:

If the square has side $x,$ Jerry walks $x+x=2x,$ while Silvia walks the diagonal $\sqrt{x^2+x^2}=x\sqrt2.$

The fraction by which Silvia's trip is shorter is $$\dfrac{2x-x\sqrt2}{2x}=1-\dfrac{\sqrt2}{2}\approx 1-0.707=0.293.$$

This is closest to $30\%.$

Thus, the correct answer is A.

Solution:

Each of the $20$ people who know each other shakes hands with only the $10$ strangers. Each of the $10$ strangers shakes hands with all $29$ other people.

Summing handshake counts and dividing by $2$ (each handshake involves two people) gives $$\dfrac{1}{2}(20\cdot10+10\cdot29)=\dfrac{1}{2}(200+290)=245.$$

Thus, the correct answer is B.

Solution:

Four lengths form a quadrilateral with positive area if and only if the longest is strictly less than the sum of the other three. With a fourth rod of length $n,$ this requires $15\lt 3+7+n$ and $n\lt 3+7+15,$ so $$5\lt n\lt 25.$$

The integers from $6$ to $24$ give $19$ values, but the rods of length $7$ and $15$ are already on the table, leaving $19-2=17$ choices.

Thus, the correct answer is B.

Solution:

Listing values: $f(1)=2,$ $f(2)=f(1)+1=3,$ $f(3)=f(1)+2=4,$ $f(4)=f(3)+1=5,$ suggesting $f(n)=n+1.$

Both rules are consistent with $f(n)=n+1:$ for even $n,$ $(n-1)+1+1=n+1,$ and for odd $n,$ $(n-2)+1+2=n+1.$ Since the recursion determines $f$ uniquely, $f(2017)=2018.$

Thus, the correct answer is B.

Solution:

Let $h=AB.$ The region is a cylinder of radius $3$ and height $h$ with a hemisphere of radius $3$ on each end.

The cylinder has volume $\pi\cdot3^2\cdot h=9\pi h,$ and the two hemispheres together form a sphere of volume $\dfrac{4}{3}\pi\cdot3^3=36\pi.$ So $$9\pi h+36\pi=216\pi,$$ giving $h=20.$

Thus, the correct answer is D.

Solution:

Consider which two of $3,$ $x+2,$ $y-4$ are the (equal) larger pair.

If $3=x+2\ge y-4:$ then $x=1$ and $y\le7,$ a downward ray from $(1,7).$ If $3=y-4\ge x+2:$ then $y=7$ and $x\le1,$ a leftward ray from $(1,7).$ If $x+2=y-4\ge3:$ then $y=x+6$ and $x\ge1,$ a ray from $(1,7)$ going up and to the right.

All three rays share the endpoint $(1,7),$ so $S$ is three rays with a common endpoint.

Thus, the correct answer is E.

Solution:

With probability $\dfrac{1}{2},$ Laurent's number lies in $[2017,4034],$ which exceeds any number Chloé could choose, so he wins for certain.

With the other probability $\dfrac{1}{2},$ Laurent's number lies in $[0,2017],$ matching Chloé's interval; by symmetry he is larger half the time. The total probability is $$\dfrac{1}{2}\cdot1+\dfrac{1}{2}\cdot\dfrac{1}{2}=\dfrac{3}{4}.$$

Thus, the correct answer is C.

Solution:

If the polygon has $n$ sides and the forgotten angle is $\alpha,$ then $(n-2)180=2017+\alpha.$ Since $0\lt\alpha\lt180,$ $$2017\lt(n-2)180\lt2197.$$

The only multiple of $180$ in this range is $2160=(14-2)180,$ so $n=14$ and $$\alpha=2160-2017=143.$$

Thus, the correct answer is D.

Solution:

Horse $k$ is at the starting point at time $t$ precisely when $k\mid t.$ So we want the smallest $t$ with at least $5$ divisors among $1,2,\ldots,10.$

Checking small values, $t=12$ is divisible by $1,2,3,4,$ and $6,$ giving exactly $5$ such horses, and no smaller $t$ reaches $5.$ Thus $T=12,$ and the sum of its digits is $1+2=3.$

Thus, the correct answer is B.

Solution:

Let the distance be $d$ miles and the usual speed $r$ mph. Since the usual trip is $3$ hours, $d=3r.$

The first $\dfrac{1}{3}$ of the drive takes $\dfrac{1}{3}\cdot180=60$ minutes at speed $r,$ so the remaining $\dfrac{2}{3}$ takes $276-60=216$ minutes $=\dfrac{18}{5}$ hours at speed $r-20.$

That final portion covers $\dfrac{2}{3}d=2r$ miles, so $$2r=(r-20)\cdot\dfrac{18}{5}.$$ Solving gives $10r=18r-360,$ so $r=45$ and $d=3\cdot45=135.$

Thus, the correct answer is B.

