Answer: 73

Small Hint:

Every resident owns the candy hearts, so add all four ownership counts and compare with $900$

Big Hint:

Counting once per item owned, residents with exactly $2,$ $3,$ or $4$ items are counted $1,$ $2,$ or $3$ extra times

Solution:

Adding the four ownership counts gives $195 + 367 + 562 + 900 = 2024$ item ownerships among the $900$ residents. Since everyone owns a bag of candy hearts, every resident owns at least one item, and a resident owning exactly $k$ items is counted $k - 1$ times beyond the first.

If $n_4$ residents own all four things, the extra counts total $$\begin{aligned} 2024 - 900 &= 437 \cdot 1 + 234 \cdot 2 \\ &\quad {}+ n_4 \cdot 3, \end{aligned}$$ so $1124 = 905 + 3 n_4,$ giving $n_4 = \frac{219}{3} = 73.$

Answer: 236

Small Hint:

An odd-length list has its median as a member, so the list has evenly many items. The mode $9$ appears at least twice.

Big Hint:

Try four items $a, b, 9, 9$ with $a + b = 12:$ the median $\frac{b + 9}{2}$ must be a whole number, so $b$ is odd

Solution:

The median is an integer that is not in the list, so the list cannot have odd length (then the median would be a member). The unique mode $9$ appears at least twice. Two items $9, 9$ sum to $18,$ not $30,$ so try four items $a \lt b \lt 9$ together with $9, 9,$ where $a$ and $b$ are distinct (a repeat would tie the mode) and $a + b = 12.$ The median $\frac{b + 9}{2}$ must be an integer, so $b$ is odd, and $a = 12 - b \lt b$ forces $b \gt 6.$ Thus $b = 7$ and $a = 5:$ the list $5, 7, 9, 9$ has median $8,$ which indeed does not appear.

No longer list works. With exactly two $9$s, six items would need four distinct other values summing to $12,$ namely $\{1, 2, 3, 6\}$ or $\{1, 2, 4, 5\},$ but both give median $4.5;$ eight or more items are impossible because six distinct positive values already sum to at least $21 \gt 12.$ With three $9$s the remaining items sum to $3,$ and every option either puts $9$ at the median or ties the mode. Four $9$s already sum to more than $30.$

The sum of squares is $25 + 49 + 81 + 81 = 236.$

Answer: 45

Small Hint:

In the row sum $999,$ each column’s two digits must add to exactly $9,$ with no carrying anywhere

Big Hint:

If the top row’s digits sum to $S,$ the bottom row’s digits sum to $27 - S;$ the column condition $10S + (27 - S) = 99$ pins down $S$

Solution:

Let the top row hold digits $a, b, c$ and the bottom row $d, e, f.$ In the sum of the two row numbers, the units digits satisfy $c + f \equiv 9 \pmod{10},$ and since $c + f \le 18$ in fact $c + f = 9$ with no carry. Repeating the argument in the tens and hundreds places gives $b + e = 9$ and $a + d = 9.$

The three column numbers add to $10(a + b + c) + (d + e + f)$ $= 99.$ Writing $S = a + b + c,$ the bottom digits sum to $27 - S,$ so $10S + 27 - S = 99$ and $S = 8.$

Conversely, any digits with $a + b + c = 8$ determine the bottom row by $d = 9 - a,$ $e = 9 - b,$ $f = 9 - c,$ and both conditions hold. The number of solutions of $a + b + c = 8$ in nonnegative digits is $\binom{10}{2} = 45.$

Answer: 33

Small Hint:

Set $a = \log_2 x,$ $b = \log_2 y,$ $c = \log_2 z$ to turn the system into three linear equations

Big Hint:

Add all three equations to find $a + b + c,$ then combine it with each equation to solve for $a,$ $b,$ $c$ one at a time

Solution:

Let $a = \log_2 x,$ $b = \log_2 y,$ $c = \log_2 z.$ The equations become $a - b - c = \frac{1}{2},$ $b - a - c = \frac{1}{3},$ $c - a - b = \frac{1}{4}.$ Adding all three gives $-(a + b + c) = \frac{13}{12}.$ Since $a - b - c = 2a - (a + b + c),$ we get $$2a = \frac{1}{2} - \frac{13}{12} = -\frac{7}{12},$$ so $a = -\frac{7}{24},$ and similarly $b = -\frac{3}{8}$ and $c = -\frac{5}{12}.$

Therefore $4a + 3b + 2c = -\frac{28}{24}$ $- \frac{27}{24}$ $- \frac{20}{24} = -\frac{75}{24}$ $= -\frac{25}{8},$ so $\left|\log_2(x^4 y^3 z^2)\right| = \frac{25}{8}$ and $m + n = 25 + 8 = 33.$

Answer: 80

Small Hint:

Each corner triangle between the hexagon and the big triangle has all three sides parallel to the big triangle’s sides, so it is similar to the big triangle

Big Hint:

