Solution:

After the eating, $444 - 5 - 9 - 25 = 405$ peanuts remain, and the three amounts form an arithmetic progression, so the middle amount, Betty's, is $\frac{405}{3} = 135.$ Hence Betty started with $135 + 9 = 144$ peanuts.

The starting amounts form a geometric progression, so they are $\frac{144}{r},$ $144,$ and $144r$ with $r \gt 1$ (Charlie had the most and Alex the least). Then $$\frac{144}{r} + 144 + 144r = 444,$$ which simplifies to $12r^2 - 25r + 12 = 0,$ with roots $r = \frac{4}{3}$ and $r = \frac{3}{4};$ since $r \gt 1,$ we take $r = \frac{4}{3}.$

So Alex initially had $144 \cdot \frac{3}{4} = 108$ peanuts. (Check: after eating, the amounts $103,$ $135,$ $167$ increase by $32$ each.)

Solution:

Let $p$ be the probability that it rains on Sunday given a dry Saturday; given a rainy Saturday it is $2p.$ The overall Sunday chance gives $$0.4(2p) + 0.6p = 0.3,$$ so $1.4p = 0.3$ and $p = \frac{3}{14}.$

The weekend is completely dry exactly when Saturday is dry and then Sunday is dry: $\frac{3}{5}\left(1 - \frac{3}{14}\right) = \frac{3}{5} \cdot \frac{11}{14} = \frac{33}{70}.$ So the probability of rain on at least one day is $1 - \frac{33}{70} = \frac{37}{70},$ which is in lowest terms, and $a + b = 37 + 70 = 107.$

Solution:

Exponentiating each equation gives $xyz - 3 + \log_5 x = 2^5$ and similarly for the others, so $$xyz + \log_5 x = 35, \qquad xyz + \log_5 y = 84, \qquad xyz + \log_5 z = 259.$$ Write $x = 5^a,$ $y = 5^b,$ $z = 5^c,$ so that $xyz = 5^{a+b+c}.$

Adding the three equations gives $3 \cdot 5^s + s = 378,$ where $s = a + b + c.$ The left side is strictly increasing in $s,$ and $s = 3$ works since $375 + 3 = 378,$ so $s = 3$ and $5^s = 125.$

Then $a = 35 - 125 = -90,$ $b = 84 - 125 = -41,$ and $c = 259 - 125 = 134,$ so $|a| + |b| + |c| = 90 + 41 + 134 = 265.$

Solution:

Each $1 \times b \times c$ layer has exactly $9$ red and $12$ green cubes, hence exactly $bc - 21$ yellow; each $a \times 1 \times c$ layer has exactly $20$ green and $25$ yellow, hence exactly $ac - 45$ red. Counting green cubes in the whole box both ways gives $12a = 20b,$ so $3a = 5b.$ Counting yellow both ways gives $a(bc - 21) = 25b = 15a,$ so $bc = 36.$ Counting red both ways gives $b(ac - 45) = 9a,$ and $\frac{9a}{b} = 15,$ so $ac = 60.$

Thus $a = \frac{60}{c}$ and $b = \frac{36}{c}$ are positive integers, so $c$ divides $\gcd(60, 36) = 12,$ and the volume is $abc = \frac{60 \cdot 36}{c} = \frac{2160}{c},$ smallest when $c = 12:$ volume $180$ with $(a, b, c) = (5, 3, 12).$

This is achievable: color every $1 \times 3 \times 12$ layer with three identical rows RRRGGGGYYYYY. Then each $1 \times 3 \times 12$ layer has $9$ red, $12$ green, and $15$ yellow cubes, and each $5 \times 1 \times 12$ layer has $15$ red, $20$ green, and $25$ yellow cubes. So the smallest possible volume is $180.$

Solution:

Let $a = BC_0,$ $b = AC_0,$ and $c = AB.$ The altitude gives $C_0C_1 = \frac{ab}{c},$ and $\triangle C_0C_1B \sim \triangle AC_0B$ with ratio $\frac{a}{c}.$ Each later altitude repeats this construction in a triangle scaled by $\frac{a}{c},$ so the segments $C_{n-1}C_n$ form a geometric series and $$\sum_{n=1}^{\infty} C_{n-1}C_n = \frac{ab/c}{1 - a/c} = \frac{ab}{c - a} = 6p = 6(a + b + c).$$

