Solution:

Adding $3$ to each number raises the sum from $S$ to $S + 6.$ Doubling each number doubles the sum, giving $$2(S + 6) = 2S + 12.$$

Thus, the correct answer is E.

Solution:

Write $N = 10a + b.$ Then $P(N) = ab$ and $S(N) = a + b,$ so $$10a + b = ab + a + b.$$ This reduces to $9a = ab.$ Since $a \neq 0,$ we can divide by $a$ to get $b = 9.$

The units digit of $N$ is $9.$

Thus, the correct answer is E.

Solution:

Let her income be $x \gt 28000$ dollars. Writing the tax with both descriptions and multiplying by $100,$ $$p \cdot 28000 + (p + 2)(x - 28000) = (p + 0.25)x.$$

Expanding, every term containing $p$ cancels, leaving $$2x - 56000 = 0.25x,$$ so $1.75x = 56000$ and $x = 32000.$

Thus, the correct answer is B.

Solution:

Let $m$ be the mean. The least number is $m - 10,$ the greatest is $m + 15,$ and the middle number is the median $5.$ Their sum is $3m,$ so $$(m - 10) + 5 + (m + 15) = 3m.$$

This gives $m = 10,$ so the sum of the three numbers is $3m = 30.$

Thus, the correct answer is D.

Solution:

The product of every integer from $1$ to $10000$ is $10000!,$ so the product of the odd ones is $10000!$ divided by the product of the even ones.

The even numbers factor as $$2 \cdot 4 \cdots 10000 = 2^{5000}(1 \cdot 2 \cdots 5000) = 2^{5000} \cdot 5000!.$$

Therefore the product of the odd integers is $$\dfrac{10000!}{2^{5000} \cdot 5000!}.$$

Thus, the correct answer is D.

Solution:

The four consecutive decreasing odd digits $GHIJ$ are either $9753$ or $7531,$ leaving one odd digit ($1$ or $9$) for $ABC.$

Since $A + B + C = 9$ and the other two digits of $ABC$ are even, the odd digit must be $1$ (a $9$ would force the two even digits to sum to $0$). So the two even digits sum to $8.$

The three consecutive decreasing even digits $DEF$ are $864,$ $642,$ or $420,$ leaving the even pairs $\{2, 0\},$ $\{8, 0\},$ or $\{8, 6\}$ for $ABC.$ Only $\{8, 0\}$ sums to $8,$ so $ABC = 810$ and $A = 8.$

Thus, the correct answer is E.

Solution:

Let $n$ tickets sell at full price $p$ dollars. Then $$np + (140 - n)\dfrac{p}{2} = 2001,$$ so $p(n + 140) = 4002 = 2 \cdot 3 \cdot 23 \cdot 29.$

Since $0 \le n \le 140,$ we need a factor of $4002$ with $140 \le n + 140 \le 280.$ The only such factor is $174 = 2 \cdot 3 \cdot 29,$ giving $n = 34$ and $p = 23.$

The full-price tickets raise $34 \cdot 23 = 782$ dollars.

Thus, the correct answer is A.

Solution:

When the sector is rolled into a cone, its radius $10$ becomes the slant height, and its arc becomes the base circle.

The arc length is $$\dfrac{252}{360}\cdot 2\pi(10) = \dfrac{7}{10}\cdot 20\pi = 14\pi,$$ so the base circumference is $14\pi$ and the base radius is $7.$

The cone therefore has base radius $7$ and slant height $10,$ which is choice C.

Thus, the correct answer is C.

Solution:

Choose $x = 500$ and $y = \dfrac{6}{5}$ so that $xy = 600.$ Then $$f(600) = \dfrac{f(500)}{6/5} = \dfrac{3}{6/5} = \dfrac{5}{2}.$$

Thus, the correct answer is C.

Solution:

The pattern repeats over a $3 \times 3$ block of nine unit squares. Four of these nine squares are not covered by pentagons; the rest of the area belongs to the pentagons.

So the pentagons enclose $$1 - \dfrac{4}{9} = \dfrac{5}{9} = 55.5\overline{5}\%,$$ which is closest to $56.$

Thus, the correct answer is D.

Solution:

Imagine continuing until all five chips are removed. The process actually stops on a white chip exactly when the whites run out before the reds, i.e. when the last chip in the full ordering is red.

The last of the five chips is equally likely to be any chip, so it is red with probability $\dfrac{3}{5}.$

Thus, the correct answer is D.

