Solution:

Let $n$ be the number of nickels, so Leah has $13-n$ pennies. One more nickel would give her $n+1$ nickels, and this equals the number of pennies: $$n+1 = 13-n.$$ Solving gives $n=6,$ so there are $6$ nickels and $7$ pennies.

The total value is $6\cdot5 + 7 = 37$ cents.

Thus, the correct answer is C.

Solution:

Under the sale, a pair of balloons costs $1 + \tfrac23 = \tfrac53$ times the regular price of one balloon.

Orvin's money buys $30$ balloons at the regular price, so he can afford $$30 \div \tfrac53 = 18$$ pairs, which is $36$ balloons.

Thus, the correct answer is C.

Solution:

The fraction of the trip on pavement is $$1 - \tfrac13 - \tfrac15 = \tfrac{7}{15}.$$

Since this equals $20$ miles, the whole trip is $$20 \div \tfrac{7}{15} = \tfrac{300}{7}$$ miles.

Thus, the correct answer is E.

Solution:

Let a muffin cost $m$ and a banana cost $b.$ Then $$2(4m+3b) = 2m+16b.$$

Expanding gives $8m+6b = 2m+16b,$ so $6m = 10b$ and $m = \tfrac53 b.$

Thus, the correct answer is B.

Solution:

Let each pane have width $2x$ and height $5x.$ The window is $4$ panes wide with $5$ vertical borders, so its width is $4(2x) + 5\cdot2 = 8x+10.$

It is $2$ panes tall with $3$ horizontal borders, so its height is $2(5x) + 3\cdot2 = 10x+6.$

Setting width equal to height gives $8x+10 = 10x+6,$ so $x=2$ and the side length is $10\cdot2+6 = 26.$

Thus, the correct answer is A.

Solution:

Let a regular lemonade hold $a$ ounces, so Ann's large holds $\tfrac32 a.$ After each drinks $\tfrac34,$ Ann has $\tfrac14\cdot\tfrac32 a = \tfrac38 a$ left, and she gives Ed $\tfrac13\cdot\tfrac38 a + 2 = \tfrac18 a + 2$ ounces.

Ed drinks his full $a$ ounces plus that gift, and Ann drinks her $\tfrac32 a$ minus the gift. Setting these equal, $$a + \left(\tfrac18 a + 2\right) = \tfrac32 a - \left(\tfrac18 a + 2\right),$$ which gives $4 = \tfrac14 a,$ so $a = 16.$

Then Ed drank $16$ ounces and Ann drank $24$ ounces, for a total of $40$ ounces.

Thus, the correct answer is D.

Solution:

Write $$\dfrac{n}{30-n} = \dfrac{30}{30-n} - 1.$$

For this to be a positive integer, $30-n$ must be a positive divisor of $30$ with $\dfrac{30}{30-n} \ge 2,$ i.e. $30-n \le 15.$

The divisors of $30$ that are at most $15$ are $1, 2, 3, 5, 6, 10, 15,$ giving $7$ values of $n$ (namely $15, 20, 24, 25, 27, 28, 29$).

Thus, the correct answer is D.

Solution:

The leftmost column shows $A+B = D$ with no carry out, so $A+B \le 9.$ Examining the tens and thousands columns (each of the form $C + \text{digit} + \text{carry}$ producing the same digit) forces $C = 0$ and eliminates all carries.

Every column then reduces to $A+B = D,$ with $A, B, C=0$ distinct. Since $A$ and $B$ are distinct positive digits, $D = A+B$ can be any value from $3$ up to $9,$ giving $7$ possibilities, for example $(A,B,C,D) = (1,2,0,3), (1,3,0,4), \ldots, (2,7,0,9).$

Thus, the correct answer is C.

Solution:

By the Pythagorean Theorem in right triangle $ABC,$ $AC = \sqrt{3^2+4^2} = 5.$

Since $5^2 + 12^2 = 13^2,$ the converse of the Pythagorean Theorem shows $\angle DAC = 90^\circ,$ so $\triangle DAC$ is right.

The area of $\triangle ABC$ is $\tfrac12\cdot3\cdot4 = 6$ and the area of $\triangle DAC$ is $\tfrac12\cdot5\cdot12 = 30.$ The quadrilateral has area $6 + 30 = 36.$

Thus, the correct answer is B.

Solution:

The distance driven is $cba - abc = 99(c-a),$ a multiple of $9.$ Driving a whole number of hours at $55$ mph makes it a multiple of $55$ too, hence a multiple of $495.$

Since the odometer difference is at most a $3$-digit number and $a \ge 1,$ the distance must be $495,$ so $c - a = 5.$

With $a \ge 1$ and $a+b+c \le 7,$ the only choice is $a=1,$ $c=6,$ $b=0.$ Then $a^2+b^2+c^2 = 1 + 0 + 36 = 37.$

Thus, the correct answer is D.

Solution:

The list sums to $11 \cdot 10 = 110.$ To maximize one entry, minimize the sum of the other ten.

