Answer: C

Solution:

Since $(-1)^k = -1$ for odd $k$ and $(-1)^k = 1$ for even $k,$ the terms alternate $-1, 1, -1, 1, \ldots$

There are $2006$ terms, forming $1003$ pairs, each equal to $(-1) + 1 = 0.$ The total is $0.$

Thus, the correct answer is C.

Answer: A

Solution:

Since $x \spadesuit y = x^2 - y^2,$ the inner value is $4 \spadesuit 5 = 16 - 25 = -9.$

Then $3 \spadesuit (-9) = 3^2 - (-9)^2 = 9 - 81 = -72.$

Thus, the correct answer is A.

Answer: A

Solution:

Let $c$ and $p$ be the Cougars' and Panthers' scores. Then $c + p = 34$ and $c - p = 14.$

Subtracting gives $2p = 20,$ so $p = 10.$

Thus, the correct answer is A.

Answer: A

Solution:

The five prices total about $8 + 5 + 3 + 2 + 1 = 19$ dollars, so the change is about $\$1.00.$

This is $$\frac{1}{20} = 5\%$$ of the twenty-dollar bill.

Thus, the correct answer is A.

Answer: A

Solution:

Bob closes the gap at a relative speed of $5 - 3 = 2$ miles per hour. To cover the $1$-mile gap takes $$\frac{1}{2} \text{ hour} = 30 \text{ minutes}.$$

Thus, the correct answer is A.

Answer: B

Solution:

The full batch weighs $100 + 100 + 400 = 600$ grams and contains $25 + 386 = 411$ calories.

Since $200$ grams is one third of the batch, it has $$\frac{411}{3} = 137$$ calories.

Thus, the correct answer is B.

Answer: B

Solution:

The driver is one of the two parents: $2$ choices.

Any of the remaining $3$ people can sit in the front passenger seat, and the last $2$ people fill the back in $2$ orders.

The total is $2 \cdot 3 \cdot 2 = 12.$

Thus, the correct answer is B.

Answer: E

Solution:

Substituting $(1, 2)$ gives $$1 = \frac{2}{4} + a \quad\Rightarrow\quad a = \frac{1}{2},$$ and $$2 = \frac{1}{4} + b \quad\Rightarrow\quad b = \frac{7}{4}.$$

Therefore $$a + b = \frac{1}{2} + \frac{7}{4} = \frac{9}{4}.$$

Thus, the correct answer is E.

Answer: B

Solution:

Let the digits be $a \lt b \lt c$ with $c$ even. Since $a \geq 1,$ no digit is zero, and $c \neq 2$ (there is no room for two smaller nonzero digits).

Once the units digit $c$ is fixed, any two distinct digits below it can be arranged in increasing order in exactly one way. So the count for each $c$ is $\binom{c-1}{2}.$

For $c = 4, 6, 8$ this gives $$\binom{3}{2} + \binom{5}{2} + \binom{7}{2} = 3 + 10 + 21 = 34.$$

Thus, the correct answer is B.

Answer: A

Solution:

Let the sides be $x,$ $3x,$ and $15.$ The triangle inequality requires $x + 3x \gt 15,$ so $x \geq 4,$ and $x + 15 \gt 3x,$ so $x \leq 7.$

The perimeter $4x + 15$ is largest when $x = 7,$ giving $7 + 21 + 15 = 43.$

Thus, the correct answer is A.

Answer: E

Solution:

Joe adds the cream last, so his cup holds all $2$ ounces of cream.

JoAnn's cup has $14$ ounces of mixture containing $2$ ounces of cream. Drinking $2$ ounces removes a fraction $\tfrac{2}{14}$ of everything, leaving $$2 \cdot \frac{12}{14} = \frac{12}{7}$$ ounces of cream.

The ratio is $$\frac{2}{\,12/7\,} = \frac{14}{12} = \frac{7}{6}.$$

Thus, the correct answer is E.

Answer: D

Solution:

The vertex form is $y = a(x - p)^2 + p.$

At $x = 0,$ $y = ap^2 + p = -p,$ so $ap^2 = -2p$ and $a = -\dfrac{2}{p}.$

Expanding, $y = a x^2 - 2ap\, x + ap^2 + p,$ so $b = -2ap = -2\left(-\dfrac{2}{p}\right)p = 4.$

Thus, the correct answer is D.

Answer: C

Solution:

Because $\angle BAD = 60^\circ$ and $AB = AD,$ triangle $ABD$ is equilateral. The diagonals $AC$ and $BD$ together with segments $BE,$ $DF$ split $ABCD$ into six congruent triangles.

Each of these triangles has area $\dfrac{24}{6} = 4.$

Rhombus $BFDE$ is the union of $\triangle BED$ and $\triangle BFD,$ two of them, so its area is $8.$

Thus, the correct answer is C.