Since $b^2 = c^2 - a^2 = (c - a)(c + a),$ we get $(c - a)(a + b + c) = (c - a)(c + a) + (c - a)b = b(b + c - a),$ so $ab = 6b(b + c - a),$ that is $7a = 6b + 6c.$ Squaring $6c = 7a - 6b$ and using $36c^2 = 36a^2 + 36b^2$ gives $36a^2 = 49a^2 - 84ab,$ hence $13a = 84b.$

Because the side lengths are pairwise relatively prime, $a = 84$ and $b = 13,$ so $c = \frac{7 \cdot 84 - 6 \cdot 13}{6} = 85,$ and indeed $13^2 + 84^2 = 85^2.$ Then $p = 84 + 13 + 85 = 182$ (and $\frac{ab}{c - a} = \frac{84 \cdot 13}{1} = 1092 = 6p$ checks).

Solution:

Every substituted power $x, x^3, x^5, x^7, x^9$ is odd, so $Q(-x) = P(-x)P(-x^3)P(-x^5)P(-x^7)P(-x^9).$ Since $P(-x) = 1 + \frac{1}{3}x + \frac{1}{6}x^2$ has only nonnegative coefficients, so does each factor $P(-x^k),$ and hence so does the product $Q(-x).$ The coefficient of $x^i$ in $Q(-x)$ is $(-1)^i a_i,$ so $|a_i| = (-1)^i a_i.$

Therefore $$\sum_{i=0}^{50} |a_i| = Q(-1) = P(-1)^5 = \left(1 + \frac{1}{3} + \frac{1}{6}\right)^5 = \left(\frac{3}{2}\right)^5 = \frac{243}{32},$$ and $m + n = 243 + 32 = 275.$

Solution:

If a square of side $t$ has its vertices on the sides of a concentric square of side $s$ and is tilted by angle $\theta,$ each side of the outer square is split into pieces $t\cos\theta$ and $t\sin\theta,$ so $s = t(\cos\theta + \sin\theta).$ This applies to $IJKL$ (side $t$) in $ABCD$ (side $s$) with some angle $\theta.$ Since $\overline{EF} \parallel \overline{AB},$ square $EFGH$ (side $u$) makes the same angle $\theta$ with $IJKL,$ so also $t = u(\cos\theta + \sin\theta).$

Hence $\frac{s}{t} = \frac{t}{u},$ so the three areas form a geometric progression: the area of $EFGH$ equals $\frac{T^2}{2016},$ where $T$ is the area of $IJKL.$ As $\theta$ ranges over $\left(0^\circ, 90^\circ\right),$ the factor $(\cos\theta + \sin\theta)^2$ takes every value in $(1, 2],$ so $T = \frac{2016}{(\cos\theta + \sin\theta)^2}$ takes every value in $[1008, 2016)$ ($\theta = 0^\circ$ is excluded because $EFGH$ is smaller than $ABCD$). For $\frac{T^2}{2016}$ to be an integer, $2016 = 2^5 \cdot 3^2 \cdot 7$ must divide $T^2,$ which forces $2^3 \cdot 3 \cdot 7 = 168$ to divide $T.$

The multiples of $168$ in $[1008, 2016)$ run from $1008$ to $1848,$ and each is attained by an appropriate $\theta,$ with the area of $EFGH$ then a positive integer less than $2016.$ The difference is $1848 - 1008 = 840.$

Solution:

Count ordered triples $(a, b, c)$ with $abc = 11 \cdot 21 \cdot 31 \cdot 41 \cdot 51 \cdot 61 = 3^2 \cdot 7 \cdot 11 \cdot 17 \cdot 31 \cdot 41 \cdot 61 = N.$ Each of the six primes $7, 11, 17, 31, 41, 61$ appears once and can go to any of the three values: $3^6$ ways. The two factors of $3$ can be split among the three values in $\binom{4}{2} = 6$ ways. That gives $6 \cdot 3^6 = 4374$ ordered triples.

If two of the three values were equal, their common value $v$ would satisfy $v^2 \mid N,$ so $v = 1$ or $v = 3.$ This produces the triples with values $\{1, 1, N\}$ and $\{3, 3, \frac{N}{9}\},$ each in $3$ orders, for $6$ ordered triples in all (all three equal is impossible). The remaining $4374 - 6 = 4368$ ordered triples have distinct entries, and each set $\{a, b, c\}$ is counted $3! = 6$ times.

