Answer: 150

Solution:

Let Tom's swimming speed be $s$ miles per hour. Then he runs at $5s$ and bicycles at $10s.$ The total time in hours is $$\frac{0.5}{s} + \frac{30}{10s} + \frac{8}{5s} = \frac{0.5 + 3 + 1.6}{s} = \frac{5.1}{s} = 4.25,$$ so $s = \frac{5.1}{4.25} = 1.2$ miles per hour.

He bicycles at $12$ miles per hour, so the ride takes $\frac{30}{12} = 2.5$ hours, which is $150$ minutes.

Answer: 200

Solution:

Since $n$ is divisible by $5,$ its last digit is $0$ or $5;$ since the first digit equals the last digit and cannot be $0,$ both are $5.$ The outer digits contribute $10$ to the digit sum, so the three middle digits must also sum to a multiple of $5.$

Choose the second and third digits freely, in $10 \cdot 10 = 100$ ways. Whatever their sum is, the fourth digit must land in a prescribed residue class modulo $5,$ and exactly $2$ of the digits $0$ through $9$ lie in each class. The count is $10 \cdot 10 \cdot 2 = 200.$

Answer: 18

Solution:

Let $AE = x$ and $EB = y,$ so the square has side $x + y$ and the two smaller squares have sides $x$ and $y.$ The condition says $$x^2 + y^2 = \frac{9}{10}(x + y)^2.$$ Multiplying by $10$ and expanding, $10x^2 + 10y^2 = 9x^2 + 18xy + 9y^2,$ so $x^2 + y^2 = 18xy.$

Dividing by $xy$ gives $$\frac{AE}{EB} + \frac{EB}{AE} = \frac{x}{y} + \frac{y}{x} = \frac{x^2 + y^2}{xy} = 18.$$

Answer: 429

Solution:

The rotation cycles the four L-shaped arms, so a symmetric coloring colors all four arms identically, and the $12$ outer squares contain $4$ copies of whatever the arm shows. The number of red squares among the outer twelve is therefore a multiple of $4.$ Since there are $8$ red squares in all, the center must be blue and each arm must contain exactly $2$ red squares and $1$ blue square.

The blue square within the arm can be chosen in $3$ ways, so exactly $3$ of the $\binom{13}{5} = 1287$ equally likely colorings are symmetric. The probability is $\frac{3}{1287} = \frac{1}{429},$ so $n = 429.$

Answer: 98

Solution:

Rewrite the equation as $9x^3 = x^3 + 3x^2 + 3x + 1 = (x + 1)^3.$ Taking real cube roots, $\sqrt[3]{9}\,x = x + 1,$ so $$x = \frac{1}{\sqrt[3]{9} - 1}.$$

Multiply numerator and denominator by $\sqrt[3]{81} + \sqrt[3]{9} + 1;$ the denominator becomes $(\sqrt[3]{9})^3 - 1 = 8,$ so $$x = \frac{\sqrt[3]{81} + \sqrt[3]{9} + 1}{8}.$$ Thus $a + b + c = 81 + 9 + 8 = 98.$

Answer: 47

Solution:

Focus on one box at a time. The box of $k$ books receives a uniformly random $k$-subset of the $12$ books, so the probability that it contains all three math books is $\binom{9}{k-3}\big/\binom{12}{k}.$ For $k = 3, 4, 5$ this gives $\frac{1}{220},$ $\frac{9}{495} = \frac{1}{55},$ and $\frac{36}{792} = \frac{1}{22}.$

The events are disjoint, so the total probability is $$\frac{1}{220} + \frac{1}{55} + \frac{1}{22} = \frac{1 + 4 + 10}{220} = \frac{15}{220} = \frac{3}{44},$$ and $m + n = 3 + 44 = 47.$

Answer: 41

Solution:

Let the height be $h$ and place the corner at the origin, so the box is $[0,12] \times [0,16] \times [0,h].$ The three faces meeting at the origin have centers $P = (6, 8, 0),$ $Q = \left(0, 8, \tfrac{h}{2}\right),$ and $R = \left(6, 0, \tfrac{h}{2}\right).$

Then $\overrightarrow{PQ} = \left(-6, 0, \tfrac{h}{2}\right)$ and $\overrightarrow{PR} = \left(0, -8, \tfrac{h}{2}\right),$ whose cross product is $(4h, 3h, 48).$ The area is $$\frac{1}{2}\sqrt{16h^2 + 9h^2 + 48^2} = \frac{1}{2}\sqrt{25h^2 + 2304} = 30,$$ so $25h^2 = 3600 - 2304 = 1296$ and $h = \frac{36}{5}.$

Therefore $m + n = 36 + 5 = 41.$

Answer: 371

Solution:

The function is defined when $-1 \le \log_m(nx) \le 1,$ that is $\frac{1}{m} \le nx \le m,$ so the domain is $\left[\frac{1}{mn}, \frac{m}{n}\right],$ with length $$\frac{m}{n} - \frac{1}{mn} = \frac{m^2 - 1}{mn} = \frac{1}{2013}.$$ Hence $n = \frac{2013(m^2 - 1)}{m}.$ Since $m$ is relatively prime to $m^2 - 1,$ $m$ must divide $2013 = 3 \cdot 11 \cdot 61.$

Because $n \approx 2013m,$ the sum $m + n$ grows with $m,$ so take the smallest factor $m = 3:$ then $n = \frac{2013 \cdot 8}{3} = 5368$ and $m + n = 5371.$

The remainder upon division by $1000$ is $371.$

Answer: 113

Solution:

Let $A'$ be the landing point, with $BA' = 9$ and $CA' = 3,$ and let the crease meet $\overline{AB}$ at $P$ and $\overline{AC}$ at $Q.$ Folding preserves distances, so $PA' = PA = x$ and $QA' = QA = y.$ In triangle $PBA',$ with $PB = 12 - x$ and $\angle B = 60^\circ,$ the law of cosines gives $$x^2 = (12 - x)^2 + 81 - 9(12 - x),$$ which simplifies to $15x = 117,$ so $x = \frac{39}{5}.$ Similarly, in triangle $QCA',$ $y^2 = (12 - y)^2 + 9 - 3(12 - y)$ gives $21y = 117,$ so $y = \frac{39}{7}.$

Finally, in triangle $APQ$ with $\angle A = 60^\circ,$ $$PQ^2 = x^2 + y^2 - xy = 39^2\left(\frac{1}{25} + \frac{1}{49} - \frac{1}{35}\right) = 39^2 \cdot \frac{49 + 25 - 35}{1225} = \frac{39^3}{1225},$$ so $PQ = \frac{39\sqrt{39}}{35}.$ Thus $m + n + p = 39 + 35 + 39 = 113.$

Answer: 80

Solution:

Since $P$ has real coefficients, $r - si$ is also a zero, and the third zero $q$ is real. The product of the zeros is $q(r^2 + s^2) = 65,$ so $q$ is a nonzero integer and $r^2 + s^2$ is a factor of $65.$ With $r, s$ nonzero, the possibilities are $r^2 + s^2 = 5 = 1^2 + 2^2$ (with $q = 13$), $13 = 2^2 + 3^2$ (with $q = 5$), and $65 = 1^2 + 8^2 = 4^2 + 7^2$ (with $q = 1$).

For each representation $\{u, v\},$ the zero $r + si$ can have $r = \pm u$ or $\pm v,$ giving $4$ distinct polynomials (the sign of $s$ changes nothing). The sum of the zeros is $p_{a,b} = q + 2r,$ and over the four choices the $2r$ terms cancel, leaving $4q$ from each representation.