Solution:

Let $X,$ $Y,$ $Z$ be the seatings where Alice-Bob, Alice-Carla, and Derek-Eric are adjacent, respectively. The answer is $5!-|X\cup Y\cup Z|.$

Treating a forbidden pair as a block gives $|X|=|Y|=|Z|=2\cdot4!=48.$ For intersections, $|X\cap Y|=2\cdot3!=12$ (Alice between Bob and Carla), $|X\cap Z|=|Y\cap Z|=2\cdot2\cdot3!=24,$ and $|X\cap Y\cap Z|=2\cdot2\cdot2!=8.$

By inclusion-exclusion, $|X\cup Y\cup Z|=(48\cdot3)-(12+24+24)+8=92,$ so the answer is $120-92=28.$

Thus, the correct answer is C.

Solution:

For $0\lt x\lt\dfrac{\pi}{2}$ all three terms are positive, so $f(x)\gt0.$ For $\dfrac{\pi}{2}\lt x\lt\pi,$ $\tan x$ is negative and dominates, keeping $f(x)\lt0.$ So no root occurs before $x=\pi.$

At $x=\pi,$ $f(\pi)=0+2(-1)+0=-2\lt0.$ At $x=\dfrac{5\pi}{4},$ $\tan x=1$ so $f=-\dfrac{\sqrt2}{2}+3\gt0.$ By the intermediate value theorem the smallest positive root lies in $\left(\pi,\dfrac{5\pi}{4}\right).$

Since $\pi\gt3$ and $\dfrac{5\pi}{4}\lt4,$ this interval sits inside $(3,4).$

Thus, the correct answer is D.

Solution:

The large semicircle has radius $3$ and center $C,$ the midpoint of $\overline{JK}.$ Placing $J$ at the origin, $A=2,$ $B=5,$ $C=3,$ $K=6$ along the base. Let $r$ be the radius of the circle at $P.$

By tangency, $PA=2+r,$ $PB=1+r,$ and $PC=3-r.$ Dropping a perpendicular from $P$ to the base at horizontal position $3+x$ with height $h,$ the Pythagorean theorem gives $$h^2=(2+r)^2-(1+x)^2=(3-r)^2-x^2=(1+r)^2-(2-x)^2.$$

These reduce to two linear equations in $r$ and $x,$ whose solution is $r=\dfrac{6}{7}$ (and $x=\dfrac{9}{7}$).

Thus, the correct answer is B.

Solution:

The $24$ solutions are the $24$th roots of unity, $z=e^{\pi i k/12}$ for $k=0,1,\ldots,23.$

Then $z^6=e^{\pi i k/2}=\cos\dfrac{k\pi}{2}+i\sin\dfrac{k\pi}{2},$ which is real exactly when $\sin\dfrac{k\pi}{2}=0,$ i.e. when $k$ is even. There are $12$ even values of $k$ in the range.

Thus, the correct answer is D.

Solution:

Adding $1$ to $n$ increases the digit sum by $1,$ except that each trailing $9$ turns into a $0,$ losing $9.$ If $n$ ends in exactly $k$ nines, then $S(n+1)=S(n)+1-9k=1275-9k.$

So the possible values are $1275,1266,1257,\ldots$ Among the choices, only $1239=1275-9\cdot4$ fits (for example, $n$ ending in four $9$s preceded by enough $1$s).

Thus, the correct answer is D.

Solution:

For the first square, the two smaller triangles it cuts off are similar to the whole triangle, giving $\dfrac{x}{3-x}=\dfrac{4-x}{x},$ so $x=\dfrac{12}{7}.$ (Equivalently, a square in the right angle has side $\dfrac{3\cdot4}{3+4}.$)

For the second square, take the hypotenuse of length $5$ as base; the altitude to it is $h=\dfrac{3\cdot4}{5}=\dfrac{12}{5}.$ A square with a side on a base $b$ and height $h$ has side $\dfrac{bh}{b+h},$ so $$y=\dfrac{5\cdot\tfrac{12}{5}}{5+\tfrac{12}{5}}=\dfrac{12}{\tfrac{37}{5}}=\dfrac{60}{37}.$$

Therefore $\dfrac{x}{y}=\dfrac{12}{7}\cdot\dfrac{37}{60}=\dfrac{37}{35}.$

Thus, the correct answer is D.

Solution:

Let $u=\log_b a.$ Since $\log_b(a^{2017})=2017\log_b a,$ the equation is $u^{2017}=2017u,$ so $u=0$ or $u^{2016}=2017.$

If $u=0,$ then $a=1,$ valid for every one of the $199$ bases. If $u^{2016}=2017,$ then $u=\pm2017^{1/2016},$ giving $2$ values of $a$ for each base, i.e. $2\cdot199=398$ pairs.

In total there are $199+398=597$ ordered pairs.

Thus, the correct answer is E.

