Answer: 400

Small Hint:

Write every number as $\frac{k}{30}$ and impose the condition $\gcd(k,30)=1$

Big Hint:

Group the eligible numerators into ten blocks of length $30$

Solution:

The numbers are $\frac{k}{30}$ for $1\leq k\lt300$ and $\gcd(k,30)=1.$ There are $\varphi(30)=8$ eligible residues in each block of $30,$ and their sum is $\frac{30\varphi(30)}{2}=120.$ Thus the sum of all eligible numerators is $$\begin{aligned}\sum_{q=0}^9(8\cdot30q+120)&=8\cdot30\cdot45\\&\quad+10\cdot120\\&=12000.\end{aligned}$$ Dividing by $30$ gives $400.$

Answer: 502

Small Hint:

Once a set of nonzero digits is chosen, their order is forced

Big Hint:

Exclude subsets of sizes $0$ and $1$ from the subsets of $\{1,2,\ldots,9\}$

Solution:

The digit $0$ cannot occur, because it would have to be the first digit and leading zeroes are not part of a decimal representation. Every subset of at least two digits from $\{1,\ldots,9\}$ gives exactly one ascending integer when written in increasing order. Hence the number is $$\begin{aligned}2^9-\binom90-\binom91&=512-1-9\\&=502.\end{aligned}$$

Answer: 164

Small Hint:

If she had $w$ wins initially, a $0.500$ ratio means she had played $2w$ matches

Big Hint:

Translate the final ratio into a strict inequality before taking the largest integer $w$

Solution:

If she initially had $w$ wins, then she had played $2w$ matches. The final condition is $$\frac{w+3}{2w+4}\gt\frac{503}{1000}.$$ Cross-multiplication gives $1000w+3000\gt1006w+2012,$ so $6w\lt988$ and $w\lt164\frac23.$ The largest possible integer is $164.$

Answer: 62

Small Hint:

Represent the three entries as $\binom nk,$ $\binom n{k+1},$ and $\binom n{k+2}$

Big Hint:

Use the ratios of consecutive binomial coefficients to obtain two linear equations in $n$ and $k$

Solution:

For three consecutive entries beginning at position $k,$ $$\begin{aligned}\frac{\binom n{k+1}}{\binom nk}&=\frac{n-k}{k+1}=\frac43,\\\frac{\binom n{k+2}}{\binom n{k+1}}&=\frac{n-k-1}{k+2}=\frac54.\end{aligned}$$ Thus $3n=7k+4$ and $4n=9k+14.$ Solving gives $k=26$ and $n=62.$

Answer: 660

Small Hint:

Every element has the form $\frac{N}{999}$, and its reduced denominator must divide $999=3^3\cdot37$

Big Hint:

Count numerators coprime to $999$, then check which additional multiples of $3$ can occur with denominator $37$

Solution:

Every element is $\frac{N}{999}$ for $1\leq N\leq998.$ Any $a$ coprime to $999$ occurs as a reduced numerator with denominator $999,$ giving $\varphi(999)=648$ values. If $a$ is divisible by $3$ but not $37,$ it can be coprime to a reduced denominator only when that denominator is $37;$ this adds the $12$ multiples of $3$ below $37.$ A numerator divisible by $37$ would need a denominator dividing $27$ and larger than it, so no further values occur. Therefore the total is $648+12=660.$

Answer: 156

Small Hint:

Write the smaller integer as $1abc$ and separate cases by the number of trailing $9$s

Big Hint:

A digit that is unchanged must be at most $4$, and the digit increased by $1$ must also pair without a carry

Solution:

Write the smaller number as $1abc.$ If $c\neq9,$ then $c\leq4$ and the unchanged digits $a$ and $b$ are each at most $4,$ giving $5^3=125$ pairs. If $c=9$ but $b\neq9,$ then $a\leq4$ and $b\leq4,$ giving $25$ pairs. If $b=c=9$ but $a\neq9,$ there are $5$ choices for $a.$ Finally, $1999+2000$ also needs no carry. The total is $125+25+5+1=156.$

Answer: 320

Small Hint:

Find the altitudes from $A$ and $D$ to the common edge $BC$

Big Hint:

The height from $D$ to plane $ABC$ is its face altitude multiplied by $\sin30^\circ$

Solution:

The altitudes to $BC$ in faces $ABC$ and $BCD$ are $\frac{2(120)}{10}=24$ and $\frac{2(80)}{10}=16,$ respectively. Because the dihedral angle is $30^\circ,$ the perpendicular height from $D$ to plane $ABC$ is $16\sin30^\circ=8.$ Using face $ABC$ as the base, the volume is $$\frac13(120)(8)=320.$$

Answer: 819

Small Hint:

A sequence with constant second difference $1$ is given by a quadratic with leading coefficient $\frac{1}{2}$

Big Hint:

Use the two zero terms to write the quadratic in factored form

Solution:

A quadratic sequence with second difference $1$ has leading coefficient $\frac{1}{2}.$ Since its values vanish at indices $19$ and $92,$ $$a_n=\frac12(n-19)(n-92).$$ Therefore $a_1=\frac12(-18)(-91)=819.$

Answer: 164

Small Hint:

Put $AB$ on the $x$-axis and compare the distances from $P$ to the two legs

Big Hint:

The common trapezoid height cancels, leaving an equation involving $AP$ and $PB$ divided by the leg lengths

Solution:

Put $A=(0,0),$ $B=(92,0),$ and let the height of the trapezoid be $h.$ If $P=(p,0),$ its perpendicular distances to legs $AD$ and $BC$ are $\frac{hp}{70}$ and $\frac{h(92-p)}{50},$ respectively. Tangency to both legs makes these equal, so $$\frac p{70}=\frac{92-p}{50}.$$ Hence $120p=6440$ and $AP=p=\frac{161}{3}.$ Therefore $m+n=161+3=164.$

Answer: 572

Small Hint:

Write $z=x+iy$; the first condition gives a square, and the second gives two circle inequalities

Big Hint:

Subtract from the $40$-by-$40$ square the union of two semicircles, accounting for their lens-shaped overlap

Solution:

Write $z=x+iy.$ The condition on $\frac{z}{40}$ gives $0\leq x\leq40$ and $0\leq y\leq40.$ Since $$\frac{40}{\overline z}=\frac{40x}{x^2+y^2}+i\frac{40y}{x^2+y^2},$$ the other condition requires $x^2+y^2\geq40x$ and $x^2+y^2\geq40y.$ Thus, within the square, we remove two semicircles of radius $20.$

Their overlap is the lens formed by two radius-$20$ circles whose centers are $20\sqrt2$ apart. Its area is $200\pi-400.$ Hence the removed union has area $400\pi-(200\pi-400)$, or $200\pi+400.$ Let $K$ denote the area of $A.$ Then $$\begin{aligned}K&=1600-(200\pi+400)\\&=1200-200\pi\\&\approx571.68.\end{aligned}$$ The nearest integer is $572.$

Answer: 945

Small Hint:

Two reflections in intersecting lines compose to a rotation through twice the angle between the lines

Big Hint:

An unoriented line returns to itself when its accumulated rotation is a multiple of $\pi$

Solution:

The composition is rotation through $$2\left(\frac{\pi}{54}-\frac{\pi}{70}\right)=\frac{8\pi}{945}.$$ A line through the origin is unchanged by a rotation exactly when the rotation angle is a multiple of $\pi.$ Thus $m$ must satisfy $\frac{8m}{945}\in\mathbb Z.$ Since $\gcd(8,945)=1,$ the least such $m$ is $945.$

Answer: 792

Small Hint:

A reachable set is determined by nonincreasing column heights between $0$ and $5$

Big Hint:

Encode the boundary of such a set as a lattice path with $5$ vertical and $7$ horizontal steps

Solution:

After any sequence of bites, the remaining squares form a lower-left order ideal: the seven column heights are nonincreasing integers between $0$ and $5.$ Conversely, every such boundary can be produced and corresponds to a lattice path across a $5$-by-$7$ rectangle. Each path consists of $5$ vertical and $7$ horizontal steps, so the number of states, including full and empty, is $$\binom{12}{5}=792.$$

Answer: 820

Small Hint:

Set $BC=40t,$ $AC=41t,$ and use the Law of Cosines with $AB=9$

Big Hint:

Express the area as a function of $\cos C$ and maximize its square

Solution:

Set $BC=40t,$ $AC=41t,$ and $x=\cos C.$ The Law of Cosines gives $$81=t^2(3281-3280x),$$ while the area is $820t^2\sqrt{1-x^2}.$ Hence it equals $$820\cdot81\,\frac{\sqrt{1-x^2}}{3281-3280x}.$$ Differentiating its logarithm shows the maximum occurs at $x=\frac{3280}{3281}.$ Then $\sqrt{1-x^2}=\frac{81}{3281}$ and $3281-3280x=\frac{6561}{3281},$ so the maximum area is $820.$

Answer: 94

Small Hint:

Let $(\alpha,\beta,\gamma)$ be normalized barycentric coordinates of $O$

Big Hint:

Write the three ratios as $\frac{1-\alpha}{\alpha}$, $\frac{1-\beta}{\beta}$, and $\frac{1-\gamma}{\gamma}$

Solution:

Let $\alpha+\beta+\gamma=1$ be the barycentric coordinates of $O.$ Then $$\begin{aligned}x=\frac{AO}{OA'}&=\frac{\beta+\gamma}{\alpha},\\y=\frac{BO}{OB'}&=\frac{\gamma+\alpha}{\beta},\\z=\frac{CO}{OC'}&=\frac{\alpha+\beta}{\gamma}.\end{aligned}$$ Expanding both sides using $\alpha+\beta+\gamma=1$ gives the standard identity $$xyz=x+y+z+2.$$ Since $x+y+z=92,$ the requested product is $94.$

Answer: 396

Small Hint:

Let $f(m)=\sum_{j\geq1}\lfloor \frac{m}{5^j}\rfloor$, the number of trailing zeroes in $m!$

Big Hint:

Every positive value attained by $f$ first appears at a multiple $5k$, where $f(5k)=k+f(k)$

Solution:

The number of trailing zeroes is $f(m)=\sum_{j\geq1}\lfloor \frac{m}{5^j}\rfloor.$ Its positive distinct values occur at the multiples $5k,$ and $$f(5k)=k+f(k)$$ is strictly increasing with $k.$ Now $$\begin{aligned}f(1595)&=319+63\\&\quad+12+2=396,\end{aligned}$$ so $f(7975)=1595+396=1991.$ For $k=1596,$ the same calculation gives $f(k)=396,$ hence $f(5k)=1992.$ Therefore exactly $1595$ positive values through $1991$ are factorial tails. Of the $1991$ positive integers below $1992,$ the number omitted is $1991-1595=396.$