Solution:

Multiples of $3$ or $4$ up to $2001$ number $$667 + 500 - 166 = 1001,$$ using $\lfloor 2001/3 \rfloor = 667,$ $\lfloor 2001/4 \rfloor = 500,$ and $\lfloor 2001/12 \rfloor = 166.$

Among these, the ones divisible by $5$ are multiples of $15$ or $20$: $$133 + 100 - 33 = 200,$$ using $\lfloor 2001/15 \rfloor = 133,$ $\lfloor 2001/20 \rfloor = 100,$ and $\lfloor 2001/60 \rfloor = 33.$

The count is $1001 - 200 = 801.$

Thus, the correct answer is B.

Solution:

The value $a + b + c$ is the first parabola at $x = 1,$ and $d + e + f$ is the reflected parabola at $x = 1.$

Reflecting the curve about $y = k$ replaces each height $y$ by $2k - y.$ So at $x = 1$ the two heights sum to $$(a + b + c) + (d + e + f) = 2k.$$

Thus, the correct answer is E.

Solution:

Each of the $\binom{9}{2} = 36$ pairs of vertices is a side of exactly two equilateral triangles, giving $72$ triangles counted with multiplicity.

The triangles $A_1 A_4 A_7,$ $A_2 A_5 A_8,$ and $A_3 A_6 A_9$ have all three vertices in the set, so each is counted three times instead of once, an overcount of $2$ apiece.

The number of distinct triangles is $72 - 3 \cdot 2 = 66.$

Thus, the correct answer is D.

Solution:

Unfold the two faces the insect crosses into the plane. They form a rhombus of side $1$ made of two equilateral triangles.

The two opposite-edge midpoints become the midpoints of opposite sides of this rhombus, which are exactly $1$ unit apart along a straight segment. Folding back preserves the length, so the shortest trip is $1.$

Thus, the correct answer is B.

Solution:

Think of the $16$ items ($8$ socks and $8$ shoes) arranged in some order: there are $16!$ arrangements.

For each leg, the sock comes before the shoe in exactly half of all arrangements. Imposing this on all eight legs independently divides by $2^8,$ giving $$\dfrac{16!}{2^8}.$$

Thus, the correct answer is D.

Solution:

$\angle APB = 90^\circ$ when $P$ is on the circle with diameter $AB,$ centered at $(2, 1)$ with radius $\dfrac{|AB|}{2} = \dfrac{\sqrt{20}}{2} = \sqrt{5}.$ The angle is obtuse when $P$ is inside this circle.

The relevant half-disk lies wholly within the pentagon, with area $\dfrac{1}{2}\pi(\sqrt{5})^2 = \dfrac{5\pi}{2}.$

The pentagon is the rectangle with corners $(0,0),$ $C,$ $D,$ $E$ minus triangle $OAB,$ so its area is $$4(2\pi + 1) - \dfrac{1}{2}(2)(4) = 8\pi.$$

The probability is $\dfrac{5\pi/2}{8\pi} = \dfrac{5}{16}.$

Thus, the correct answer is C.

Solution:

When two mutually tangent circles of radii $r$ and $s$ both rest on a line, the distance between their points of tangency is $2\sqrt{rs}.$

The big circles' contact points are $2\sqrt{1 \cdot 4} = 4$ apart. Placing the small circle of radius $x$ between them, its two tangent distances add up: $$2\sqrt{1 \cdot x} + 2\sqrt{4 \cdot x} = 4.$$

Then $6\sqrt{x} = 4,$ so $\sqrt{x} = \dfrac{2}{3}$ and $x = \dfrac{4}{9}.$

Thus, the correct answer is D.

Solution:

The $y$-intercept is $P(0) = c = 2.$ By Vieta's formulas the product of the zeros is $-c = -2,$ the mean of the zeros is $-\dfrac{a}{3},$ and the sum of the coefficients is $P(1) = 1 + a + b + c.$

All three are equal to $-2.$ From $-\dfrac{a}{3} = -2$ we get $a = 6.$

Then $1 + a + b + c = -2$ becomes $1 + 6 + b + 2 = -2,$ so $b = -11.$

Thus, the correct answer is A.

Solution:

The midpoints are $M = (2, 5)$ of $\overline{AB}$ and $N = (3, 2)$ of $\overline{BC}.$

For the midpoint quadrilateral to be a square, consecutive sides are perpendicular and equal. With $\overrightarrow{NM} = \langle -1, 3 \rangle,$ the side $\overrightarrow{MQ}$ to the midpoint $Q$ of $\overline{DA}$ must be $\langle 3, 1 \rangle,$ so $Q = (5, 6).$

Since $Q$ is the midpoint of $\overline{DA}$ and $A = (3, 9),$ we get $D = 2Q - A = (7, 3).$ The sum of its coordinates is $7 + 3 = 10.$

Thus, the correct answer is C.