Solution:

Using $10x+10,$ the root $-1$ enters $S.$ Then $1$ enters as a root of $-x^{10}-x^9-\cdots-x+10,$ and $-10$ enters from $x+10.$

Now $x^3+x-10$ has root $2,$ and $x+2$ gives $-2;$ then $2x-10$ and $2x+10$ give $\pm5.$ At this point $S=\{0,\pm1,\pm2,\pm5,\pm10\}.$

No further integer can appear: by the Rational Root Theorem any integer root divides the constant term, which is always a factor of $10.$ So $S$ has $9$ elements.

Thus, the correct answer is D.

Solution:

By symmetry, group the relevant interior points into three types: $C=\{(0,0)\},$ the "axis" points $A=\{(\pm1,0),(0,\pm1)\},$ and the "diagonal" points $I=\{(\pm1,\pm1)\}.$ Let $a,c,i$ be the probabilities of eventually hitting a corner starting from a point of type $A,C,I.$

Reading off the transition probabilities (a point in $A$ goes to $A$ with prob $\tfrac28,$ to $C$ with $\tfrac18,$ to $I$ with $\tfrac28,$ and to a side interior with $\tfrac38,$ etc.) gives $$a=\tfrac28 a+\tfrac18 c+\tfrac28 i,\quad c=\tfrac48 a+\tfrac48 i,\quad i=\tfrac28 a+\tfrac18 c+\tfrac18.$$

Solving yields $a=\dfrac{1}{14},$ $c=\dfrac{4}{35},$ $i=\dfrac{11}{70}.$ The required probability is $c=\dfrac{4}{35},$ so $m+n=4+35=39.$

Thus, the correct answer is E.

Solution:

Since $g$ has three distinct roots all shared by the quartic $f,$ we can write $f(x)=(x-q)g(x)$ for some remaining root $q.$ Expanding, $$f(x)=x^4+(a-q)x^3+(1-qa)x^2+(10-q)x-10q.$$

Matching the $x$ coefficient, $10-q=100,$ so $q=-90.$ Matching the $x^3$ coefficient, $a-q=1,$ so $a=-89.$

Then $g(1)=1+a+1+10=12-89=-77$ and $1-q=91,$ so $$f(1)=(1-q)g(1)=91\cdot(-77)=-7007.$$

Thus, the correct answer is C.

Solution:

Because $YE\parallel AD$ and $EF\parallel AC,$ we get $\triangle XEY\sim\triangle XAD$ and $\triangle XEF\sim\triangle XAC,$ giving $\dfrac{XY}{XE}=\dfrac{XD}{XA}$ and $\dfrac{XF}{XE}=\dfrac{XC}{XA}.$ Hence $\dfrac{XC}{XD}=\dfrac{XF}{XY},$ so $XF\cdot XD=XC\cdot XY.$

Power of a Point at $X$ gives $XC\cdot XG=XD\cdot XB,$ and combining yields $XF\cdot XG=XB\cdot XY.$ With $d=BD,$ $DX=\dfrac14 d$ and $BY=\dfrac{11}{36}d,$ so $$XF\cdot XG=\left(d-\tfrac14 d\right)\left(d-\tfrac14 d-\tfrac{11}{36}d\right)=\dfrac34 d\cdot\dfrac49 d=\dfrac{d^2}{3}.$$

Since $ABCD$ is cyclic, $\angle BAD$ and $\angle BCD$ are supplementary. The Law of Cosines on $\triangle ABD$ and $\triangle CBD$ gives $\dfrac{73-d^2}{48}=\dfrac{d^2-40}{24},$ so $d^2=51.$ Therefore $XF\cdot XG=\dfrac{51}{3}=17.$

Thus, the correct answer is A.

Solution:

Let $A=\{\sqrt2 i,-\sqrt2 i\}$ (each of magnitude $\sqrt2$) and $B$ be the other four elements (each of magnitude $\dfrac12$). Since $|P|=(\sqrt2)^{\#A}\left(\tfrac12\right)^{\#B}=1$ forces $\#A=8$ and $\#B=4,$ exactly $8$ factors must come from $A$ and $4$ from $B.$

A product of $8$ elements of $A$ equals $\pm16$ (real), and a product of $4$ elements of $B$ equals one of $\pm\tfrac{1}{16},\pm\tfrac{i}{16}.$ Their product is one of $\pm1,\pm i,$ each equally likely, so exactly $\tfrac14$ of these configurations give $P=-1.$

The chance of landing in the $8$-from-$A$, $4$-from-$B$ pattern is $\binom{12}{4}\left(\tfrac13\right)^8\left(\tfrac23\right)^4=\dfrac{880}{3^{10}}.$ Multiplying by $\tfrac14$ gives $$P=\dfrac{1}{4}\cdot\dfrac{880}{3^{10}}=\dfrac{220}{3^{10}}=\dfrac{2^2\cdot5\cdot11}{3^{10}}.$$

Thus, the correct answer is E.