Answer: C

Solution:

Inside the parentheses, $2^0-1+5^2+0 = 1-1+25+0 = 25.$

Then $$(25)^{-1}\times 5 = \dfrac{5}{25} = \dfrac{1}{5}.$$

Thus, the correct answer is C.

Answer: E

Solution:

By the Triangle Inequality, the third side $s$ satisfies $20-15 \lt s \lt 20+15,$ that is $5 \lt s \lt 35.$

The perimeter is $35 + s,$ so it lies strictly between $40$ and $70.$ Among the choices, only $72$ falls outside this range.

Thus, the correct answer is E.

Answer: E

Solution:

The sum of the $14$ other scores was $14\cdot 80 = 1120.$ The sum of all $15$ scores was $15\cdot 81 = 1215.$

Therefore Payton's score was $1215 - 1120 = 95.$

Thus, the correct answer is E.

Answer: B

Solution:

Let $x$ and $y$ be the numbers with $x \gt y \gt 0.$ Then $x+y = 5(x-y).$

Expanding gives $x+y = 5x-5y,$ so $6y = 4x$ and $$\dfrac{x}{y} = \dfrac{3}{2}.$$

Thus, the correct answer is B.

Answer: D

Solution:

To make $\dfrac{a}{b}$ larger, round the numerator $a$ up and the denominator $b$ down. To make $-c$ larger, round $c$ down.

Only choice $\text{(D)}$ does all three: it rounds $a$ up while rounding $b$ and $c$ down, so every change pushes the estimate above the exact value. In the other choices at least one change works the wrong way, so the estimate is not guaranteed to be larger.

Thus, the correct answer is D.

Answer: B

Solution:

Let $p$ and $c$ be Pete's and Claire's current ages. Then $p-2 = 3(c-2)$ and $p-4 = 4(c-4).$

Solving these gives $p = 20$ and $c = 8,$ so Pete is $12$ years older than Claire.

The ratio is $2 : 1$ when Claire is $12,$ which is $12 - 8 = 4$ years from now.

Thus, the correct answer is B.

Answer: D

Solution:

Let $r,h$ and $R,H$ be the radii and heights of the first and second cylinders. The volumes are equal, so $\pi r^2 h = \pi R^2 H,$ and $R = 1.1r.$

Then $\pi r^2 h = \pi(1.1r)^2 H = \pi(1.21 r^2) H.$ Dividing by $\pi r^2$ yields $h = 1.21H,$ so the first height is $21\%$ more than the second.

Thus, the correct answer is D.

Answer: C

Solution:

Let the sides of the rectangle be $4a$ and $3a.$ By the Pythagorean Theorem the diagonal is $5a = d,$ so $a = \dfrac{d}{5}.$

The area is $4a\cdot 3a = 12a^2 = 12\left(\dfrac{d}{5}\right)^2 = \dfrac{12}{25}d^2,$ so $k = \dfrac{12}{25}.$

Thus, the correct answer is C.

Answer: C

Solution:

Because the marbles left for Cheryl are determined at random, her two marbles are equally likely to be any pair. Fixing her first marble, the second is equally likely to be any of the $5$ remaining marbles.

Exactly one of those $5$ matches the first marble in color, so the probability is $\dfrac{1}{5}.$

Thus, the correct answer is C.

Answer: E

Solution:

Adding $1$ to both sides and factoring gives $$(x+1)(y+1) = 81 = 3^4.$$

Because $x$ and $y$ are distinct positive integers with $x \gt y,$ the only possibility is $x+1 = 3^3 = 27$ and $y+1 = 3.$ Therefore $x = 26.$

Thus, the correct answer is E.

Answer: D

Solution:

The number of common tangent lines depends on the relative position of the two circles:

If the smaller circle is inside the larger, there are $0$ tangents. If it is internally tangent, there is $1.$ If the circles intersect at two points, there are $2.$ If they are externally tangent, there are $3.$ If they are separated, there are $4.$

Thus $k$ can be any of $0, 1, 2, 3, 4,$ which gives $5$ possible values.