Solution:

Adding $1$ to each equation factors the left sides: $$(a + 1)(b + 1) = 525 = 3 \cdot 5^2 \cdot 7, \quad (b + 1)(c + 1) = 147 = 3 \cdot 7^2, \quad (c + 1)(d + 1) = 105 = 3 \cdot 5 \cdot 7.$$

Since $525$ has a factor of $25$ while $147$ is not divisible by $5,$ the factor $a + 1$ must carry the $25.$ Among divisors of $525,$ only $a + 1 = 25$ makes $a = 24$ divide $8! = 40320.$

Then $b + 1 = 21,$ $c + 1 = 7,$ and $d + 1 = 15,$ giving $b = 20,$ $c = 6,$ $d = 14.$ (Indeed $24 \cdot 20 \cdot 6 \cdot 14 = 40320 = 8!.$)

So $a - d = 24 - 14 = 10.$

Thus, the correct answer is D.

Solution:

Triangle $EFG$ has base $FG = \dfrac{1}{3}AB$ and height equal to the rectangle's height, so its area is $\dfrac{1}{6}(70) = \dfrac{35}{3}.$

Because $EC \parallel AF,$ triangles $AFH$ and $CEH$ are similar with ratio $\dfrac{EC}{AF} = \dfrac{3}{2},$ so $\dfrac{EH}{EF} = \dfrac{3}{5}.$ Likewise $\dfrac{EJ}{EG} = \dfrac{3}{7}.$

Then $\dfrac{[EHJ]}{[EFG]} = \dfrac{EH}{EF}\cdot\dfrac{EJ}{EG} = \dfrac{3}{5}\cdot\dfrac{3}{7} = \dfrac{9}{35},$ giving $$[EHJ] = \dfrac{9}{35}\cdot\dfrac{35}{3} = 3.$$

Thus, the correct answer is C.

Solution:

Writing $P(x) = (x - r)(x - s)(x^2 + \alpha x + \beta)$ with integer roots $r, s,$ matching coefficients forces $\alpha$ and $\beta$ to be integers.

The other two zeros are $$-\dfrac{\alpha}{2} \pm \dfrac{i\sqrt{4\beta - \alpha^2}}{2}.$$ A real part of $\dfrac{1}{2}$ requires $\alpha = -1,$ making the imaginary part $\dfrac{\sqrt{4\beta - 1}}{2}.$

Choice A needs $\sqrt{4\beta - 1} = \sqrt{11},$ i.e. $\beta = 3,$ an integer, so it works. The other choices force a non-integer $\beta$ (for example choice E needs $\beta = 3.5,$ and choice D needs $\alpha = -2$ with $\beta = \tfrac{5}{4}$).

Thus, the correct answer is A.

Solution:

Let $E$ be the foot of the perpendicular from $C$ to line $AD.$ The exterior angle of $\triangle ADB$ gives $\angle ADC = 15^\circ + 45^\circ = 60^\circ,$ so $\triangle CDE$ is a $30$-$60$-$90$ triangle with $DE = \tfrac{1}{2}CD = BD.$

Then $\triangle BDE$ is isosceles with $\angle EBD = \angle BED = 30^\circ,$ and since $\angle ECB = 30^\circ$ too, $\triangle BEC$ is isosceles with $BE = EC.$

Also $\angle ABE = 45^\circ - 30^\circ = 15^\circ = \angle EAB,$ so $\triangle ABE$ is isosceles with $AE = BE.$ Hence $AE = BE = EC,$ making right triangle $AEC$ isosceles with $\angle ECA = 45^\circ.$

Therefore $\angle ACB = \angle ECA + \angle ECD = 45^\circ + 30^\circ = 75^\circ.$

Thus, the correct answer is D.

Solution:

If $a, b, c$ are consecutive terms then $b = ac - 1,$ so $c = \dfrac{1 + b}{a}.$ Applying this repeatedly, the first five terms are $$a,\ b,\ \dfrac{1 + b}{a},\ \dfrac{1 + a + b}{ab},\ \dfrac{1 + a}{b},$$ after which $a$ and $b$ recur, so the sequence is periodic with period $5.$

Here $b = 2000$ is the second term. The value $2001$ can be placed in any one of the other four of the five distinct positions, and each choice determines $x = a$ uniquely and yields a valid sequence of positive reals.

So there are $4$ values of $x.$

Thus, the correct answer is D.