Answer: D

Solution:

The total cost in cents is $N(4B + 5J) = 253 = 11 \cdot 23.$ Since $N \gt 1,$ the value of $N$ is $11,$ $23,$ or $253.$

If $N = 253$ then $4B + 5J = 1,$ and if $N = 23$ then $4B + 5J = 11;$ neither has a positive integer solution.

So $N = 11$ and $4B + 5J = 23,$ whose only positive solution is $B = 2,$ $J = 3.$

The jam costs $N \cdot J \cdot 5$¢ $= 11 \cdot 3 \cdot 5 = 165$ cents, or $\$1.65.$

Thus, the correct answer is D.

Answer: B

Solution:

The circles are externally tangent, so $OP = 2 + 4 = 6.$ In quadrilateral $AOPD,$ both $OA = 2$ and $PD = 4$ are perpendicular to the tangent line $AD,$ making it a right trapezoid.

Drawing the line through $O$ parallel to $AD$ creates a right triangle with hypotenuse $OP = 6$ and one leg $PD - OA = 2,$ so $AD = \sqrt{6^2 - 2^2} = \sqrt{32} = 4\sqrt2.$

The trapezoid $AOPD$ has area $$\frac{1}{2}(2 + 4)(4\sqrt2) = 12\sqrt2.$$

By symmetry the hexagon $AOBCPD$ is made of two such trapezoids, so its area is $2 \cdot 12\sqrt2 = 24\sqrt2.$

Thus, the correct answer is B.

Answer: C

Solution:

The distance is $AC = \sqrt{7^2 + 1^2} = \sqrt{50}.$ In a regular hexagon with side $s,$ the distance between vertices two apart is $s\sqrt3,$ so $s^2 \cdot 3 = 50,$ giving $s^2 = \dfrac{50}{3}.$

The hexagon's area is $$\frac{3\sqrt3}{2}s^2 = \frac{3\sqrt3}{2} \cdot \frac{50}{3} = 25\sqrt3.$$

Thus, the correct answer is C.

Answer: C

Solution:

Since the weights sum to $21,$ the probability of rolling $k$ is $\dfrac{k}{21}.$

A total of $7$ comes from $(1,6), (2,5), \ldots, (6,1),$ so the probability is $$\frac{1 \cdot 6 + 2 \cdot 5 + 3 \cdot 4 + 4 \cdot 3 + 5 \cdot 2 + 6 \cdot 1}{21^2} = \frac{56}{441} = \frac{8}{63}.$$

Thus, the correct answer is C.

Answer: B

Solution:

Each step changes the coordinate sum by $1,$ so after $10$ steps the endpoint $(a, b)$ has $a + b$ even, and $|a| + |b| \le 10.$ Any such point is reachable: walk $|a| + |b|$ steps to it, then use the remaining even number of steps going out and back.

The reachable points lie on the lines $a + b = 2k$ for $-5 \le k \le 5.$ Each such line meets the diamond in exactly $11$ lattice points.

With $11$ lines and $11$ points each, there are $121$ points.

Thus, the correct answer is B.

Answer: B

Solution:

The number has the form $aabb,$ $abab,$ or $baab.$ Divisibility by $9$ means $2(a + b)$ is a multiple of $9,$ so $a + b = 9.$

The children include a $4$- or $8$-year-old, so the number is divisible by $4.$ The possibilities become $1188, 2772, 3636, 5544, 6336, 7272, 9900.$

Since the last two digits are Mr. Jones's age, $9900$ is impossible, and none of the others is a multiple of $5.$ So the children's ages cannot include $5.$ Indeed $5544$ is divisible by $1, 2, 3, 4, 6, 7, 8, 9.$

Thus, the correct answer is B.

Answer: C

Solution:

The equation says $\lfloor \log_{10} x \rfloor = \lfloor \log_{10} 4x \rfloor,$ i.e. $x$ and $4x$ lie in the same interval $[10^n, 10^{n+1}).$

This holds exactly when $10^n \le x$ and $4x \lt 10^{n+1},$ that is $10^n \le x \lt \dfrac{10^{n+1}}{4}.$

Within $[10^n, 10^{n+1}),$ the favorable fraction is $$\frac{10^{n+1}/4 - 10^n}{10^{n+1} - 10^n} = \frac{10/4 - 1}{10 - 1} = \frac{1}{6}.$$

Since this fraction is the same on every such interval, the overall probability is $\dfrac{1}{6}.$

Thus, the correct answer is C.