The total is $4 \cdot 13 + 4 \cdot 5 + 4 \cdot 1 + 4 \cdot 1 = 80.$

Answer: 148

Solution:

Divisibility by $16,$ $15,$ and $14$ means $N = 1680m$ where $1680 = \operatorname{lcm}(14, 15, 16) = 2^4 \cdot 3 \cdot 5 \cdot 7.$ Every positive integer less than $14$ divides $1680$ except $9,$ $11,$ and $13,$ and a divisor of $N$ leaves remainder $0,$ not $3.$ So necessarily $\{x, y, z\} = \{9, 11, 13\},$ and we need $1680m \equiv 3$ modulo each of $9,$ $11,$ $13.$

Since $1680 \equiv 6 \pmod 9,$ the first congruence is $6m \equiv 3 \pmod 9,$ i.e. $m \equiv 2 \pmod 3.$ Since $1680 \equiv 8 \pmod{11},$ we need $8m \equiv 3 \pmod{11},$ i.e. $m \equiv 10 \pmod{11}.$ Since $1680 \equiv 3 \pmod{13},$ we need $m \equiv 1 \pmod{13}.$ By the Chinese remainder theorem these combine to $m \equiv 131 \pmod{429},$ so the least $m$ is $131.$

Then $N = 1680 \cdot 131 = 2^4 \cdot 3 \cdot 5 \cdot 7 \cdot 131,$ and since $131$ is prime, the sum of the distinct prime divisors is $2 + 3 + 5 + 7 + 131 = 148.$

Answer: 21

Solution:

Note $\angle R = 45^\circ.$ Because the hexagon's interior angles are $120^\circ,$ segments $\overline{BC}$ cut off a corner triangle at $Q$ with two $60^\circ$ base angles, so triangle $BQC$ is equilateral and $QB = QC = 1.$ Put $Q$ at the origin with $QR$ along the positive $x$-axis. Then $C = (1, 0),$ $D = (2, 0),$ and the hexagon's vertices are $B = \left(\tfrac{1}{2}, \tfrac{\sqrt{3}}{2}\right),$ $A = (1, \sqrt{3}),$ $F = (2, \sqrt{3}),$ $E = \left(\tfrac{5}{2}, \tfrac{\sqrt{3}}{2}\right).$

Since $\angle R = 45^\circ,$ line $RP$ has slope $-1.$ If it passed through $E,$ it would be $x + y = \tfrac{5 + \sqrt{3}}{2},$ which puts $F$ (with $x + y = 2 + \sqrt{3}$) outside the triangle; so the vertex on $\overline{RP}$ is $F,$ and $RP$ is the line $x + y = 2 + \sqrt{3}.$ It meets the $x$-axis at $R = (2 + \sqrt{3},\ 0)$ and the line $y = \sqrt{3}\,x$ (line $QP$) where $x(1 + \sqrt{3}) = 2 + \sqrt{3},$ giving $P$ height $$y = \frac{\sqrt{3}(2 + \sqrt{3})}{1 + \sqrt{3}} = \frac{3 + \sqrt{3}}{2}.$$

The area is $\frac{1}{2} \cdot QR \cdot y = \frac{1}{2}(2 + \sqrt{3}) \cdot \frac{3 + \sqrt{3}}{2} = \frac{9 + 5\sqrt{3}}{4},$ so $a + b + c + d = 9 + 5 + 3 + 4 = 21.$

Answer: 961

Solution:

By Heron's formula with $s = 27,$ the area of $\triangle AB_0C_0$ is $\sqrt{27 \cdot 15 \cdot 10 \cdot 2} = 90.$ In the similarity $\triangle B_0C_1C_0 \sim \triangle AB_0C_0,$ side $B_0C_0$ corresponds to $AC_0,$ so the ratio is $r = \frac{17}{25},$ and $C_1C_0$ (corresponding to $B_0C_0$) equals $17r.$ Hence $$\frac{AC_1}{AC_0} = \frac{25 - 17r}{25} = 1 - r^2,$$ which is the similarity ratio of $\triangle AB_1C_1$ to $\triangle AB_0C_0.$

Segments $\overline{B_1C_1}$ and $\overline{B_0C_1}$ split $\triangle AB_0C_0$ into the three pieces, so $$[B_0C_1B_1] = 90\left(1 - r^2 - (1 - r^2)^2\right) = 90\,r^2(1 - r^2).$$ Each successive stage repeats the construction inside $\triangle AB_nC_n,$ scaling all areas by $(1 - r^2)^2,$ and the triangles $B_{n-1}C_nB_n$ have disjoint interiors.

