Solution:

The troop buys $1000 \div 5 = 200$ groups of five bars, costing $200 \cdot 2 = 400$ dollars.

They sell $1000 \div 2 = 500$ pairs of bars, earning $500 \cdot 1 = 500$ dollars.

The profit is $500 - 400 = 100$ dollars.

Thus, the correct answer is A.

Solution:

The statement translates to $$\dfrac{x}{100}\cdot x = 4,$$ so $x^2 = 400.$

Since $x$ is positive, $x = 20.$

Thus, the correct answer is D.

Solution:

Buying all the CDs costs three times what one third of them cost, namely $3\cdot\dfrac15 = \dfrac35$ of her money.

She has $1 - \dfrac35 = \dfrac25$ of her money left.

Thus, the correct answer is C.

Solution:

Lisa needs an A on at least $0.8 \cdot 50 = 40$ quizzes.

She has $22$ already, so she needs $40 - 22 = 18$ more of the remaining $20$ quizzes.

She can earn a lower grade on at most $20 - 18 = 2$ of them.

Thus, the correct answer is B.

Solution:

The four quarter circles in a tile together form one full circle of radius $\dfrac12,$ with area $\pi\left(\dfrac12\right)^2 = \dfrac{\pi}{4}.$

So each tile has shaded area $1 - \dfrac{\pi}{4}$ square feet.

There are $8 \cdot 10 = 80$ tiles, so the total shaded area is $$80\left(1 - \dfrac{\pi}{4}\right) = 80 - 20\pi.$$

Thus, the correct answer is A.

Solution:

Let $H$ be the foot of the altitude from $C$ to line $AB.$ Since $\triangle ABC$ is isosceles with $AC = BC,$ $H$ is the midpoint of $AB,$ so $AH = HB = 1.$

Then $CH^2 = 7^2 - 1^2 = 48.$ Applying the Pythagorean Theorem to $\triangle CHD$ with $HD = HB + BD = 1 + BD$ gives $$8^2 = 48 + (1 + BD)^2,$$ so $(1 + BD)^2 = 16.$

Therefore $1 + BD = 4,$ which means $BD = 3.$

Thus, the correct answer is A.

Solution:

Setting $y = 0$ gives $|3x| = 12,$ so $x = \pm 4.$ Setting $x = 0$ gives $|4y| = 12,$ so $y = \pm 3.$

The graph is a rhombus with vertices $(\pm 4, 0)$ and $(0, \pm 3),$ so its diagonals have lengths $8$ and $6.$

Its area is $\dfrac12 \cdot 8 \cdot 6 = 24.$

Thus, the correct answer is D.

Solution:

The vertex of the parabola $y = x^2 + a^2$ is $(0, a^2).$

The line $y = x + a$ passes through it exactly when $a^2 = 0 + a,$ that is $a^2 - a = 0.$

This gives $a = 0$ or $a = 1,$ so there are $2$ values.

Thus, the correct answer is C.

Solution:

The percentage scoring $95$ is $100 - 10 - 25 - 20 - 15 = 30.$

The mean is $$0.10(70) + 0.25(80) + 0.20(85) + 0.15(90) + 0.30(95) = 86.$$

Cumulatively, $10\%$ are below $80,$ $35\%$ are at or below $80,$ and $55\%$ are at or below $85.$ The middle scores fall at $85,$ so the median is $85.$

The difference is $86 - 85 = 1.$

Thus, the correct answer is B.

Solution:

The sequence begins $2005, 133, 55, 250, 133, \ldots$ since $2^3 + 0^3 + 0^3 + 5^3 = 133,$ $1^3 + 3^3 + 3^3 = 55,$ $5^3 + 5^3 = 250,$ and $2^3 + 5^3 + 0^3 = 133.$

After the initial $2005,$ the terms cycle through $133, 55, 250$ with period $3.$

Term $n$ for $n \ge 2$ is the $((n-2)\bmod 3)$th entry of $133, 55, 250.$ Since $2005 - 2 = 2003 \equiv 2 \pmod 3,$ the $2005$th term is $250.$

Thus, the correct answer is E.

Solution:

There are $\binom{8}{2} = 28$ equally likely pairs of bills.

The sum is $\$20$ or more in these cases: both twenties ($1$ way), one twenty with one of the six smaller bills ($2 \cdot 6 = 12$ ways), or both tens ($1$ way).

That is $1 + 12 + 1 = 14$ favorable pairs, so the probability is $\dfrac{14}{28} = \dfrac12.$

Thus, the correct answer is D.

Solution:

Let $r_1$ and $r_2$ be the roots of $x^2 + px + m = 0,$ so $m = r_1 r_2$ and $p = -(r_1 + r_2).$

The roots of $x^2 + mx + n = 0$ are $2r_1$ and $2r_2,$ so $n = 4r_1 r_2$ and $m = -2(r_1 + r_2).$

Then $n = 4m$ and $m = 2p,$ which gives $p = \dfrac{m}{2},$ so $$\dfrac{n}{p} = \dfrac{4m}{\tfrac{m}{2}} = 8.$$

Thus, the correct answer is D.

