Solution:

Since $6^{2002} = 2^{2002}\cdot 3^{2002},$ the ratio becomes $$\dfrac{2^{2001}\cdot 3^{2003}}{2^{2002}\cdot 3^{2002}} = \dfrac{3}{2}.$$

Thus, the correct answer is E.

Solution:

Substituting directly, $$(2, 4, 6) = \dfrac{2\cdot4\cdot6}{2+4+6} = \dfrac{48}{12} = 4.$$

Thus, the correct answer is C.

Solution:

The mean is $$\dfrac{1}{9}\left(9 + 99 + 999 + \cdots + 999999999\right) = 1 + 11 + 111 + \cdots + 111111111.$$

Adding these nine repunits column by column gives $M = 123456789.$

The only digit missing from $M$ is $0.$

Thus, the correct answer is A.

Solution:

Factoring $3x-2$ from the first two terms, $$(3x - 2)\big[(4x + 1) - 4x\big] + 1 = (3x - 2)(1) + 1 = 3x - 1.$$

At $x = 4,$ this equals $3\cdot4 - 1 = 11.$

Thus, the correct answer is D.

Solution:

The two small circles line up along a diameter of the big circle, so that diameter is $2\cdot3 + 2\cdot2 = 10$ and the large radius is $5.$

The shaded region is the large disk with the two small disks removed: $$\pi(5^2) - \pi(3^2) - \pi(2^2) = \pi(25 - 9 - 4) = 12\pi.$$

Thus, the correct answer is E.

Solution:

Factor as $n^2 - 3n + 2 = (n-1)(n-2).$

For $n \ge 4,$ both factors exceed $1,$ so the product is composite. For $n = 1$ and $n = 2$ the value is $0,$ and for $n = 3$ the value is $(2)(1) = 2,$ which is prime.

So exactly one value of $n$ works.

Thus, the correct answer is B.

Solution:

The sum $\dfrac12 + \dfrac13 + \dfrac17 + \dfrac1n$ is greater than $0$ and less than $\dfrac12 + \dfrac13 + \dfrac17 + 1 \lt 2,$ so as an integer it must equal $1.$

Since $\dfrac12 + \dfrac13 + \dfrac17 = \dfrac{41}{42},$ we need $\dfrac1n = \dfrac{1}{42},$ so $n = 42.$

Then $2,$ $3,$ $6,$ and $7$ all divide $42,$ but $n = 42$ is not greater than $84.$ So the false statement is $n \gt 84.$

Thus, the correct answer is E.

Solution:

A $31$-day month is $4$ weeks plus $3$ extra days, so exactly the weekdays of the $1$st, $2$nd, and $3$rd of the month occur five times.

For July to have five Mondays, Monday must be one of July $1,$ $2,$ or $3.$ In all three cases August $1$ lands on a Wednesday, Thursday, or Friday, and the common weekday among the resulting five-time days is Thursday.

More directly, since July has $31$ days, August $1$ is the same weekday as July $4.$ With Monday on July $1,$ $2,$ or $3,$ the three five-time weekdays of August always include Thursday.

Thus, the correct answer is D.

Solution:

The alphabetical order of the letters is $A, M, O, S, U.$ Words beginning with $A,$ $M,$ $O,$ or $S$ fill positions $1$ through $96$ (four choices of first letter, $24$ each).

Words beginning with $U$ occupy positions $97$–$120.$ Listing them alphabetically, $USAMO$ is the $19$th such word, so it occupies position $96 + 19 = 115.$

Thus, the correct answer is D.

Solution:

Since the roots are $a$ and $b,$ Vieta's formulas give $a + b = -a$ and $ab = b.$

From $ab = b$ with $b \ne 0,$ we get $a = 1.$ Then $a + b = -a$ gives $1 + b = -1,$ so $b = -2.$

Thus $(a, b) = (1, -2),$ and the correct answer is C.