Thus, the correct answer is D.

Answer: B

Solution:

The $y$-intercepts of the two parabolas are $-2$ and $4.$ To intersect the $x$-axis, the first parabola opens upward and the second opens downward, so their $x$-intercepts are $\pm t$ for some $t \gt 0.$

The kite has one diagonal of length $4 - (-2) = 6$ along the $y$-axis and the other of length $2t.$ Its area is $\dfrac{1}{2}\cdot 6\cdot 2t = 6t = 12,$ so $t = 2.$

Thus the $x$-intercepts are $\pm 2.$ For the first parabola, $0 = a(2)^2 - 2$ gives $a = \dfrac{1}{2};$ for the second, $0 = 4 - b(2)^2$ gives $b = 1.$ Therefore $a + b = 1.5.$

Thus, the correct answer is B.

Answer: E

Solution:

Each of the $12$ teams plays $11$ games, so $\dfrac{12\cdot 11}{2} = 66$ games are played, and each game adds $2$ points to the list. The total of all scores is $66\cdot 2 = 132.$

If every game is a draw, each team scores $11,$ so the highest score need not reach $12;$ thus statement $\text{(E)}$ can fail. The other statements always hold: the sum $132 \ge 100,$ the sum being even forces an even number of odd scores and hence an even number of even scores, and two teams cannot both score $0$ because their mutual game gives at least one of them a point.

Thus, the correct answer is E.

Answer: D

Solution:

By the change-of-base formula, $\dfrac{1}{\log_b a} = \log_a b.$ Therefore $$1 = \log_a 2 + \log_a 3 + \log_a 4 = \log_a 24.$$

It follows that $a = 24.$

Thus, the correct answer is D.

Answer: C

Solution:

The numerator and denominator share no common factors. To write the fraction as a decimal, rewrite it with a power of $10$ in the denominator; the smallest that works is $10^{26}:$ $$\dfrac{123456789}{2^{26}\cdot 5^4} = \dfrac{123456789\cdot 5^{22}}{10^{26}}.$$

Since the numerator $123456789\cdot 5^{22}$ is not divisible by $10,$ the decimal has exactly $26$ places after the point.

Thus, the correct answer is C.

Answer: C

Solution:

Triangles $ABC$ and $ABD$ are $3$-$4$-$5$ right triangles with area $6$ and common hypotenuse $AB.$ Let $E$ be the foot of the altitude from $C$ to $AB;$ then $CE = \dfrac{3\cdot 4}{5} = \dfrac{12}{5}.$ Likewise the altitude from $D$ meets $AB$ at the same point $E$ with $DE = \dfrac{12}{5}.$

Triangle $CDE$ has sides $\dfrac{12}{5},$ $\dfrac{12}{5},$ and $CD = \dfrac{12}{5}\sqrt{2},$ so it is an isosceles right triangle with the right angle at $E.$ Thus $DE \perp CE$ and $DE \perp AB,$ making $DE$ perpendicular to the plane of $ABC.$

The tetrahedron's volume is $\dfrac{1}{3}\cdot [ABC]\cdot DE = \dfrac{1}{3}\cdot 6\cdot \dfrac{12}{5} = \dfrac{24}{5}.$

Thus, the correct answer is C.

Answer: A

Solution:

There are $2^8 = 256$ equally likely outcomes. Count the arrangements of standers (heads) with no two adjacent around the circle of $8$ seats, grouped by how many people stand.

The number of ways to choose $k$ non-adjacent seats from a circle of $n$ is $\dfrac{n}{n-k}\dbinom{n-k}{k}.$ For $n = 8$ this gives $$1,\; 8,\; 20,\; 16,\; 2$$ for $k = 0,1,2,3,4,$ and more than $4$ standers is impossible without an adjacency.

