Solution:

Each of the nine numbers is $10^k-1,$ so their sum is $9+99+\cdots+999{,}999{,}999.$ Dividing by $9,$ $$M=1+11+111+\cdots+111{,}111{,}111=123{,}456{,}789.$$ Its digits are $1$ through $9,$ so the missing digit is $0.$

Thus, the correct answer is A.

Solution:

Factor $3x-2$ out of the first two terms: $$(3x-2)(4x+1)-(3x-2)4x+1=(3x-2)(4x+1-4x)+1=3x-1.$$ At $x=4$ this equals $3\cdot4-1=11.$

Thus, the correct answer is D.

Solution:

Factor $n^2-3n+2=(n-1)(n-2).$ For $n\ge4$ both factors exceed $1,$ so the value is composite. Checking $n=1,2,3$ gives $0,$ $0,$ and $2.$ Only $n=3$ yields a prime, so there is exactly one such $n.$

Thus, the correct answer is B.

Solution:

Since $\dfrac12+\dfrac13+\dfrac17=\dfrac{41}{42},$ the sum $\dfrac{41}{42}+\dfrac1n$ lies strictly between $0$ and $2,$ so it must equal $1.$ Then $\dfrac1n=\dfrac1{42},$ giving $n=42.$

Now $2,$ $3,$ $6,$ and $7$ all divide $42,$ but $n\gt84$ is false. The untrue statement is $n\gt84.$

Thus, the correct answer is E.

Solution:

The five interior angles sum to $540^\circ.$ Writing the sequence as $x-2d,$ $x-d,$ $x,$ $x+d,$ $x+2d,$ the sum is $5x=540,$ so $x=108.$

Thus, the correct answer is D.

Solution:

Since $a$ and $b$ are the roots, $x^2+ax+b=(x-a)(x-b)=x^2-(a+b)x+ab.$ Matching coefficients gives $a+b=-a$ and $ab=b.$

As $b\neq0,$ the second equation gives $a=1,$ and then $a+b=-a$ gives $b=-2.$ So $(a,b)=(1,-2).$

Thus, the correct answer is C.

Solution:

Let the integers be $n-1,$ $n,$ $n+1.$ Then $(n-1)n(n+1)=8\cdot3n,$ so $n^2-1=24$ and $n=5.$

The three integers $4,$ $5,$ $6$ have squares summing to $16+25+36=77.$

Thus, the correct answer is B.

Solution:

Since July has $31=4\cdot7+3$ days, the weekday of July $1$ occurs five times, so Monday falls on July $1,$ $2,$ or $3.$ The days that occur five times in August are those of August $1,$ $2,$ $3,$ and August $1$ is three weekdays after July $1.$

Testing the three cases, the August weekdays occurring five times are $\{\text{Thu},\text{Fri},\text{Sat}\},$ $\{\text{Wed},\text{Thu},\text{Fri}\},$ and $\{\text{Tue},\text{Wed},\text{Thu}\}.$ Thursday appears in every case.

Thus, the correct answer is D.

Solution:

Let $b=a+r,$ $c=a+2r,$ $d=a+3r.$ The geometric condition $b^2=ad$ gives $(a+r)^2=a(a+3r),$ i.e. $r^2=ar,$ so $r=a.$

Then $d=a+3a=4a$ and $\dfrac{a}{d}=\dfrac14.$

Thus, the correct answer is C.

Solution:

Every element is one more than a multiple of $3,$ so any sum of three of them is a multiple of $3.$ The smallest sum is $1+4+7=12$ and the largest is $13+16+19=48,$ and every multiple of $3$ between them is attainable.

There are $13$ multiples of $3$ from $12$ to $48.$

Thus, the correct answer is A.

Solution:

$A-B$ and $A+B$ have the same parity; being prime, both are odd, so one of $A,B$ is even. Since $A$ lies between the two odd primes $A-B$ and $A+B,$ it is the odd one, forcing $B=2.$

Then $A-2,$ $A,$ $A+2$ are three primes, which must be $3,$ $5,$ $7.$ Their sum together with $2$ is $2+3+5+7=17,$ a prime.

Thus, the correct answer is E.

Solution:

Set $\dfrac{n}{20-n}=k^2.$ Solving, $n=\dfrac{20k^2}{k^2+1}.$ Since $k^2$ and $k^2+1$ are coprime, $k^2+1$ must divide $20,$ which happens only for $k=0,1,2,3.$

These give $n=0,$ $10,$ $16,$ $18,$ which is four values.

Thus, the correct answer is D.

Solution:

The sum of $n,$ $n+1,\ldots,$ $n+17$ is $18n+\dfrac{17\cdot18}{2}=9(2n+17).$ Since $9$ is a perfect square, $2n+17$ must be one too.

The smallest positive integer $n$ making $2n+17$ a perfect square is $n=4,$ giving $2n+17=25$ and a sum of $9\cdot25=225.$

Thus, the correct answer is B.

Solution:

Each pair of circles meets in at most $2$ points, and there are $\binom{4}{2}=6$ pairs, giving at most $6\cdot2=12$ intersection points.

A configuration of four circles achieving all $12$ points is possible.

Thus, the correct answer is D.

Solution:

Let $a$ be the leading digit and $x$ the three-digit number after removing it, so $N=1000a+x.$ The condition $N=9x$ gives $1000a=8x,$ i.e. $x=125a.$

For $a=1,\ldots,7$ this makes $x$ a three-digit number, while $a=8$ gives $x=1000.$ So there are $7$ such numbers.