Solution:

From $4^{x_1} = 5$ we get $x_1 = \log_4 5,$ and in general $x_k = \log_{k+3}(k+4).$

The product telescopes: $$x_1 x_2 \cdots x_{124} = \log_4 5 \cdot \log_5 6 \cdots \log_{127} 128 = \log_4 128.$$

Since $128 = 2^7$ and $4 = 2^2,$ this equals $\dfrac{7\log 2}{2\log 2} = \dfrac72.$

Thus, the correct answer is D.

Solution:

Since the circle is tangent to $y = 6$ and its center $(0, k)$ is above that line, the radius is $r = k - 6.$

The distance from $(0, k)$ to the line $x - y = 0$ is $\dfrac{|0 - k|}{\sqrt2} = \dfrac{k}{\sqrt2},$ and this must also equal $r.$

Setting $\dfrac{k}{\sqrt2} = k - 6$ gives $k = \dfrac{6\sqrt2}{\sqrt2 - 1} = 6\sqrt2\,(\sqrt2+1) = 12 + 6\sqrt2.$

Then $r = k - 6 = 6 + 6\sqrt2.$

Thus, the correct answer is E.

Solution:

The eight digits are distinct and chosen from $1$ through $9,$ whose total is $45.$ So the eight used digits sum to between $45 - 9 = 36$ and $45 - 1 = 44.$

Let the four units digits sum to $U$ and the four tens digits sum to $T.$ Then $10T + U = 221,$ so $U$ ends in $1.$ Since $1+2+3+4 = 10 \le U \le 6+7+8+9 = 30,$ we have $U = 11$ or $U = 21.$

If $U = 11,$ then $10T = 210,$ so $T = 21$ and the eight digits sum to $32,$ which is below $36.$ So $U = 21,$ giving $T = 20$ and total $41.$

The missing digit is $45 - 41 = 4.$ For example, $13 + 25 + 86 + 97 = 221.$

Thus, the correct answer is D.

Solution:

A sphere of radius $1$ tangent to the three coordinate planes in one octant has its center at a point like $(1, 1, 1),$ at distance $$\sqrt{1^2 + 1^2 + 1^2} = \sqrt3$$ from the origin.

The farthest point of that sphere from the origin is at distance $\sqrt3 + 1,$ so the containing sphere has radius $1 + \sqrt3.$

Thus, the correct answer is D.

Solution:

The equation is equivalent to $\log_{10}\left(2^a 3^b 5^c 7^d\right) = 2005,$ so $$2^a 3^b 5^c 7^d = 10^{2005} = 2^{2005} \cdot 5^{2005}.$$

Clearing the denominators of $a, b, c, d$ with a common integer multiplier and using the uniqueness of prime factorization, the exponents must match: $a = 2005,$ $b = 0,$ $c = 2005,$ and $d = 0.$

So there is exactly $1$ such four-tuple.

Thus, the correct answer is B.

Solution:

Line $AB$ has slope $1.$ For $\angle A$ to be acute, $C$ must lie beyond the line through $A$ perpendicular to $AB;$ in the first quadrant that line runs between $P(4, 0)$ and $Q(0, 4).$ For $\angle B$ to be acute, $C$ must lie before the line through $B$ perpendicular to $AB,$ between $S(14, 0)$ and $T(0, 14).$

For $\angle C$ to be acute, $C$ must lie outside the circle $U$ with diameter $AB,$ whose radius is $\dfrac{AB}{2} = \dfrac{5\sqrt2}{2}.$

The region is the large right triangle $OST$ minus the small right triangle $OPQ$ minus the semicircle-equivalent area of $U$ inside the strip: $$\dfrac12 \cdot 14^2 - \dfrac12 \cdot 4^2 - \pi\left(\dfrac{5\sqrt2}{2}\right)^2 = 98 - 8 - \dfrac{25\pi}{2} = 90 - \dfrac{25\pi}{2} \approx 51.$$

Thus, the correct answer is C.

Solution:

Let $x = 10a + b$ and $y = 10b + a$ with $a \gt b.$ Then $$x^2 - y^2 = (10a+b)^2 - (10b+a)^2 = 99(a^2 - b^2) = 99(a+b)(a-b).$$

Since $99 = 9 \cdot 11,$ for this to be a perfect square we need $11 \mid (a+b)(a-b).$ As $a + b \le 17$ and $a - b \le 8,$ the only multiple of $11$ available is $a + b = 11.$

Then $x^2 - y^2 = 9 \cdot 11^2 (a - b),$ which is a perfect square exactly when $a - b$ is a perfect square. Taking $a - b = 1$ with $a + b = 11$ gives $(a, b) = (6, 5).$

So $x = 65,$ $y = 56,$ and $m = \sqrt{65^2 - 56^2} = \sqrt{1089} = 33.$ Thus $x + y + m = 65 + 56 + 33 = 154.$

Thus, the correct answer is E.