Sorted, the sixth number must be $9$ (the median), and $8$ must appear more often than any other value. Trying $8$ three times, the smallest possible ten numbers are $$1,1,8,8,8,9,9,10,10,11,$$ which sum to $75$ and keep $8$ the unique mode.

The largest entry is then $110 - 75 = 35.$

Thus, the correct answer is E.

Solution:

Write each triangle by its side lengths in nonincreasing order. Only one equilateral triangle is allowed (all are similar), and of the similar pair $2,2,1$ and $4,4,2$ only one may appear.

The remaining valid, pairwise non-similar triangles are $$4\,4\,3,\ 4\,4\,1,\ 4\,3\,3,\ 4\,3\,2,\ 3\,3\,2,\ 3\,3\,1,\ 3\,2\,2,$$ seven in all. Together with one equilateral and one of the similar pair, $S$ has at most $9$ elements.

Thus, the correct answer is B.

Solution:

Since $b$ is the largest of $1, a, b,$ no such triangle exists exactly when $b \ge a+1.$ Since $1$ is the largest of $\tfrac1b, \tfrac1a, 1,$ no such triangle exists exactly when $1 \ge \tfrac1a + \tfrac1b,$ that is $a \le \tfrac{b}{b-1}.$

Both conditions hold with $b$ smallest when $a+1 = b$ and $a = \tfrac{b}{b-1}$ meet, giving $b - 1 = \tfrac{b}{b-1},$ or $b^2 - 3b + 1 = 0.$

The root larger than $1$ is $b = \dfrac{3+\sqrt5}{2}.$

Thus, the correct answer is C.

Solution:

Let the edges be $x, y, z.$ Then $xy+yz+zx = 47$ and $x+y+z = 12.$ Therefore $$x^2+y^2+z^2 = (x+y+z)^2 - 2(xy+yz+zx) = 144 - 94 = 50.$$

Each of the $4$ interior diagonals has length $\sqrt{x^2+y^2+z^2} = \sqrt{50} = 5\sqrt2,$ so their total length is $4 \cdot 5\sqrt2 = 20\sqrt2.$

Thus, the correct answer is D.

Solution:

Since $k \ln k = \ln(k^k),$ the sum gives $p = \ln\left(\prod_{k=1}^{6} k^k\right),$ so $$e^p = 1^1 \cdot 2^2 \cdot 3^3 \cdot 4^4 \cdot 5^5 \cdot 6^6.$$

The factors of $2$ come from $2^2$ (giving $2$), $4^4 = 2^8$ (giving $8$), and $6^6 = 2^6 \cdot 3^6$ (giving $6$). In total the exponent of $2$ is $2 + 8 + 6 = 16.$

Thus, the correct answer is C.

Solution:

Since $P(0) = k,$ write $P(x) = ax^3 + bx^2 + cx + k.$

Then $P(1) = a+b+c+k = 2k$ and $P(-1) = -a+b-c+k = 3k.$ Adding these gives $2b + 2k = 5k,$ so $2b = 3k.$

The odd-power terms cancel in the sum: $$P(2)+P(-2) = (8a+4b+2c+k) + (-8a+4b-2c+k) = 8b + 2k.$$ Since $8b = 4(2b) = 12k,$ this equals $12k + 2k = 14k.$

Thus, the correct answer is E.

Solution:

The line through $Q$ is $y = m(x-20) + 14.$ Substituting into $y = x^2$ gives $$x^2 - mx + (20m - 14) = 0.$$

There is no intersection exactly when this has no real root, i.e. when the discriminant $m^2 - 4(20m-14) = m^2 - 80m + 56$ is negative. That happens between the two roots $r$ and $s$ of $m^2 - 80m + 56 = 0.$

By Vieta's formulas, $r + s = 80.$

Thus, the correct answer is E.

Solution:

Any single number covers sums $1$ through $5.$ If a consecutive block sums to $n,$ the remaining numbers form a consecutive block summing to $15 - n,$ so sums $10$ through $14$ are automatically covered as well. Thus an arrangement is bad only if it fails to produce $6$ or $7.$

If $6$ cannot be formed, then $1$ and $5$ are not adjacent, and working through the cases forces the arrangement $14352.$ If $7$ cannot be formed, then $2$ and $5$ are not adjacent, forcing $23154.$

These are the only two bad arrangements up to rotation and reflection.

Thus, the correct answer is B.

Solution:

Let the top radius be $1,$ the bottom radius $r,$ and the sphere radius $a.$ The sphere touches both bases, so the cone's height is $2a,$ and applying the Pythagorean Theorem to the side profile gives $r = a^2.$

The frustum volume is $\tfrac13 \pi (r^2 + r + 1)(2a).$ Setting it equal to twice the sphere volume $\tfrac43 \pi a^3$ and using $r = a^2$ yields $$a^4 - 3a^2 + 1 = 0,$$ that is $r^2 - 3r + 1 = 0.$

The positive root is $r = \dfrac{3+\sqrt5}{2}.$

Thus, the correct answer is E.

