Solution:

There are $36$ red balls, and this count stays fixed. For the red balls to be $72\%$ of the urn, the urn must contain $36 \div 0.72 = 50$ balls. Since $100 - 50 = 50$, exactly $50$ blue balls are removed.

Thus, the correct answer is D.

Solution:

The rocks are worth $$2.80,$ $$2.75,$ and $$2$ per pound respectively, so the $1$-pound rocks are never worthwhile. Testing how many $5$-pound rocks to take and filling the rest with $4$-pound rocks: three $5$-pound rocks plus three $1$-pound rocks give $$42 + \$6 = \$48;$ two $5$-pound and two $4$-pound rocks use all $18$ pounds for $$28 + \$22 = \$50;$ one $5$-pound, three $4$-pound, and one $1$-pound rock give $$49.$

The maximum value is $$50.$ Thus, the correct answer is C.

Solution:

The choices of three non-consecutive periods are $\{1,3,5\},$ $\{1,3,6\},$ $\{1,4,6\},$ and $\{2,4,6\},$ a total of $4.$ The three distinct courses can be placed into any such set in $3! = 6$ orders, giving $4 \cdot 6 = 24$ schedules.

Thus, the correct answer is E.

Solution:

Negating the three false statements gives $d \lt 6,$ $d \gt 5,$ and $d \gt 4.$ The intersection of these conditions is $5 \lt d \lt 6,$ that is, the interval $(5, 6).$

Thus, the correct answer is D.

Solution:

Since $x^2 - 3x + 2 = (x-1)(x-2),$ its roots are $1$ and $2.$ If $1$ is a shared root then $1 - 5 + k = 0,$ so $k = 4.$ If $2$ is a shared root then $4 - 10 + k = 0,$ so $k = 6.$ The sum of possible values is $4 + 6 = 10.$

Thus, the correct answer is E.

Solution:

Because $m + 10 \lt n + 1,$ the six numbers are already increasing, so the median is the average of the middle two: $\frac{(m+10)+(n+1)}{2} = n,$ giving $m = n - 11.$ The mean condition is $$\frac{(n-11)+(n-7)+(n-1)+(n+1)+(n+2)+2n}{6} = n,$$ so $7n - 16 = 6n$ and $n = 16.$ Then $m = 5,$ and $m + n = 21.$

Thus, the correct answer is B.

Solution:

Since $4000 = 2^5 \cdot 5^3,$ the expression equals $2^{5+n} \cdot 5^{3-n}.$ This is an integer exactly when both $5 + n \ge 0$ and $3 - n \ge 0,$ that is, $-5 \le n \le 3.$ There are $3 - (-5) + 1 = 9$ such integers.

Thus, the correct answer is E.

Solution:

The base $DE$ of $\triangle ADE$ is $4$ times the base of a smallest triangle, so by the square scaling of similar areas, $[ADE] = 4^2 \cdot 1 = 16.$ The trapezoid $DBCE$ is what remains of $\triangle ABC,$ so its area is $40 - 16 = 24.$

Thus, the correct answer is E.

Solution:

For $0 \le x \le \pi$ and $0 \le y \le \pi$ we have $\sin x \ge 0,$ $\sin y \ge 0,$ $\cos x \le 1,$ and $\cos y \le 1.$ Hence $$\sin(x+y) = \sin x \cos y + \cos x \sin y \le \sin x + \sin y.$$ The inequality therefore holds for every $y$ with $0 \le y \le \pi.$

Thus, the correct answer is E.

Solution:

The second equation gives $|x| - |y| = \pm 1,$ equivalently $x = \pm y \pm 1.$ Substituting into $x + 3y = 3:$

If $x = y + 1,$ then $(x, y) = \left(\tfrac32, \tfrac12\right).$ If $x = y - 1,$ then $(x, y) = (0, 1).$ If $x = -y + 1,$ then again $(x, y) = (0, 1).$ If $x = -y - 1,$ then $(x, y) = (-3, 2).$

The distinct solutions are $(-3, 2),$ $(0, 1),$ and $\left(\tfrac32, \tfrac12\right),$ all of which check, so there are $3.$

Thus, the correct answer is C.

