Answer: 59

Solution:

A three-letter arrangement is a palindrome exactly when the third letter matches the first, so the probability of a letter palindrome is $\frac{1}{26}.$ Similarly, the probability of a digit palindrome is $\frac{1}{10},$ and the two events are independent.

By inclusion-exclusion, the probability of at least one palindrome is $$\frac{1}{26} + \frac{1}{10} - \frac{1}{26} \cdot \frac{1}{10} = \frac{10 + 26 - 1}{260} = \frac{35}{260} = \frac{7}{52}.$$ Thus $m + n = 7 + 52 = 59.$

Answer: 154

Solution:

Let $r$ be the common radius. The longer side holds a row of seven circles, so it equals $14r.$ The centers of three mutually tangent circles in adjacent rows form an equilateral triangle with side $2r,$ whose height is $r\sqrt{3},$ so the two gaps between rows of centers contribute $2r\sqrt{3},$ and the shorter side is $r + 2r\sqrt{3} + r = 2r + 2r\sqrt{3}.$

The ratio is $$\frac{14r}{2r\left(1 + \sqrt{3}\right)} = \frac{7}{1 + \sqrt{3}} = \frac{7\left(\sqrt{3} - 1\right)}{2} = \frac{1}{2}\left(\sqrt{147} - 7\right),$$ so $p = 147,$ $q = 7,$ and $p + q = 154.$

Answer: 25

Solution:

In $n$ years Jane's age is $25 + n,$ and Dick's age is its digit reversal. If Jane's future age is $10a + b,$ Dick's is $10b + a,$ which is larger exactly when $b \gt a.$ Conversely, every two-digit value of $25 + n$ with tens digit less than units digit yields exactly one valid pair: $n = 10a + b - 25$ and $d = 10b + a - n = 25 + 9(b - a) \gt 25,$ so Dick is indeed older than Jane now.

So we count two-digit numbers that are at least $26$ and have tens digit less than units digit: $4$ starting with $2$ (namely $26$ through $29$), then $6,$ $5,$ $4,$ $3,$ $2,$ $1$ starting with $3$ through $8.$ The total is $4 + 6 + 5 + 4 + 3 + 2 + 1 = 25.$

Answer: 840

Solution:

Since $\frac{1}{k^2 + k} = \frac{1}{k(k + 1)} = \frac{1}{k} - \frac{1}{k + 1},$ the sum telescopes: $$a_m + a_{m+1} + \cdots + a_{n-1} = \frac{1}{m} - \frac{1}{n} = \frac{1}{29}.$$

Multiplying through by $29mn$ gives $29n - 29m = mn,$ which rearranges to $(29 - m)(29 + n) = 29^2.$ Since $29$ is prime and $29 + n \gt 29,$ the only factorization with $m$ a positive integer is $29 - m = 1$ and $29 + n = 841,$ so $m = 28$ and $n = 812.$

Therefore $m + n = 28 + 812 = 840.$

Answer: 183

Solution:

Each of the $\binom{12}{2} = 66$ pairs of vertices determines exactly three squares: two having the pair as a side (one on each side of the segment) and one having it as a diagonal. That counts $3 \cdot 66 = 198$ squares.

A square is overcounted only if it has more than two vertices among the $A_i.$ If three vertices of a square lie on the circumcircle, the square's own circumcircle shares three points with it and hence coincides with it, and an inscribed square's vertices are spaced $90^\circ$ apart — three steps of the dodecagon — so the fourth vertex is also an $A_i.$ The fully inscribed squares are exactly $A_1A_4A_7A_{10},$ $A_2A_5A_8A_{11},$ and $A_3A_6A_9A_{12},$ and each is generated by all $\binom{4}{2} = 6$ of its vertex pairs, so each is counted $6$ times instead of once.