The union's area is the geometric series $$\frac{90\,r^2(1 - r^2)}{1 - (1 - r^2)^2} = \frac{90(1 - r^2)}{2 - r^2} = 90 \cdot \frac{336/625}{961/625} = \frac{90 \cdot 336}{961}.$$ Since $961 = 31^2$ shares no factor with $90 \cdot 336 = 30240,$ the answer is $q = 961.$

Answer: 36

Solution:

The signs and the alternation between sines and cosines suggest powers of $i:$ indeed $$Q + iP = 1 + \frac{1}{2}ie^{i\theta} + \frac{1}{4}i^2e^{2i\theta} + \frac{1}{8}i^3e^{3i\theta} + \cdots = \frac{1}{1 - \frac{ie^{i\theta}}{2}} = \frac{2}{2 - ie^{i\theta}}.$$ Since $2 - ie^{i\theta} = (2 + \sin\theta) - i\cos\theta,$ multiplying by the conjugate gives $$Q + iP = \frac{2(2 + \sin\theta) + 2i\cos\theta}{5 + 4\sin\theta},$$ so $\frac{P}{Q} = \frac{\cos\theta}{2 + \sin\theta}.$

Setting $\frac{\cos\theta}{2 + \sin\theta} = \frac{2\sqrt{2}}{7}$ and squaring, $49(1 - \sin^2\theta) = 8(2 + \sin\theta)^2,$ which rearranges to $$57\sin^2\theta + 32\sin\theta - 17 = (3\sin\theta - 1)(19\sin\theta + 17) = 0.$$

Since $\pi \le \theta \lt 2\pi$ forces $\sin\theta \le 0,$ we get $\sin\theta = -\frac{17}{19}$ (and then $\cos\theta = \frac{6\sqrt{2}}{19} \gt 0,$ consistent with the positive ratio). Thus $m + n = 17 + 19 = 36.$

Answer: 272

Solution:

Let $d$ be the common difference of $(b, a, c),$ so $a - b = c - a = d \gt 0$ and $c = b + 2d \lt p,$ whence $0 \lt 2d \lt p.$ Let $D \gt 0$ be the common difference of $(A, B, C).$ Reducing mod $p,$ we get $D = B - A \equiv b - a = -d$ and $D = C - B \equiv c - b = 2d,$ so $p \mid 3d.$ Since $0 \lt d \lt p,$ the prime $p$ cannot divide $d,$ so $p = 3;$ then $2d \lt 3$ gives $d = 1$ and $(b, a, c) = (0, 1, 2).$

So the valid triples are exactly the increasing arithmetic progressions in $[0, 99]$ with $A \equiv 1,$ $B \equiv 0,$ $C \equiv 2 \pmod 3.$ Write $A = 1 + 3j$ with $j \ge 0;$ the difference satisfies $D \equiv -1 \equiv 2 \pmod 3,$ so $D = 2 + 3k$ with $k \ge 0.$ The constraint is $C = A + 2D = 5 + 3j + 6k \le 99,$ i.e. $j + 2k \le 31,$ and every such pair $(j, k)$ works.

For each $k = 0, 1, \ldots, 15$ there are $32 - 2k$ choices of $j,$ so $$N = \sum_{k=0}^{15} (32 - 2k) = 16 \cdot 32 - 2 \cdot \frac{15 \cdot 16}{2} = 512 - 240 = 272.$$