The total is $1 + 8 + 20 + 16 + 2 = 47,$ so the probability is $\dfrac{47}{256}.$

Thus, the correct answer is A.

Answer: C

Solution:

Let the integer zeros be $p$ and $q.$ By Vieta's formulas $p + q = a$ and $pq = 2a,$ so $pq = 2(p+q).$ Rearranging gives $$(p-2)(q-2) = 4.$$

The integer factor pairs of $4$ are $(1,4),(2,2),(4,1),(-1,-4),(-2,-2),(-4,-1),$ which yield $(p,q)$ pairs summing to $a = 9, 8, -1,$ and $0.$

The distinct possible values of $a$ are $9, 8, 0, -1,$ whose sum is $16.$

Thus, the correct answer is C.

Answer: B

Solution:

In every such quadrilateral $CD \ge AB.$ Let $E$ be the foot of the perpendicular from $A$ to $\overline{CD};$ then $CE = 2$ and $AE = BC.$ Let $x = AE$ and $y = DE,$ so $AD = 2 + y.$

By the Pythagorean Theorem $x^2 + y^2 = (2+y)^2,$ so $x^2 = 4 + 4y$ and $x$ is even. Writing $x = 2z$ gives $y = z^2 - 1,$ and the perimeter is $$x + 2y + 6 = 2z^2 + 2z + 4.$$

Increasing values $z = 1, 2, 3, \dots$ give the required quadrilaterals with increasing perimeter. For $z = 31$ the perimeter is $1988,$ and for $z = 32$ it is $2116.$ Therefore there are $31$ possible values of $p \lt 2015.$

Thus, the correct answer is B.

Answer: A

Solution:

The altitude of $T$ to its base of length $8$ is $\sqrt{5^2 - 4^2} = 3,$ so $T$ has area $\dfrac{1}{2}\cdot 8\cdot 3 = 12$ and perimeter $18.$

For $T'$ we need $2a + b = 18$ and area $\dfrac{1}{4}b\sqrt{4a^2 - b^2} = 12.$ Substituting $a = \dfrac{18 - b}{2}$ and squaring leads to $$(b - 8)(b^2 - b - 8) = 0.$$

Since $T$ and $T'$ are not congruent, $b \ne 8,$ so $b^2 - b - 8 = 0$ and $b = \dfrac{1 + \sqrt{33}}{2}.$ Because $25 \lt 33 \lt 36,$ this is between $3$ and $3.5,$ so the closest integer is $3.$

Thus, the correct answer is A.

Answer: D

Solution:

The ellipse $\dfrac{x^2}{16} + y^2 = 1$ has semi-axes $4$ and $1,$ so $c^2 = 16 - 1 = 15$ and the foci are $(\pm\sqrt{15}, 0).$

A circle through both foci has its center on the $y$-axis, say $(0, k),$ with radius $\sqrt{k^2 + 15}.$ Its top point always lies outside the ellipse. For four intersection points, its bottom point $(0, k - \sqrt{k^2 + 15})$ must be below $y = -1,$ which happens exactly when $0 \le k \lt 7.$

As $k$ ranges over $[0, 7),$ the radius $\sqrt{k^2 + 15}$ ranges over $[\sqrt{15}, 8),$ so $a + b = \sqrt{15} + 8.$

Thus, the correct answer is D.

Answer: D

Solution:

Note $S(1) = 2,$ $S(2) = 4,$ $S(3) = 8.$ Every valid sequence ends in a run of one, two, or three equal letters; removing that run leaves a valid sequence of length $n-1,$ $n-2,$ or $n-3.$ Thus $$S(n) = S(n-1) + S(n-2) + S(n-3).$$

Modulo $3,$ the sequence $S(n)$ is periodic with period $13.$ Since $2015 = 13\cdot 155,$ $S(2015) \equiv S(13) \equiv 2 \pmod 3.$ Modulo $4,$ it is periodic with period $4,$ and $2015 = 4\cdot 503 + 3,$ so $S(2015) \equiv S(3) \equiv 0 \pmod 4.$

Writing $S(2015) = 4k,$ the condition $4k \equiv 2 \pmod 3$ gives $k \equiv 2 \pmod 3,$ so $S(2015) \equiv 8 \pmod{12}.$

Thus, the correct answer is D.