The number of distinct squares is $198 - 3 \cdot 5 = 183.$

Answer: 12

Solution:

Let $p = \log_{225} x$ and $q = \log_{64} y,$ so $\log_x 225 = \frac{1}{p}$ and $\log_y 64 = \frac{1}{q}.$ The system becomes $p + q = 4$ and $\frac{1}{p} - \frac{1}{q} = 1.$ Substituting $q = 4 - p$ into the second equation and clearing denominators gives $4 - 2p = p(4 - p),$ that is, $p^2 - 6p + 4 = 0.$

The two solutions of the system correspond to the two roots of this quadratic, so by Vieta's formulas $p_1 + p_2 = 6,$ and then $q_1 + q_2 = 8 - 6 = 2.$ Hence $$x_1 y_1 x_2 y_2 = 225^{p_1 + p_2} \cdot 64^{q_1 + q_2} = 225^6 \cdot 64^2 = 15^{12} \cdot 2^{12} = 30^{12},$$ so $\log_{30}\left(x_1 y_1 x_2 y_2\right) = 12.$

Answer: 428

Solution:

Apply the expansion with $x = 10^{2002},$ $y = 1,$ and $r = \frac{10}{7}:$ $$\left(10^{2002} + 1\right)^{10/7} = 10^{2860} + \frac{10}{7} \cdot 10^{858} + \frac{\frac{10}{7} \cdot \frac{3}{7}}{2} \cdot 10^{-1144} + \cdots.$$ The first term is an integer, and the third and later terms are far smaller than $10^{-1000},$ too small to affect the leading decimal digits. So those digits come from the fractional part of $\frac{10}{7} \cdot 10^{858} = \frac{10^{859}}{7}.$

That fractional part is $\frac{10^{859} \bmod 7}{7}.$ Since $10^6 \equiv 1 \pmod{7}$ and $859 \equiv 1 \pmod{6},$ we get $10^{859} \equiv 10 \equiv 3 \pmod{7},$ so the fractional part is $\frac{3}{7} = 0.428571\ldots$

The first three digits to the right of the decimal point are $428.$

Answer: 748

Solution:

Iterating the recurrence gives $a_9 = 13a_1 + 21a_2,$ and the sequence is nondecreasing exactly when $0 \lt a_1 \le a_2$ (all later terms then take care of themselves). So we need the smallest $k$ for which $13x + 21y = k$ has two solutions with $0 \lt x \le y.$

Suppose $13x + 21y = 13u + 21v$ with $x \lt u.$ Then $13(u - x) = 21(y - v),$ so $u - x$ is a positive multiple of $21.$ Hence $u \ge x + 21 \ge 22,$ and since $u \le v,$ also $v \ge 22,$ giving $k = 13u + 21v \ge 34 \cdot 22 = 748.$

Conversely $k = 748$ works: $(x, y) = (1, 35)$ and $(22, 22)$ give the sequences $1, 35, 36, 71, 107, 178, 285, 463, 748$ and $22, 22, 44, 66, 110, 176, 286, 462, 748.$ The answer is $k = 748.$

Answer: 757

Solution:

The three progressions $\{1, 1 + h, \ldots\},$ $\{2, 2 + t, \ldots\},$ $\{3, 3 + u, \ldots\}$ must partition the positive integers. If $h = 2,$ Harold paints picket $3,$ which Ulysses also paints, so $h \ge 3.$ If $h \ge 5,$ consider picket $4:$ Harold's next picket is $1 + h \ge 6,$ and Ulysses cannot paint it (that would need $u = 1,$ repainting everything from $3$ on), so Tanya must, forcing $t = 2.$ Then picket $5$ is unpainted unless $u = 2,$ but then Tanya and Ulysses together cover every picket from $2$ on, and Harold's picket $1 + h$ is painted twice. So $h = 3$ or $h = 4.$