Solution:

The crease lies along the perpendicular bisector of $AB,$ meeting $AC$ at $E$ because $AC \gt BC.$ Let $D$ be the midpoint of $AB,$ so $AD = \tfrac52$ and $\triangle ADE$ is right-angled at $D.$ Since $\triangle ADE \sim \triangle ACB,$ we have $\tfrac{DE}{AD} = \tfrac{CB}{AC} = \tfrac34,$ so $$DE = \frac52 \cdot \frac34 = \frac{15}{8}.$$

Thus, the correct answer is D.

Solution:

Partition $\{1, \ldots, 12\}$ into the six divisibility chains $\{1,2,4,8\},$ $\{3,6,12\},$ $\{5,10\},$ $\{7\},$ $\{9\},$ $\{11\}.$ Since no element of $S$ may divide another, at most one comes from each chain; needing $6$ elements forces exactly one from each, so $7, 9, 11 \in S.$

Because $9 \in S,$ $3 \notin S,$ so the second chain contributes $6$ or $12,$ and then neither $1$ nor $2$ can be chosen from the first chain (they divide $6$ and $12$). Taking $4$ from the first chain works: $S = \{4, 5, 6, 7, 9, 11\}$ has the property. Hence the least possible element is $4.$

Thus, the correct answer is C.

Solution:

Adding $1$ to every $a_i$ gives a bijection between these expressions and the base-$3$ numerals for $0$ through $3^8 - 1,$ so exactly $3^8 = 6561$ distinct integers occur. They are symmetric about $0$ (negating all $a_i$ negates the value), so besides $0$ itself, half are positive: $$1 + \tfrac12(6561 - 1) = 3281$$ nonnegative integers, namely $0$ through $3280.$

Thus, the correct answer is D.

Solution:

Writing both logarithms in base $2:$ $\tfrac{2}{\log_2 3x} = \tfrac{3}{\log_2 2x},$ so $2 \log_2 2x = 3 \log_2 3x,$ i.e. $(2x)^2 = (3x)^3.$ Then $4x^2 = 27x^3,$ giving $x = \tfrac{4}{27}.$ Since $\gcd(4, 27) = 1,$ we get $p + q = 4 + 27 = 31.$

Thus, the correct answer is D.

Solution:

Under the symmetry group of the square, the $49$ cells break into orbits, and every cell in an orbit must have the same color. Classifying cells by distance from the center yields exactly $10$ orbits that can be colored independently. Each orbit is black or white, giving $2^{10}$ colorings, but the all-black and all-white grids are excluded. So there are $2^{10} - 2 = 1022$ symmetric scanning codes.

Thus, the correct answer is B.

Solution:

Substituting $x^2 = y + a$ into $x^2 + y^2 = a^2$ gives $y^2 + y + (a - a^2) = 0,$ which factors as $(y + 1 - a)(y + a) = 0,$ so $y = a - 1$ or $y = -a.$ These correspond to $x^2 = 2a - 1$ and $x^2 = 0.$

The equation $x^2 = 0$ always gives the single point $(0, -a),$ the vertex of the parabola. The equation $x^2 = 2a - 1$ gives two more points exactly when $2a - 1 \gt 0,$ i.e. $a \gt \tfrac12.$ So there are $3$ intersection points precisely when $a \gt \tfrac12.$

Thus, the correct answer is E.

Solution:

Place the right angle at the origin with legs on the axes, so the vertices are $(4, 0),$ $(0, 3),$ $(0, 0),$ and the square $S$ is $[0, s] \times [0, s].$ The hypotenuse is $3x + 4y - 12 = 0,$ and the distance from its nearest corner $(s, s)$ is $$\frac{|3s + 4s - 12|}{\sqrt{3^2 + 4^2}} = \frac{|7s - 12|}{5} = 2.$$ This gives $s = \tfrac{22}{7}$ or $s = \tfrac27;$ only $s = \tfrac27$ keeps the square inside the triangle.