Solution:

Let the integers be $n - 1,$ $n,$ $n + 1.$ Their product is $n(n^2 - 1)$ and their sum is $3n,$ so $$n(n^2 - 1) = 8(3n) = 24n.$$

Since $n \ne 0,$ we get $n^2 - 1 = 24,$ so $n^2 = 25$ and $n = 5.$

The three integers are $4, 5, 6,$ and $4^2 + 5^2 + 6^2 = 16 + 25 + 36 = 77.$

Thus, the correct answer is B.

Solution:

Cross multiplying gives $(x - 1)(x - 6) = (x - 2)(x - k),$ which expands to $$x^2 - 7x + 6 = x^2 - (2 + k)x + 2k.$$

Cancelling $x^2$ leaves $(k - 5)x = 2k - 6,$ so $x = \dfrac{2k - 6}{k - 5}.$ This has no solution exactly when $k = 5,$ since then the coefficient of $x$ is $0$ while the right side is nonzero.

Thus, the correct answer is E.

Solution:

Grouping and factoring, $$8xy - 12y + 2x - 3 = 4y(2x - 3) + (2x - 3) = (4y + 1)(2x - 3).$$

For this to equal $0$ for all $y,$ the factor that depends on $y$ cannot be forced to zero, so we need $2x - 3 = 0,$ giving $x = \dfrac32.$

Thus, the correct answer is D.

Solution:

Since $25 = 5^2$ and $64 = 2^6,$ we have $25^{64}\cdot 64^{25} = 5^{128}\cdot 2^{150},$ so $$N = \sqrt{5^{128}\cdot 2^{150}} = 5^{64}\cdot 2^{75}.$$

Writing $2^{75} = 2^{64}\cdot 2^{11},$ we get $N = (5\cdot 2)^{64}\cdot 2^{11} = 10^{64}\cdot 2048.$

So $N$ is $2048$ followed by $64$ zeros, and its digit sum is $2 + 0 + 4 + 8 = 14.$

Thus, the correct answer is B.

Solution:

The numbers $A - B$ and $A + B$ differ by $2B,$ so they have the same parity. Being prime, they must both be odd, which forces $A$ and $B$ to have opposite parity.

Since $2$ is the only even prime, $B = 2.$ Then $A - 2,$ $A,$ $A + 2$ are three primes forming an arithmetic progression of odd numbers, which must be $3, 5, 7.$

The four primes are $2, 3, 5, 7,$ and their sum is $17,$ which is prime.

Thus, the correct answer is E.

Solution:

Suppose $\dfrac{n}{20 - n} = k^2$ for some integer $k \ge 0.$ Solving, $$n = \dfrac{20k^2}{k^2 + 1}.$$

Since $k^2$ and $k^2 + 1$ share no common factor, $k^2 + 1$ must divide $20.$ This happens only for $k = 0, 1, 2, 3,$ giving $k^2 + 1 = 1, 2, 5, 10.$

The corresponding values $n = 0, 10, 16, 18$ are all integers, so there are $4$ such $n.$

Thus, the correct answer is D.

Solution:

Set the octagon on coordinate axes with the axis-aligned sides of length $2$ and each slanted side spanning $\sqrt2$ horizontally and $\sqrt2$ vertically. Then $$A = (\sqrt2, 0), \quad D = (2 + 2\sqrt2, 2 + \sqrt2), \quad G = (0, 2 + \sqrt2).$$

Since $D$ and $G$ share the height $2 + \sqrt2,$ segment $DG$ is horizontal with length $2 + 2\sqrt2,$ and the height from $A$ up to that level is $2 + \sqrt2.$

Therefore $$[\triangle ADG] = \tfrac12(2 + 2\sqrt2)(2 + \sqrt2) = (1 + \sqrt2)(2 + \sqrt2) = 4 + 3\sqrt2.$$

Thus, the correct answer is C.

Solution:

Any two distinct circles intersect in at most $2$ points. There are $\binom{4}{2} = 6$ pairs of circles, giving at most $6\cdot 2 = 12$ intersection points.