Solution:

The logarithms are defined only when $x - 40 \gt 0$ and $60 - x \gt 0,$ so $40 \lt x \lt 60.$

Within this range the inequality becomes $(x-40)(60-x) \lt 100,$ which expands to $x^2 - 100x + 2500 \gt 0,$ i.e. $(x-50)^2 \gt 0.$ This holds for every $x \ne 50.$

The integers strictly between $40$ and $60$ except $50$ are $41, \ldots, 49$ and $51, \ldots, 59,$ which is $18$ values.

Thus, the correct answer is B.

Solution:

Let $x = BE = GH = CF$ and $\theta = \angle DHG = \angle AGJ = \angle FKH,$ with $AD = GJ = HK = 1.$ In right triangle $GDH,$ $x \sin\theta = DG = 1 - \cos\theta,$ so $x = \dfrac{1 - \cos\theta}{\sin\theta}.$

Along side $CD,$ $$1 = CF + FH + HD = x + \sin\theta + x\cos\theta.$$ Substituting for $x$ gives $$1 = \dfrac{(1-\cos\theta)(1+\cos\theta)}{\sin\theta} + \sin\theta = \dfrac{\sin^2\theta}{\sin\theta} + \sin\theta = 2\sin\theta.$$

Hence $\sin\theta = \tfrac12,$ so $\theta = 30^\circ$ and $$x = \dfrac{1 - \frac{\sqrt3}{2}}{\frac12} = 2 - \sqrt3.$$

Thus, the correct answer is C.

Solution:

Let $p_j$ be the probability of eventually reaching pad $10$ starting from pad $j.$ By the symmetry of the jump rule at the center, $p_5 = \tfrac12.$

Each interior pad satisfies $p_j = \tfrac{10-j}{10}\,p_{j+1} + \tfrac{j}{10}\,p_{j-1},$ which gives $$p_4 = \tfrac25 p_3 + \tfrac35 p_5,\quad p_3 = \tfrac{3}{10} p_2 + \tfrac{7}{10} p_4,$$ $$p_2 = \tfrac15 p_1 + \tfrac45 p_3,\quad p_1 = \tfrac{9}{10} p_2.$$

Substituting downward from $p_5 = \tfrac12$ and solving yields $p_1 = \dfrac{63}{146}.$

Thus, the correct answer is C.

Solution:

Working modulo $2017,$ the identity $n \cdot k! \cdot (2014-k)! = 2014!$ together with $2016 \cdot 2015 \cdots (2015-k) \equiv (-1)^k (k+2)!$ leads to $$2\binom{2014}{k} \equiv (-1)^k (k+2)(k+1) \pmod{2017},$$ so $\binom{2014}{k} \equiv (-1)^k \binom{k+2}{2}.$

Then $$S \equiv \sum_{k=0}^{62} (-1)^k \binom{k+2}{2} = 1 + \sum_{k=1}^{31}\left[\binom{2k+2}{2} - \binom{2k+1}{2}\right] = 1 + \sum_{k=1}^{31}(2k+1).$$

The remaining sum is $3 + 5 + \cdots + 63 = 1023,$ so $S \equiv 1 + 1023 = 1024 \pmod{2017}.$

Thus, the correct answer is C.

Solution:

Because arcs $AB, CD$ are equal and arcs $BC, DE$ are equal, the chords $AC, BD, CE$ are all equal; let $x = AC = BD = CE,$ $y = AD,$ and $z = BE.$

Ptolemy's theorem on $ABCD,$ $BCDE,$ and $ABDE$ gives $$10y + 9 = x^2,\quad 100 + 3z = x^2, \quad 30 + 14x = yz.$$ Solving the first two for $y$ and $z$ and substituting into the third yields $$x^3 - 109x - 420 = 0 = (x-12)(x+5)(x+7).$$

So $x = 12,$ $y = \tfrac{135}{10} = \tfrac{27}{2},$ and $z = \tfrac{44}{3}.$ The five diagonals are $AC, BD, CE, AD, BE,$ summing to $$3x + y + z = 36 + \tfrac{27}{2} + \tfrac{44}{3} = \tfrac{385}{6}.$$

Thus $m + n = 385 + 6 = 391,$ and the correct answer is D.

Solution:

Let $x = \tfrac{\pi y}{2}.$ Dividing by $2$ and using $\tfrac12(1 - \cos(2\pi y)) = \sin^2(\pi y),$ the equation simplifies to $$\cos(\pi y)\cos\left(\dfrac{4028\pi}{y}\right) = 1.$$

Both cosines must equal $1$ or both equal $-1,$ so $y$ and $\tfrac{4028}{y}$ are integers of the same parity. Since $4028 = 2^2 \cdot 19 \cdot 53$ is even, both must be even, so $y = 2a$ with $a$ a positive odd divisor of $2014 = 2 \cdot 19 \cdot 53,$ giving $a \in \{1, 19, 53, 19 \cdot 53\}.$

Each such $a$ gives $x = \tfrac{\pi y}{2} = \pi a,$ so the sum of solutions is $$\pi(1 + 19 + 53 + 19\cdot53) = \pi(19+1)(53+1) = 1080\pi.$$

Thus, the correct answer is D.