If $h = 3,$ Harold paints $1, 4, 7, \ldots.$ Ulysses cannot paint picket $5$ (then $u = 2$ and he would repaint $7$), so Tanya does: $t = 3,$ covering $2, 5, 8, \ldots.$ What remains is exactly $3, 6, 9, \ldots,$ so $u = 3,$ giving $333.$ If $h = 4,$ Harold paints $1, 5, 9, \ldots;$ picket $4$ again forces $t = 2,$ and the leftover pickets $3, 7, 11, \ldots$ force $u = 4,$ giving $424.$

The sum of the paintable integers is $333 + 424 = 757.$

Answer: 148

Solution:

Here $AB = 3 + 9 = 12,$ $AC = 10 + 27 = 37,$ and angle $B$ is right, so $BC = \sqrt{37^2 - 12^2} = 35$ and $[ABC] = \frac{1}{2} \cdot 12 \cdot 35 = 210.$ The quadrilateral is triangle $ADC$ with triangle $AGF$ removed, where $G$ is the intersection of $\overline{AD}$ and $\overline{EF}.$

By the angle bisector theorem in triangle $ABC,$ $BD : DC = AB : AC = 12 : 37,$ so $[ADC] = \frac{37}{49} \cdot 210 = \frac{1110}{7}.$ In triangle $AEF,$ ray $AG$ bisects the same angle, so $EG : GF = AE : AF = 3 : 10$ and $[AGF] = \frac{10}{13}\,[AEF].$ Also $[AEF] = \frac{AE}{AB} \cdot \frac{AF}{AC}\,[ABC] = \frac{3}{12} \cdot \frac{10}{37} \cdot 210 = \frac{525}{37}.$

Therefore $$[DCFG] = \frac{1110}{7} - \frac{10}{13} \cdot \frac{525}{37} = \frac{1110}{7} - \frac{5250}{481} \approx 158.57 - 10.92 = 147.66,$$ and the closest integer is $148.$

Answer: 230

Solution:

Place $A = (0, 0, 0)$ with the cube $[0, 12]^3$ and $P = (12, 7, 5)$ on the face $x = 12.$ Reflecting the cube across the relevant face at each bounce straightens the reflected path into the straight ray from $A$ through $P:$ each crossing of a plane $x = 12k,$ $y = 12k,$ or $z = 12k$ corresponds to a bounce, and the beam reaches a vertex of the cube exactly when all three coordinates are simultaneously multiples of $12.$

The ray consists of the points $(12t, 7t, 5t).$ Since $7$ and $5$ are relatively prime to $12,$ the coordinates $7t$ and $5t$ are first divisible by $12$ when $t = 12,$ at the point $(144, 84, 60).$ The path length equals the straight-line distance $$\sqrt{144^2 + 84^2 + 60^2} = 12\sqrt{12^2 + 7^2 + 5^2} = 12\sqrt{218}.$$

Since $218 = 2 \cdot 109$ is squarefree, $m + n = 12 + 218 = 230.$

Answer: 275

Solution:

Composing the map with itself, $$F(F(z)) = \frac{\frac{z+i}{z-i} + i}{\frac{z+i}{z-i} - i} = \frac{(z + i) + i(z - i)}{(z + i) - i(z - i)} = \frac{(1 + i)(z + 1)}{(1 - i)(z - 1)} = i\,\frac{z + 1}{z - 1},$$ and applying $F$ once more gives $$F(F(F(z))) = \frac{i\,\frac{z+1}{z-1} + i}{i\,\frac{z+1}{z-1} - i} = \frac{(z + 1) + (z - 1)}{(z + 1) - (z - 1)} = z,$$ so the sequence $z_0, z_1, z_2, \ldots$ is periodic with period $3.$

Since $2002 = 3 \cdot 667 + 1,$ we have $z_{2002} = z_1 = F(z_0) = \frac{z_0 + i}{z_0 - i} = \frac{\frac{1}{137} + 2i}{\frac{1}{137}} = 1 + 274i.$ Thus $a + b = 1 + 274 = 275.$

Answer: 63

Solution:

Since $E$ is the midpoint of $\overline{AB},$ $AE = EB = 12.$ Let $P$ be the centroid, which trisects the medians: $AP = \frac{2}{3} \cdot 18 = 12$ and $PE = \frac{1}{3} \cdot 27 = 9.$ By the power of the point $E$ with respect to the circumcircle, $EF \cdot EC = EA \cdot EB = 144,$ so $EF = \frac{144}{27} = \frac{16}{3}.$

Triangle $AEP$ is isosceles with $AE = AP = 12$ and base $PE = 9,$ so the altitude from $A$ to $\overline{PE}$ is $\sqrt{144 - \frac{81}{4}} = \frac{3\sqrt{55}}{2},$ giving $[AEP] = \frac{1}{2} \cdot 9 \cdot \frac{3\sqrt{55}}{2} = \frac{27\sqrt{55}}{4}.$ Since $F$ and $P$ both lie on line $CE,$ triangles $AEF$ and $AEP$ share the apex $A$ and have collinear bases, so $$[AEF] = \frac{EF}{EP}\,[AEP] = \frac{16/3}{9} \cdot \frac{27\sqrt{55}}{4} = 4\sqrt{55}.$$

Finally, since $E$ is the midpoint of $\overline{AB},$ $[AFB] = 2\,[AFE] = 8\sqrt{55},$ and $m + n = 8 + 55 = 63.$

Answer: 30

Solution:

Let $\mathcal{S}$ have $n$ elements with sum $S.$ The condition says $\frac{S - x}{n - 1}$ is an integer for every $x \in \mathcal{S},$ which means every element is congruent to $S$ modulo $n - 1.$ In particular all elements are congruent to each other, and since $1 \in \mathcal{S},$ every element is $1$ more than a multiple of $n - 1.$

Then $2002 \equiv 1 \pmod{n - 1},$ so $n - 1$ divides $2001 = 3 \cdot 23 \cdot 29.$ Moreover the $n$ distinct elements run from $1$ up to $2002$ in steps that are multiples of $n - 1,$ so $2002 \ge 1 + (n - 1)^2,$ forcing $n - 1 \le 44.$ The largest divisor of $2001$ that is at most $44$ is $29,$ so $n \le 30.$

Thirty is attainable: take the $29$ numbers $1, 30, 59, \ldots, 813$ together with $2002.$ All are $\equiv 1 \pmod{29},$ and the sum of all $30$ is $\equiv 30 \equiv 1 \pmod{29},$ so every deleted mean is an integer. The answer is $30.$

Answer: 163

Solution:

Place $D = (0, 0, 0),$ $C = (12, 0, 0),$ $B = (12, 12, 0),$ $A = (0, 12, 0),$ and $E = (x, y, 12),$ using the given distance from $E$ to face $ABCD.$ From $CE = DE$ we get $x = 6,$ and then $DE = 14$ gives $36 + y^2 + 144 = 196,$ so $y = 4$ and $E = (6, 4, 12).$

In trapezoid $ABFG,$ $\overline{GF}$ is parallel to $\overline{AB}$ with $GF = 6$ and $AG = BF,$ so $G$ and $F$ are symmetric about the plane $x = 6:$ $G = (3, y_2, z_2)$ and $F = (9, y_2, z_2).$ Face $ADEG$ is planar, and the plane through $A,$ $D,$ $E$ contains the entire $y$-axis direction (both $A$ and $D$ have $x = z = 0$), so it is the plane $z = 2x,$ which indeed contains $E.$ Hence $z_2 = 6.$ Now $AG = 8$ gives $3^2 + (y_2 - 12)^2 + 6^2 = 64,$ so $y_2 = 12 \pm \sqrt{19}.$

Then $$EG^2 = 3^2 + (y_2 - 4)^2 + 6^2 = 45 + \left(8 \pm \sqrt{19}\right)^2 = 128 \pm 16\sqrt{19},$$ and the stated form $p - q\sqrt{r}$ corresponds to $128 - 16\sqrt{19}.$ Thus $p + q + r = 128 + 16 + 19 = 163.$