Answer: C

Solution:

Let the rectangle's sides be $x$ and $y.$ Point $A$ is on the ellipse with foci $B$ and $D,$ so $x + y = AB + AD = 2a.$ The distance between the foci is the diagonal, so $\sqrt{x^2 + y^2} = 2\sqrt{a^2 - b^2}.$

Then $2xy = (x + y)^2 - (x^2 + y^2) = 4a^2 - (4a^2 - 4b^2) = 4b^2,$ so $xy = 2b^2.$ The area gives $2b^2 = 2006,$ hence $b^2 = 1003.$

The ellipse area gives $\pi ab = 2006\pi,$ so $ab = 2006$ and $a = \dfrac{2006}{\sqrt{1003}} = 2\sqrt{1003}.$

The perimeter is $2(x + y) = 4a = 8\sqrt{1003}.$

Thus, the correct answer is C.

Answer: B

Solution:

Since factors of $2$ are more plentiful than factors of $5,$ $n$ equals the number of factors of $5$ in $a!\,b!\,c!,$ namely $$n = \sum_{k \ge 1}\left(\left\lfloor \tfrac{a}{5^k}\right\rfloor + \left\lfloor \tfrac{b}{5^k}\right\rfloor + \left\lfloor \tfrac{c}{5^k}\right\rfloor\right).$$

For each $k,$ $\lfloor a/5^k \rfloor + \lfloor b/5^k \rfloor + \lfloor c/5^k \rfloor \ge \lfloor 2006/5^k \rfloor - 2.$ Summing over $k = 1, 2, 3, 4$ (as $2006 \lt 5^5$) gives $$n \ge (401 + 80 + 16 + 3) - 4 \cdot 2 = 492.$$

Equality is attainable, for example with $a = b = 624$ and $c = 758.$ So the minimum is $492.$

Thus, the correct answer is B.

Answer: E

Solution:

Rotate $\triangle ABC$ by $90^\circ$ about $C,$ sending $A$ to $B$ and $P$ to $P'.$ Then $CP' = CP = 6$ and $\angle PCP' = 90^\circ,$ so $\triangle PCP'$ is an isosceles right triangle with $PP' = 6\sqrt2.$

Also $BP' = AP = 11.$ Since $(6\sqrt2)^2 + 7^2 = 72 + 49 = 121 = 11^2,$ triangle $BPP'$ has a right angle at $P.$ Hence $\angle BPC = \angle BPP' + \angle P'PC = 90^\circ + 45^\circ = 135^\circ.$

By the Law of Cosines in $\triangle BPC,$ $$BC^2 = 6^2 + 7^2 - 2 \cdot 6 \cdot 7 \cos 135^\circ = 85 + 42\sqrt2.$$

So $s^2 = 85 + 42\sqrt2,$ giving $a = 85,$ $b = 42,$ and $a + b = 127.$

Thus, the correct answer is E.

Answer: C

Solution:

Fixing $y,$ solve $\sin^2 x - \sin x \sin y + \sin^2 y = \dfrac34$ as a quadratic in $\sin x:$ $$\sin x = \frac{1}{2}\sin y \pm \frac{\sqrt3}{2}\cos y = \sin\!\left(y \pm \frac{\pi}{3}\right).$$

Within $S,$ $\sin x = \sin\!\left(y - \tfrac{\pi}{3}\right)$ gives the line $x = y - \tfrac{\pi}{3},$ while $\sin x = \sin\!\left(y + \tfrac{\pi}{3}\right)$ gives $x = y + \tfrac{\pi}{3}$ for $y \le \tfrac{\pi}{6}$ and $x = -y + \tfrac{2\pi}{3}$ for $y \ge \tfrac{\pi}{6}.$

These lines split $S$ into regions; testing the corners shows the inequality holds only in the middle band. Its area is $$\left(\frac{\pi}{2}\right)^2 - \frac{1}{2}\left(\frac{\pi}{3}\right)^2 - 2 \cdot \frac{1}{2} \left(\frac{\pi}{6}\right)^2 = \frac{\pi^2}{6}.$$

Thus, the correct answer is C.

Answer: B

Solution:

The rule gives $a_n \equiv a_{n+3} \pmod 2,$ so $a_2$ has the same parity as $a_{2006} = 1;$ thus $a_2$ is odd.

Every term is a multiple of $\gcd(a_1, a_2),$ and $a_{2006} = 1$ forces $\gcd(999, a_2) = 1.$ Since $999 = 3^3 \cdot 37,$ we need $a_2$ not divisible by $3$ or $37.$

Among the odd integers in $[1, 998]$ there are $499;$ removing the $166$ multiples of $3$ and $13$ multiples of $37,$ then adding back the $4$ multiples of $111,$ leaves $$499 - 166 - 13 + 4 = 324.$$

Each such $a_2$ works: with $\gcd(a_1, a_2) = 1$ the sequence eventually cycles through $1, 1, 0,$ and the parity condition makes $a_{2006} = 1.$

Thus, the correct answer is B.