Separate the cases according to whether the thousands digit is even or odd

After choosing the thousands and units digits, count the choices for the two middle digits in order

The thousands digit is $4,$ $5,$ or $6.$ If it is $4$ or $6,$ the units digit has $4$ choices among $0,$ $2,$ $4,$ $6,$ and $8,$ after which the hundreds and tens digits have $8$ and $7$ choices. These two cases contribute $2\cdot4\cdot8\cdot7=448.$ If the thousands digit is $5,$ all $5$ even units digits are available, contributing $5\cdot8\cdot7=280.$ Thus the total is $448+280=728.$

Group the $40$ days into ten four-day cycles and sum horizontal and vertical displacements separately

For cycle index $k,$ compare $(4k+1)^2$ with $(4k+3)^2$, and similarly compare the other pair

Index the ten cycles by $k=0,1,\ldots,9.$ The horizontal displacement is $$\begin{aligned}&\frac12\sum_{k=0}^9\left((4k+1)^2-(4k+3)^2\right)\\&\quad=\sum_{k=0}^9(-8k-4)\\&\quad=-400.\end{aligned}$$ Similarly, the vertical displacement is $$\begin{aligned}&\frac12\sum_{k=0}^9\left((4k+2)^2-(4k+4)^2\right)\\&\quad=\sum_{k=0}^9(-8k-6)\\&\quad=-420.\end{aligned}$$ Therefore the distance from the start is $\sqrt{400^2+420^2}=580.$

Let $N$ be the total number of contestants and $T$ the total number of fish

Use the two averages by first subtracting the known groups with fewer than $3$ fish and with more than $12$ fish

The $9+5+7=21$ contestants below $3$ fish caught $5+14=19$ fish. Thus condition (b) gives $T-19=6(N-21),$ or $T=6N-107.$ The $5+2+1=8$ contestants above $12$ fish caught $65+28+15=108$ fish, so condition (c) gives $T-108=5(N-8),$ or $T=5N+68.$ Hence $N=175$ and $T=943.$

The first equation implies that $b-a=d-c$

Set $x=b-a=d-c$ and factor $bc-ad$ in terms of $x$ and $c-a$

Let $x=b-a=d-c>0$ and $y=c-a.$ Then $$\begin{aligned}bc-ad&=(a+x)c-a(c+x)\\&=x(c-a)=xy=93.\end{aligned}$$ Since $b<c,$ we need $x<y.$ The positive factor pairs are $(x,y)=(1,93)$ and $(3,31).$ For the first, $d=a+94<500$ gives $405$ choices for $a.$ For the second, $d=a+34<500$ gives $465$ choices. The total is $405+465=870.$

Collapse the repeated shifts to write $P_{20}$ directly in terms of $P_0$

Only collect the linear terms after substituting $x-210$

The accumulated shift is $1+2+\cdots+20=210,$ so $P_{20}(x)=P_0(x-210).$ The coefficient of $x$ is therefore $$\begin{aligned}&3(210)^2-2(313)(210)-77\\&\quad=132300-131460-77\\&\quad=763.\end{aligned}$$

A sum of an odd number of consecutive integers is divisible by the number of terms

A sum of ten consecutive integers is congruent to $5\pmod {10}$

The sums of $9$ and $11$ consecutive integers are divisible by $9$ and $11,$ so the desired number is a multiple of $99.$ A sum of $10$ consecutive integers has the form $10a+45,$ hence is congruent to $5\pmod {10}.$ The first multiple of $99$ ending in $5$ is $5\cdot99=495,$ and each of the three required representations then exists.

Condition on the six selected values and record only whether each belongs to the brick or the box in increasing order

The brick fits exactly when every prefix of this six-letter word contains at least as many $a$’s as $b$’s

After the six distinct values are fixed and sorted, each of the $\binom63=20$ assignments of three values to the brick is equally likely. The sorted brick dimensions fit the sorted box dimensions exactly when, in every prefix of the resulting word of three $a$’s and three $b$’s, the number of $a$’s is at least the number of $b$’s. There are $C_3=5$ such words. Thus $p=\frac{5}{20}=\frac{1}{4},$ and the requested sum is $1+4=5.$

For an ordered pair, each element can lie in the first subset only, the second only, or both

When the two subsets are swapped, identify the one ordered pair that remains fixed

For an ordered pair $(A,B)$ with $A\cup B=S,$ each element has three possible memberships: $A$ only, $B$ only, or both. This gives $3^6=729$ ordered pairs. Swapping $A$ and $B$ fixes only the pair $A=B=S.$ Therefore the number of unordered pairs is $$\frac{729+1}{2}=365.$$

Measure every label’s clockwise displacement from the point labeled $1$

Reduce the resulting quadratic congruence separately modulo $32$ and modulo $125$

Label $j$ is displaced $$2+3+\cdots+j=\frac{j(j+1)}2-1$$ points clockwise from label $1.$ Thus it shares the point of label $1993$ exactly when $j(j+1)\equiv1993\cdot1994$ modulo $4000,$ or equivalently when $j(j+1)\equiv2042\pmod {4000}.$ Modulo $125,$ the solutions are $j\equiv6$ and $j\equiv118,$ and modulo $32,$ they are $j\equiv9$ and $j\equiv22.$ Combining these by the Chinese Remainder Theorem gives $$\begin{aligned}j\equiv{}&118,1993,\\&2006,3881\pmod {4000}.\end{aligned}$$ The smallest positive possibility is $118.$

Let $x$ be the number of triangular faces and count face-edge and face-vertex incidences

Use Euler’s formula to express $x$ in terms of $V$, then obtain two divisibility conditions on $V$

Let $x$ be the number of triangular faces, so there are $32-x$ pentagons and $$E=\frac{3x+5(32-x)}2=80-x.$$ Euler’s formula gives $V+x=50.$ Counting face-vertex incidences yields $$\begin{aligned}TV&=3x=150-3V,\\PV&=5(32-x)=5V-90.\end{aligned}$$ Hence $V(T+3)=150$ and $V(5-P)=90.$ Also $18\leq V\leq50,$ so the only common divisor of $150$ and $90$ in that range is $V=30.$ Then $T=2$ and $P=2,$ giving $100P+10T+V=250.$

Compute Alfred’s chance to win a single game when he goes first and when Bonnie goes first

If $p_r$ is Alfred’s chance to win game $r$, express $p_{r+1}$ in terms of $p_r$

Alfred wins a game with probability $\frac{2}{3}$ when he starts and $\frac{1}{3}$ when Bonnie starts. Since the loser starts the next game, $$\begin{aligned}p_{r+1}&=\frac23(1-p_r)+\frac13p_r\\&=\frac23-\frac13p_r,\qquad p_1=\frac23.\end{aligned}$$ Thus $p_r-\frac12=\frac{(-\frac{1}{3})^{r-1}}{6}.$ In particular, $p_6=\frac12-\frac1{1458}=\frac{364}{729}.$ Therefore $m+n=1093,$ whose last three digits are $093.$