So the number of sets is $\frac{4368}{6} = 728.$

Solution:

Write $a_n = 1 + (n-1)d$ and $b_n = r^{n-1}$ with integers $d \ge 1$ and $r \ge 2.$ Since $c_1 = 2 \lt 100,$ we have $k \ge 3,$ and the two conditions read $$(k-2)d + r^{k-2} = 99, \qquad kd + r^k = 999.$$

Subtracting, $2d + r^{k-3}(r-1)r(r+1) = 900.$ The product of three consecutive integers is divisible by $3,$ so $3 \mid 2d,$ hence $3 \mid d.$ Then $(k-2)d + r^{k-2} = 99$ forces $3 \mid r^{k-2},$ so $3 \mid r.$ The bounds $r^{k-2} \le 98$ and $r^k \le 998$ leave only $(r, k) = (3, 3),$ $(3, 4),$ $(3, 5),$ $(3, 6),$ $(6, 3),$ $(9, 3).$

Testing each against $(k-2)d = 99 - r^{k-2}$ and $kd = 999 - r^k,$ only $(r, k) = (9, 3)$ gives a consistent value, $d = 90.$ Then $c_3 = 1 + 2 \cdot 90 + 9^2 = 262.$

Solution:

By Power of a Point at $Q$ in $\omega,$ $QC \cdot QT = QA \cdot QB = 7 \cdot 6 = 42.$ Extend $\overline{AB}$ beyond $B$ to the point $R$ with $BR = 8,$ so that $QR = QB + BR = 14$ and $QP \cdot QR = 3 \cdot 14 = 42 = QC \cdot QT.$ By the converse of Power of a Point, $C,$ $P,$ $T,$ and $R$ are concyclic.

In circle $CPTR,$ $\angle BRT = \angle PRT = \angle PCT,$ and in $\omega,$ $\angle PCT = \angle SCT = \angle SAT$ (both subtend arc $ST$). Also $ASTB$ is cyclic, so the exterior angle of the quadrilateral at $B$ equals the opposite interior angle: $\angle RBT = \angle AST.$ Hence $\triangle AST \sim \triangle RBT.$

Therefore $\frac{ST}{BT} = \frac{AS}{RB},$ so $ST = 5 \cdot \frac{7}{8} = \frac{35}{8},$ and $m + n = 35 + 8 = 43.$

Solution:

If $a = p_1^{m_1} \cdots p_t^{m_t},$ then $a^k$ has $(km_1 + 1)(km_2 + 1) \cdots (km_t + 1)$ positive divisors, and each factor is $\equiv 1 \pmod{k},$ so the product is too. Conversely, if $N = km + 1,$ then $a = p^m$ gives $a^k = p^{km}$ with exactly $N$ divisors. So $N$ is $k$-nice exactly when $N \equiv 1 \pmod{k}.$

Among $1, 2, \ldots, 999$ there are $143$ integers $\equiv 1 \pmod{7}$ (namely $1, 8, \ldots, 995$) and $125$ integers $\equiv 1 \pmod{8}$ (namely $1, 9, \ldots, 993$). Since $\operatorname{lcm}(7, 8) = 56,$ there are $18$ integers $\equiv 1 \pmod{56}$ (namely $1, 57, \ldots, 953$). By inclusion-exclusion, $143 + 125 - 18 = 250$ of them are $7$-nice or $8$-nice.

Hence $999 - 250 = 749$ positive integers less than $1000$ are neither.

Solution:

Let $P_n$ be the number of valid paintings of a ring of $n$ sections. Cutting a ring open between two adjacent sections shows that ring paintings correspond exactly to rows of $n$ sections with adjacent colors different and the two end colors different. A row of $n$ sections with adjacent colors different can be painted in $4 \cdot 3^{n-1}$ ways, and the rows whose end colors match correspond, by merging the two end sections into one, to ring paintings of $n - 1$ sections. Hence $$P_n + P_{n-1} = 4 \cdot 3^{n-1}.$$

Three mutually adjacent sections give $P_3 = 4 \cdot 3 \cdot 2 = 24,$ so $P_4 = 108 - 24 = 84,$ then $P_5 = 324 - 84 = 240,$ and finally $P_6 = 972 - 240 = 732.$

Solution:

There are $6! = 720$ arrangements, and every score lies between $2$ and $7.$ Let $b_n$ be the number of arrangements with score at least $n.$ Since each score $s$ satisfies $s = 2 + \#\{n \ge 3 : s \ge n\},$ the total of all $720$ scores is $$2 \cdot 720 + b_3 + b_4 + b_5 + b_6 + b_7.$$