Solution:

The elements sum to $8.$ If $a + b + c + d = x,$ then $e + f + g + h = 8 - x,$ so $$x^2 + (8 - x)^2 = 2(x - 4)^2 + 32.$$

This is minimized when $x = 4,$ giving $32.$ But $13$ must lie in one group, and no three of the remaining elements add with $13$ to make $4$ (that would need three of them summing to $-9,$ which is impossible here). So $x = 4$ is unattainable and $(x - 4)^2 \ge 1.$

The minimum is $2(1) + 32 = 34,$ achieved for instance by $\{-7, -5, 2, 13\}$ (sum $3$) and $\{-3, -2, 4, 6\}$ (sum $5$).

Thus, the correct answer is C.

Solution:

Write $n = 7^k Q$ where $7 \nmid Q,$ and let $d$ be the number of divisors of $Q.$ Then $n$ has $(k + 1)d = 60$ divisors and $7n = 7^{k+1}Q$ has $(k + 2)d = 80$ divisors.

Dividing, $\dfrac{k + 2}{k + 1} = \dfrac{80}{60} = \dfrac43,$ so $3(k + 2) = 4(k + 1),$ giving $k = 2.$

Thus, the correct answer is C.

Solution:

Because $|z_0| = 1,$ every $|z_n| = 1,$ so $\overline{z_n} = \dfrac{1}{z_n}$ and $$z_{n+1} = \dfrac{i z_n}{\overline{z_n}} = i z_n^2.$$

Iterating, $z_1 = i z_0^2,$ $z_2 = i(i z_0^2)^2 = -i z_0^4,$ and in general for $n \ge 2,$ $z_n = (\text{constant of modulus }1)\cdot z_0^{2^n}.$

The condition $z_{2005} = 1$ becomes an equation of the form $z_0^{2^{2005}} = c$ for a fixed constant $c$ with $|c| = 1.$ Every nonzero complex equation $z_0^{N} = c$ has exactly $N$ distinct solutions, all on the unit circle.

Here $N = 2^{2005},$ so there are $2^{2005}$ possible values for $z_0.$

Thus, the correct answer is E.

Solution:

The conditions give $x + y = 10^z$ and $x^2 + y^2 = 10 \cdot 10^z.$ Then $$2xy = (x+y)^2 - (x^2+y^2) = 10^{2z} - 10 \cdot 10^z,$$ so $xy = \dfrac12\left(10^{2z} - 10 \cdot 10^z\right).$

Using $x^3 + y^3 = (x+y)^3 - 3xy(x+y),$ $$x^3 + y^3 = 10^{3z} - \dfrac32\left(10^{2z} - 10 \cdot 10^z\right)10^z = -\dfrac12 \cdot 10^{3z} + 15 \cdot 10^{2z}.$$

So $a = -\dfrac12$ and $b = 15,$ giving $a + b = \dfrac{29}{2}.$

Thus, the correct answer is B.

Solution:

For vertices $(a, a^2), (b, b^2), (c, c^2),$ the slope of a side is $\dfrac{b^2 - a^2}{b - a} = a + b.$ Adding the three side slopes, $$(a+b) + (b+c) + (c+a) = 2(a + b + c) = 2 \cdot \dfrac{m}{n}.$$

One side has slope $2 = \tan\theta.$ Because the triangle is equilateral, its sides make angles $\theta$ and $\theta \pm 60^\circ,$ so the other two slopes are $$\tan(\theta \pm 60^\circ) = \dfrac{2 \pm \sqrt3}{1 \mp 2\sqrt3} = -\dfrac{8 \pm 5\sqrt3}{11}.$$

The sum of the three slopes is $2 - \dfrac{8 + 5\sqrt3}{11} - \dfrac{8 - 5\sqrt3}{11} = \dfrac{22 - 16}{11} = \dfrac{6}{11}.$

Thus $a + b + c = \dfrac12 \cdot \dfrac{6}{11} = \dfrac{3}{11},$ so $m + n = 3 + 11 = 14.$

Thus, the correct answer is A.

Solution:

There are $4^6$ equally likely combinations of moves. Label the vertices $A, B, C, A', B', C',$ where primed vertices are opposite the corresponding unprimed ones. An ant cannot move to its own vertex or the opposite one, so a valid outcome is a permutation $f$ with $f(A) \notin \{A, A'\},$ and similarly for each pair.

There are $4 \cdot 3 = 12$ ordered choices for $(f(A), f(A')).$ Of these, $f(A)$ and $f(A')$ are opposite in $4$ cases and adjacent in $8.$

If $f(A), f(A')$ are opposite, say $B, B',$ then $\{f(C), f(C')\} = \{A, A'\}$ and $\{f(B), f(B')\} = \{C, C'\},$ giving $4 \cdot 2 \cdot 2 = 16$ valid combinations.

If $f(A), f(A')$ are adjacent, say $B, C,$ then one of $f(B), f(B')$ must be $C'$ and there are $4$ ordered choices for $(f(B), f(B')),$ each leaving $2$ for $(f(C), f(C')):$ that is $8 \cdot 4 \cdot 2 = 64$ valid combinations.

Hence the probability is $$\dfrac{16 + 64}{4^6} = \dfrac{80}{4096} = \dfrac{5}{256}.$$

Thus, the correct answer is A.