Answer: A

Solution:

The second point is on the same side as the first with probability $\dfrac14,$ on the opposite side with probability $\dfrac14,$ and on an adjacent side with probability $\dfrac12.$

Opposite sides: the distance is at least $1 \ge \dfrac12$ always, probability $1.$

Same side: for points $(a, 0)$ and $(b, 0),$ the condition $|a - b| \ge \dfrac12$ has probability $\dfrac14.$

Adjacent sides: for points $(a, 0)$ and $(0, b),$ the condition $\sqrt{a^2 + b^2} \ge \dfrac12$ is the region outside a quarter-circle of radius $\dfrac12,$ with probability $1 - \dfrac14\pi\left(\dfrac12\right)^2 = 1 - \dfrac{\pi}{16}.$

The total probability is $$\dfrac14\cdot 1 + \dfrac14\cdot\dfrac14 + \dfrac12\left(1 - \dfrac{\pi}{16}\right) = \dfrac{26 - \pi}{32}.$$ Thus $a + b + c = 26 + 1 + 32 = 59.$

Thus, the correct answer is A.

Answer: D

Solution:

There are $20$ possible values for each of $a$ and $b,$ namely the reduced fractions in $[0, 2)$ with denominator dividing into $1 \le d \le 5.$

Writing $x = \cos(a\pi)$ and $y = \sin(b\pi),$ the fourth power $(x + iy)^4$ is real if and only if $x = 0,$ $y = 0,$ or $x = \pm y.$

The case $x = 0$ means $a \in \left\{\dfrac12, \dfrac32\right\},$ giving $2\cdot 20 = 40$ pairs; the case $y = 0$ means $b \in \{0, 1\},$ giving another $40$ pairs, of which $4$ were already counted. The remaining condition $\cos(a\pi) = \pm\sin(b\pi)$ with neither zero contributes $20$ more pairs.

In all there are $40 + 40 - 4 + 20 = 96$ valid pairs out of $400,$ so the probability is $\dfrac{96}{400} = \dfrac{6}{25}.$

Thus, the correct answer is D.

Answer: D

Solution:

If a circle of radius $r$ is tangent to the $x$-axis and nestled in the crevice between two circles of radii $r_1$ and $r_2$ that are also tangent to the axis and to each other, then $$\dfrac{1}{\sqrt{r}} = \dfrac{1}{\sqrt{r_1}} + \dfrac{1}{\sqrt{r_2}}.$$

Let $x = \dfrac{1}{\sqrt{70^2}} + \dfrac{1}{\sqrt{73^2}} = \dfrac{1}{70} + \dfrac{1}{73},$ which is the sum over $L_0.$ The single circle of $L_1$ also contributes $x.$ For $k \ge 2,$ each new circle contributes the sum of its two neighbors, and every earlier circle is counted twice except the two circles of $L_0;$ this yields a sum of $3^{k-1}x$ over $L_k.$

Therefore $$\sum_{C \in S} \dfrac{1}{\sqrt{r(C)}} = x + \sum_{k=1}^{6} 3^{k-1}x = x\left(1 + \dfrac{3^6 - 1}{2}\right) = x\cdot\dfrac{3^6 + 1}{2} = 365x.$$

Since $x = \dfrac{1}{70} + \dfrac{1}{73} = \dfrac{143}{70\cdot 73} = \dfrac{143}{5110},$ the sum is $365\cdot\dfrac{143}{5110} = \dfrac{143}{14}.$

Thus, the correct answer is D.