The field has area $\tfrac12 \cdot 3 \cdot 4 = 6$ and the unplanted square has area $\left(\tfrac27\right)^2 = \tfrac{4}{49}.$ The planted fraction is $$1 - \frac{4/49}{6} = 1 - \frac{2}{147} = \frac{145}{147}.$$

Thus, the correct answer is D.

Solution:

Since $D$ and $E$ are midpoints, $\triangle ADE$ has $\tfrac14$ the area of $\triangle ABC,$ namely $30,$ so trapezoid $EDBC$ has area $120 - 30 = 90.$

By the Angle Bisector Theorem, $G$ divides $BC$ with $BG = \tfrac{AB}{AB + AC} \cdot BC = \tfrac56 BC,$ and likewise $F$ divides $DE$ so that $DF = \tfrac56 DE.$ Because $FDBG$ and $EDBC$ share the same height, the area of $FDBG$ is $\tfrac56$ of the area of $EDBC:$ $\tfrac56 \cdot 90 = 75.$

Thus, the correct answer is D.

Solution:

Each element of $A$ is uniquely $2^i 3^j 5^k$ with $i, j, k \ge 0,$ so summing all reciprocals factors as $$\left(\sum_{i \ge 0} \tfrac{1}{2^i}\right) \left(\sum_{j \ge 0} \tfrac{1}{3^j}\right) \left(\sum_{k \ge 0} \tfrac{1}{5^k}\right) = \frac{1}{1 - \frac12} \cdot \frac{1}{1 - \frac13} \cdot \frac{1}{1 - \frac15}.$$ This equals $2 \cdot \tfrac32 \cdot \tfrac54 = \tfrac{15}{4}.$ With $\gcd(15, 4) = 1,$ $m + n = 15 + 4 = 19.$

Thus, the correct answer is C.

Solution:

Since $\triangle ABC$ is an isosceles right triangle, $CM = BM = \tfrac32 \sqrt2$ and the base angles at $B, C$ are $45^\circ.$ As $AIME$ is cyclic with right angle at $A,$ angle $\angle IME = 90^\circ.$ Let $x = CI$ and $y = BE.$ By the Law of Cosines in $\triangle MCI,$ $$IM^2 = x^2 + \tfrac92 - 2 \cdot x \cdot \tfrac32\sqrt2 \cdot \cos 45^\circ = x^2 - 3x + \tfrac92,$$ and similarly $ME^2 = y^2 - 3y + \tfrac92.$

The Pythagorean Theorem in right triangles $EMI$ and $IAE$ gives $IM^2 + ME^2 = (3-x)^2 + (3-y)^2,$ which simplifies to $x + y = 3.$ The area condition $\tfrac12 IM \cdot ME = 2$ means $IM^2 \cdot ME^2 = 16.$ Substituting $y = 3 - x$ makes $ME^2 = x^2 - 3x + \tfrac92 = IM^2,$ so $\left(x^2 - 3x + \tfrac92\right)^2 = 16,$ hence $x^2 - 3x + \tfrac92 = 4,$ i.e. $x^2 - 3x + \tfrac12 = 0.$

Since $AI \gt AE$ forces $y \gt x,$ we take the smaller root $x = \tfrac{3 - \sqrt{7}}{2}.$ Then $a + b + c = 3 + 7 + 2 = 12.$

Thus, the correct answer is D.

Solution:

Each polynomial in choices A–D has no positive root and exactly one negative root, which lies in $(-1, 0)$ (it is positive at $0$ and negative at $-1$) and is increasing there. On the interval $(-1, 0),$ $x^{19} \lt x^{17}$ and $x^{13} \lt x^{11}.$ Increasing a term makes the polynomial larger, which pushes its root to the left (smaller). So the smallest exponents give the greatest root, favoring choice B ($x^{17} + 2018x^{11} + 1$) over A, C, and D.