This maximum is achievable by a configuration where every pair of circles crosses twice, so the answer is $12.$

Thus, the correct answer is D.

Solution:

Let $d = a_2 - a_1.$ Then $a_{k+100} = a_k + 100d,$ so the second block sum is the first plus $100\cdot 100 d:$ $$a_{101} + \cdots + a_{200} = (a_1 + \cdots + a_{100}) + 10000d.$$

Therefore $200 = 100 + 10000d,$ giving $d = \dfrac{100}{10000} = 0.01.$

Thus, the correct answer is C.

Solution:

Rewrite the equations as $a + 8c = 4 + 7b$ and $8a - c = 7 - 4b.$ Squaring both and adding, $$(a + 8c)^2 + (8a - c)^2 = (4 + 7b)^2 + (7 - 4b)^2.$$

The left side expands to $65a^2 + 65c^2$ (the $ac$ terms cancel), and the right side expands to $65 + 65b^2$ (the $b$ terms cancel). So $$65(a^2 + c^2) = 65(1 + b^2),$$ giving $a^2 - b^2 + c^2 = 1.$

Thus, the correct answer is B.

Solution:

Let Andy's lawn have area $A,$ so Beth's is $\dfrac{A}{2}$ and Carlos' is $\dfrac{A}{3}.$ Let Carlos mow at rate $R,$ so Beth mows at $2R$ and Andy at $3R.$

The times are $$\text{Andy: } \dfrac{A}{3R}, \quad \text{Beth: } \dfrac{A/2}{2R} = \dfrac{A}{4R}, \quad \text{Carlos: } \dfrac{A/3}{R} = \dfrac{A}{3R}.$$

Since $\dfrac{A}{4R}$ is the smallest, Beth finishes first.

Thus, the correct answer is B.

Solution:

Let $OM = a$ and $ON = b,$ so $OX = 2a$ and $OY = 2b.$ The right angle at $O$ gives $$19^2 = (2a)^2 + b^2 \quad\text{and}\quad 22^2 = a^2 + (2b)^2.$$

Adding these, $5(a^2 + b^2) = 19^2 + 22^2 = 845,$ so $a^2 + b^2 = 169$ and $MN = \sqrt{a^2 + b^2} = 13.$

Since $\triangle XOY \sim \triangle MON$ with ratio $2,$ we have $XY = 2\cdot MN = 26.$

Thus, the correct answer is B.

Solution:

Setting $n = 1,$ we get $a_{m+1} = a_m + a_1 + m = a_m + (m + 1),$ so $a_{m+1} - a_m = m + 1.$

Summing from $m = 1$ to $11,$ $$a_{12} - a_1 = 2 + 3 + \cdots + 12 = \dfrac{12\cdot 13}{2} - 1 = 77.$$

Therefore $a_{12} = 1 + 77 = 78.$

Thus, the correct answer is D.

Solution:

Put the center $O$ at height $20.$ The bottom $A$ is at height $0,$ and the rider reaches height $10,$ which is $10$ feet below the center.

The horizontal from the center down to the rider's level forms a right triangle where the vertical leg is $10$ and the hypotenuse (the radius) is $20.$ That leg is half the hypotenuse, so the radius to the rider makes $60^\circ$ with the downward vertical.

The wheel turns $360^\circ$ in $60$ seconds, so turning $60^\circ$ takes $\dfrac{60}{360}\cdot 60 = 10$ seconds.

Thus, the correct answer is D.

Solution:

Let the original list have $n$ integers with mean $m,$ so its sum is $mn.$ Appending $15$ gives $$(m + 2)(n + 1) = mn + 15 \implies m + 2n = 13.$$

Appending $1$ to that enlarged list gives $$(m + 1)(n + 2) = mn + 16 \implies 2m + n = 14.$$

Solving $m + 2n = 13$ and $2m + n = 14$ yields $m = 5$ and $n = 4.$

Thus, the correct answer is A.