Thus, the correct answer is D.

Solution:

The product is a multiple of $3$ if and only if at least one die shows $3$ or $6.$ The octahedral die avoids $3,6$ with probability $\dfrac68=\dfrac34,$ and the six-sided die avoids them with probability $\dfrac46=\dfrac23.$

So neither shows a multiple of $3$ with probability $\dfrac34\cdot\dfrac23=\dfrac12,$ and the answer is $1-\dfrac12=\dfrac12.$

Thus, the correct answer is C.

Solution:

Let Andy's lawn have area $A;$ then Beth's is $\dfrac A2$ and Carlos' is $\dfrac A3.$ With Carlos' rate $R,$ Beth mows at $2R$ and Andy at $3R.$

Their times are $\dfrac{A}{3R},$ $\dfrac{A/2}{2R}=\dfrac{A}{4R},$ and $\dfrac{A/3}{R}=\dfrac{A}{3R}$ respectively. Beth's time $\dfrac{A}{4R}$ is the smallest, so Beth finishes first.

Thus, the correct answer is B.

Solution:

The points closer to $(0,0)$ than to $(3,1)$ lie on the origin side of the perpendicular bisector of that segment, the line $3x+y=5.$

Within the rectangle, this region is a trapezoid whose parallel sides have lengths $\dfrac53$ (at $y=0$) and $\dfrac43$ (at $y=1$), so its area is $\dfrac12\left(\dfrac53+\dfrac43\right)=\dfrac32.$ The rectangle has area $2,$ so the probability is $\dfrac{3/2}{2}=\dfrac34.$

Thus, the correct answer is C.

Solution:

Adding the three equations gives $2(ab+bc+ca)=484,$ so $ab+bc+ca=242.$ Subtracting each original equation from this yields $bc=90,$ $ca=80,$ and $ab=72.$

Multiplying, $(abc)^2=90\cdot80\cdot72=720^2,$ and since $abc\gt0,$ we get $abc=720.$

Thus, the correct answer is D.

Solution:

Let $OM=a$ and $ON=b.$ Since $M,N$ are midpoints, $XN^2=(2a)^2+b^2=361$ and $YM^2=a^2+(2b)^2=484.$

Adding, $5(a^2+b^2)=845,$ so $a^2+b^2=169.$ Then $MN=\sqrt{a^2+b^2}=13,$ and since $MN$ is the midsegment joining $M$ and $N,$ we have $XY=2\,MN=26.$

Thus, the correct answer is B.

Solution:

Since $2002=11\cdot13\cdot14$ with $11,$ $13,$ $14$ pairwise coprime, $a_n=11$ when $182\mid n,$ $a_n=13$ when $154\mid n,$ and $a_n=14$ when $143\mid n$ (and no $n\lt2002$ is divisible by all three).

For $n\le2001$ there are $10$ multiples of $182,$ $12$ of $154,$ and $13$ of $143.$ So the sum is $11\cdot10+13\cdot12+14\cdot13=110+156+182=448.$

Thus, the correct answer is A.

Solution:

By change of base, $a_n=\dfrac{1}{\log_n 2002}=\log_{2002} n.$ So $$b-c=\log_{2002}\frac{2\cdot3\cdot4\cdot5}{10\cdot11\cdot12\cdot13\cdot14}.$$

The fraction equals $\dfrac{120}{240240}=\dfrac{1}{2002},$ so $b-c=\log_{2002}\dfrac{1}{2002}=-1.$

Thus, the correct answer is B.

Solution:

Let $M$ be the midpoint of $BC,$ set $AM=2a,$ and let $\theta=\angle AMB,$ so $\angle AMC=180^\circ-\theta.$ With $BM=CM=a,$ so that $BC=2a,$ the Law of Cosines in $\triangle ABM$ and $\triangle AMC$ gives $$a^2+4a^2-4a^2\cos\theta=1,\qquad a^2+4a^2+4a^2\cos\theta=4.$$

Adding, $10a^2=5,$ so $a=\dfrac{\sqrt2}{2}$ and $BC=2a=\sqrt2.$

Thus, the correct answer is C.

Solution:

For any quadrilateral, the area is at most $\tfrac12\,d_1 d_2$ where $d_1,d_2$ are the diagonals, with equality exactly when they are perpendicular. Here $$2002=\text{Area}\le\tfrac12\,AC\cdot BD\le\tfrac12(PA+PC)(PB+PD)=\tfrac12\cdot52\cdot77=2002.$$

Equality forces the diagonals to be perpendicular and to intersect at $P.$ Then $$AB=\sqrt{24^2+32^2}=40,\quad BC=\sqrt{28^2+32^2}=4\sqrt{113},$$ $$CD=\sqrt{28^2+45^2}=53,\quad DA=\sqrt{45^2+24^2}=51.$$

The perimeter is $144+4\sqrt{113}=4\left(36+\sqrt{113}\right).$

Thus, the correct answer is E.

Solution:

Completing the square, $$f(x)+f(y)=(x+3)^2+(y+3)^2-16,$$ so the first condition is the disk of radius $4$ centered at $(-3,-3).$

Also $$f(x)-f(y)=(x-y)(x+y+6),$$ so the second condition $(x-y)(x+y+6)\le0$ describes two half-planes bounded by the perpendicular lines through $(-3,-3)$ of slopes $1$ and $-1.$ These cut the disk into two equal halves.

Thus $R$ has area $\tfrac12\pi\cdot4^2=8\pi\approx25.13,$ which is closest to $25.$

Thus, the correct answer is E.