The linear choice E has root $-\tfrac{2018}{2019} = -\left(1 - \tfrac{1}{2019}\right),$ very close to $-1;$ evaluating the polynomial of choice B there gives a negative value, so B's root lies to the right of E's. Hence B has the greatest real root.

Thus, the correct answer is B.

Solution:

Writing $z = a + bi$ with $(a + bi)^2 = 4 + 4\sqrt{15}\,i$ gives $a^2 - b^2 = 4$ and $2ab = 4\sqrt{15}.$ Then $a^4 - 4a^2 - 60 = 0,$ so $(a^2 - 10)(a^2 + 6) = 0,$ yielding $a = \pm\sqrt{10},$ $b = \pm\sqrt{6}.$ The vertices from the first equation are $\pm(\sqrt{10} + \sqrt{6}\,i).$ The same method on $z^2 = 2 + 2\sqrt{3}\,i$ gives $\pm(\sqrt{3} + i).$

Applying the shoelace formula to $(\sqrt{10}, \sqrt{6}),$ $(\sqrt{3}, 1),$ $(-\sqrt{10}, -\sqrt{6}),$ $(-\sqrt{3}, -1)$ gives area $6\sqrt{2} - 2\sqrt{10}.$ Thus $p + q + r + s = 6 + 2 + 2 + 10 = 20.$

Thus, the correct answer is A.

Solution:

Extend $PN$ through $N$ to $Q$ with $PN = NQ.$ Since $N$ is the midpoint of $UG$ and of $PQ,$ the quadrilateral $UPGQ$ is a parallelogram, so $GQ \parallel PT$ and $GQ = PU = 1 = AG.$ Then $\angle QGA = 180^\circ - \angle T = \angle P + \angle A = 36^\circ + 56^\circ = 92^\circ,$ and the isosceles triangle $QGA$ gives $\angle QAG = \tfrac12(180^\circ - 92^\circ) = 44^\circ.$

Because $M, N$ are midpoints, $MN$ is a midline of $\triangle QPA,$ so $MN \parallel AQ$ and $$\angle NMP = \angle QAP = \angle QAG + \angle GAP = 44^\circ + 56^\circ = 100^\circ.$$ The acute angle between line $MN$ and $PA$ is therefore $180^\circ - 100^\circ = 80^\circ.$

Thus, the correct answer is E.

Solution:

If $c \le \tfrac12,$ Carol beats Bob automatically, so she wins only if Alice is below $c,$ probability $c \le \tfrac12.$ If $c \ge \tfrac23,$ she wins with probability $1 - c \le \tfrac13.$ Neither case exceeds $\tfrac12.$

For $\tfrac12 \lt c \lt \tfrac23,$ the chance Bob's number exceeds $c$ is $\frac{2/3 - c}{2/3 - 1/2} = 4 - 6c,$ so the probability Carol is above Alice and below Bob is $c(4 - 6c);$ the reverse ordering has probability $(1 - c)(6c - 3).$ Adding, $$c(4 - 6c) + (1 - c)(6c - 3) = -12c^2 + 13c - 3.$$ This downward parabola is maximized at $c = \tfrac{13}{24},$ which lies in $\left(\tfrac12, \tfrac23\right),$ and its value exceeds $\tfrac12.$

Thus, the correct answer is B.

Solution:

Using $A_n = a \cdot \tfrac{10^n - 1}{9},$ $B_n = b \cdot \tfrac{10^n - 1}{9},$ and $C_n = c \cdot \tfrac{10^{2n} - 1}{9},$ the equation $C_n - B_n = A_n^2$ becomes, after dividing by $10^n - 1$ and clearing fractions, $$(9c - a^2) \cdot 10^n = 9b - 9c - a^2.$$ For this to hold at two different $n,$ the coefficient of $10^n$ must be zero, so $9c = a^2$ and hence $9b - 9c - a^2 = 0.$