Score $\ge n$ means no rook occupies a square with row + column $\lt n.$ Place the rooks row by row. For $b_3,$ only $(1,1)$ is banned: $5 \cdot 5! = 600.$ For $b_4,$ row $1$ has $4$ allowed columns, then row $2$ has $4$ (column $1$ and the used column are excluded): $4 \cdot 4 \cdot 4! = 384.$ Similarly $b_5 = 3 \cdot 3 \cdot 3 \cdot 3! = 162,$ $b_6 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2! = 32,$ and $b_7 = 1$ (all rooks on the anti-diagonal).

The total is $1440 + 600 + 384 + 162 + 32 + 1 = 2619,$ so the average is $\frac{2619}{720} = \frac{291}{80},$ and $p + q = 291 + 80 = 371.$

Solution:

Since $PA = PB = PC$ and $QA = QB = QC,$ both $P$ and $Q$ lie on the line through the center $H$ of $\triangle ABC$ perpendicular to its plane, on opposite sides. Any point equidistant from $A,$ $B,$ $C$ also lies on that line, so $O$ is on it, and $OP = OQ = d$ makes $O$ the midpoint of $\overline{PQ},$ with $PQ = 2d.$ Let $D$ be the midpoint of $\overline{AB}$ and $a = 600;$ then $DH = \frac{a\sqrt{3}}{6}$ and $CH = \frac{a\sqrt{3}}{3}.$ Since $\overline{PD} \perp \overline{AB}$ and $\overline{QD} \perp \overline{AB},$ the dihedral angle is $\angle PDQ = 120^\circ;$ write $x = \angle PDH$ and $y = \angle QDH,$ so $x + y = 120^\circ.$

Right triangles $PDH$ and $QDH$ give $PH = DH \tan x$ and $QH = DH \tan y,$ so $$2d = PQ = PH + QH = \frac{a\sqrt{3}}{6}(\tan x + \tan y).$$ Since $OC = OP = OQ = d,$ point $C$ lies on the circle with diameter $\overline{PQ},$ so $\angle PCQ = 90^\circ,$ and $H$ is the foot of the altitude from $C$ to the hypotenuse $\overline{PQ}.$ Thus $CH^2 = PH \cdot QH,$ which gives $\tan x \tan y = \frac{CH^2}{DH^2} = 4.$

By the tangent addition formula, $-\sqrt{3} = \tan 120^\circ = \frac{\tan x + \tan y}{1 - \tan x \tan y} = \frac{\tan x + \tan y}{-3},$ so $\tan x + \tan y = 3\sqrt{3}.$ Then $2d = \frac{a\sqrt{3}}{6} \cdot 3\sqrt{3} = \frac{3a}{2},$ so $d = \frac{3a}{4} = 450.$

Solution:

Since $\sum x_i = 1,$ we have $2\sum_{i \lt j} x_i x_j = 1 - \sum x_i^2.$ Doubling the given equation and rearranging, $$1 - \sum_{i=1}^{216} x_i^2 = \frac{214}{215} + \sum_{i=1}^{216} \frac{a_i x_i^2}{1 - a_i}, \qquad \text{so} \qquad \frac{1}{215} = \sum_{i=1}^{216}\left(1 + \frac{a_i}{1 - a_i}\right)x_i^2 = \sum_{i=1}^{216} \frac{x_i^2}{1 - a_i}.$$

Now $\sum a_i = \left(\frac{1}{2} + \cdots + \frac{1}{2^{215}}\right) + \frac{1}{2^{215}} = 1,$ so $\sum (1 - a_i) = 216 - 1 = 215.$ By the Cauchy-Schwarz inequality, $$1 = \left(\sum_{i=1}^{216} x_i\right)^2 \le \left(\sum_{i=1}^{216} \frac{x_i^2}{1 - a_i}\right)\left(\sum_{i=1}^{216} (1 - a_i)\right) = \frac{1}{215} \cdot 215 = 1.$$

Equality holds, so $x_i$ is proportional to $1 - a_i,$ forcing $x_i = \frac{1 - a_i}{215}.$ The only, hence maximum, possible value of $x_2$ is $\frac{1 - \frac{1}{4}}{215} = \frac{3}{860},$ and $m + n = 3 + 860 = 863.$