Writing the side containing $\overline{AB}$ as $s$ plus two corner pieces leads to $\frac{s}{200} + \frac{s}{240} + \frac{s}{300} = 1$

Solution:

Let $s$ be the hexagon’s side length, and let the triangle formed by lines $AB,$ $CD,$ $EF$ have sides of lengths $P,$ $Q,$ $R$ along those three lines, respectively. The corner triangle cut off at the vertex $X$ where lines $AB$ and $CD$ meet has third side $BC,$ and since $BC \parallel EF,$ all three of its sides are parallel to sides of the big triangle. So it is similar to the big triangle with ratio $\frac{BC}{R} = \frac{s}{R},$ and its side along line $AB$ has length $P \cdot \frac{s}{R}.$ Likewise the corner at $AB \cap EF$ contains $FA \parallel CD$ and cuts off $P \cdot \frac{s}{Q}$ from the $P$-side.

The $P$-side therefore decomposes as corner piece, $AB,$ corner piece: $$P = P \cdot \frac{s}{R} + s + P \cdot \frac{s}{Q},$$ and dividing by $P$ gives $1 = s\left(\frac{1}{P} + \frac{1}{Q} + \frac{1}{R}\right),$ symmetric in the three sides.

Hence $s = \frac{1}{\frac{1}{200} + \frac{1}{240} + \frac{1}{300}} = \frac{1200}{6 + 5 + 4} = 80.$

Answer: 55

Small Hint:

Group Bob’s sets by their maximum: if the maximum is $a,$ the rest of the set is any subset of $\{1, \ldots, a - 1\}$

Big Hint:

So the list has $\sum_{a \in A} 2^{a-1}$ sets; binary representations are unique, so write $2024$ in base $2$

Solution:

For a fixed $a \in A,$ the sets $B$ with maximum element $a$ consist of $a$ together with an arbitrary subset of $\{1, \ldots, a - 1\},$ so there are $2^{a-1}$ of them, and every set on Bob’s list is counted exactly once by its maximum. Hence $$\sum_{a \in A} 2^{a-1} = 2024.$$

Since $2024 = 2^{10} + 2^9 + 2^8 + 2^7$ $+ 2^6 + 2^5 + 2^3$ and binary representations are unique, $A = \{11, 10, 9, 8, 7, 6, 4\}.$ The sum of the elements of $A$ is $11 + 10 + 9 + 8 + 7 + 6$ $+ 4 = 55.$

Answer: 699

Small Hint:

Changing a digit to $1$ subtracts $(\text{digit} - 1)$ times a power of $10,$ so $N \equiv 6(a-1)$ $\equiv 2(b-1)$ $\equiv 3(c-1)$ $\equiv d - 1 \pmod 7$ for digits $a, b, c, d$

Big Hint:

Let $k = N \bmod 7:$ each digit is then determined mod $7,$ and consistency of $N$ with its own digits forces a single value of $k$

Solution:

Write $N$ with digits $a, b, c, d.$ Changing the thousands digit to $1$ produces $N - (a - 1) \cdot 1000,$ so $N \equiv 1000(a-1) \pmod 7,$ and similarly for the other digits. Since $1000 \equiv 6,$ $100 \equiv 2,$ and $10 \equiv 3 \pmod 7,$ $$\begin{aligned} N &\equiv 6(a-1) \\ &\equiv 2(b-1) \equiv 3(c-1) \\ &\equiv d - 1 \pmod 7. \end{aligned}$$

Let $k = N \bmod 7.$ Using $6 \equiv -1$ and the inverses $2^{-1} \equiv 4,$ $3^{-1} \equiv 5,$ the digits satisfy $a \equiv 1 - k,$ $b \equiv 1 + 4k,$ $c \equiv 1 + 5k,$ $d \equiv 1 + k \pmod 7.$ But also $k \equiv N$ $\equiv 6a + 2b + 3c + d \pmod 7;$ substituting gives $k \equiv 12 + 18k \equiv 5 + 4k,$ so $3k \equiv 2$ and $k \equiv 3 \pmod 7.$

Then $a \equiv 5,$ $b \equiv 6,$ $c \equiv 2,$ $d \equiv 4 \pmod 7,$ and taking the largest digit in each class gives $a = 5$ (the class $\{5, 12, \ldots\}$ has no larger digit), $b = 6,$ $c = 9,$ $d = 4:$ $N = 5694.$ Indeed $1694, 5194, 5614, 5691$ are all multiples of $7.$ Finally $Q = 5,$ $R = 694,$ and $Q + R = 699.$

Answer: 127

Small Hint:

Slice with a plane through the axis: the tube is a circle of radius $3$ centered $6$ from the axis, and the sphere’s center lies on the axis

Big Hint:

Tangency puts the tube’s center at distance $11 - 3$ or $11 + 3$ from the sphere’s center; scale the tube center’s distance from the axis by $\frac{11}{8}$ or $\frac{11}{14}$

Solution:

By symmetry the axis of the torus passes through the center $O$ of the sphere. Work in a plane through the axis: there the torus appears as a circle of radius $3$ (the tube) whose center sits at distance $6$ from the axis, and the sphere appears as a circle of radius $11$ centered at $O.$ The two surfaces are tangent along the circle swept by the tangency point of these cross-sections, which lies on the ray from $O$ through the tube’s center. For internal tangency the tube’s center is at distance $11 - 3 = 8$ from $O;$ for external tangency, $11 + 3 = 14.$

The tangency point lies at distance $11$ from $O$ along that ray, so it is the tube center scaled by $\frac{11}{8}$ (resp. $\frac{11}{14}$) from $O,$ and its distance from the axis is the same multiple of the tube center’s distance $6:$ $$r_i = \frac{11}{8} \cdot 6 = \frac{33}{4},$$ $$r_o = \frac{11}{14} \cdot 6 = \frac{33}{7}.$$

Then $r_i - r_o = \frac{33 \cdot 3}{28} = \frac{99}{28},$ which is in lowest terms, so $m + n = 99 + 28 = 127.$

Answer: 902

Small Hint:

A maximal placement has no empty row or column: a chip matching the column’s color could always be added inside an empty row

Big Hint:

So every row and column carries a color, chips fill exactly the cells where the two colors agree, and rows and columns must use the same set of colors

Solution:

In a valid placement, each nonempty row has a single color, and likewise each column. If some row were empty, choose any cell of it: a chip of the color of that cell’s column (either color if the column is also empty) could legally be added, contradicting the third condition. So every row and every column is nonempty, and we may speak of its color.

A chip at a cell forces its row and column colors to agree; conversely, if a row and a column share a color but their common cell is empty, a chip of that color could be added. Hence chips occupy exactly the cells whose row color equals the column color. For every row to be nonempty, each row’s color must appear among the column colors, and vice versa — the rows and the columns use the same set of colors. Any such coloring conversely yields a valid maximal placement (at most $25$ cells hold chips of each color, so the supply suffices), and distinct colorings give distinct placements.

Counting the colorings: all rows and columns white, all black, or both colors used by the rows and by the columns: $1 + 1 + (2^5 - 2)^2 = 2 + 900$ $= 902.$

Answer: 468

Small Hint:

The right angle gives $IA^2 = R^2 - OI^2,$ and Euler’s formula $OI^2 = R^2 - 2Rr$ turns this into $IA^2 = 2Rr$

Big Hint:

Combine $IA = \frac{r}{\sin(\frac{A}{2})}$ with $s - a = r \cot\frac{A}{2},$ $a = 2R \sin A,$ and $rs = \frac{1}{2} bc \sin A$

Solution:

Since $\angle OIA = 90^\circ,$ the Pythagorean theorem in triangle $OIA$ gives $IA^2 = OA^2 - OI^2 = R^2 - OI^2,$ and Euler’s formula $OI^2 = R^2 - 2Rr$ yields $$IA^2 = 2Rr = 2 \cdot 13 \cdot 6 = 156.$$ Combining with $IA = \frac{r}{\sin(\frac{A}{2})}$ gives $\sin^2\frac{A}{2} = \frac{36}{156} = \frac{3}{13},$ so $\cos^2\frac{A}{2} = \frac{10}{13}.$

Then $\sin A = 2 \sin\frac{A}{2}\cos\frac{A}{2} = \frac{2\sqrt{30}}{13},$ so $a = BC = 2R \sin A = 4\sqrt{30},$ while $s - a = r \cot\frac{A}{2} = 6\sqrt{\frac{10}{3}}$ $= 2\sqrt{30}.$ Hence the semiperimeter is $s = 6\sqrt{30}.$

Equating the two area formulas $[ABC] = rs = \frac{1}{2}\, bc \sin A,$ $$\begin{aligned} bc = \frac{2rs}{\sin A} &= \frac{2 \cdot 6 \cdot 6\sqrt{30}}{\frac{2\sqrt{30}}{13}} \\ &= 36 \cdot 13 = 468. \end{aligned}$$

Answer: 601

Small Hint:

With $a + b + c = 300,$ the given sum equals $300(ab + bc + ca) - 3abc$

Big Hint:

Expand $(100 - a)(100 - b)(100 - c)$ using $a + b + c = 300:$ the condition says exactly that this product vanishes

Solution:

The left side is the symmetric sum $(a + b + c)(ab + bc + ca)$ $- 3abc = 300q - 3p,$ where $q = ab + bc + ca$ and $p = abc.$ So the condition is $100q - p = 2{,}000{,}000.$ Now expand $$\begin{gathered} (100 - a)(100 - b)(100 - c) \\ = 10^6 - 10^4 (a + b + c) \\ \quad {}+ 100q - p \\ = (100q - p) - 2 \cdot 10^6, \end{gathered}$$ using $a + b + c = 300.$ The condition holds exactly when this product is $0,$ that is, when at least one of $a, b, c$ equals $100.$

If $a = 100,$ then $b + c = 200,$ giving $201$ triples, and likewise for $b$ and $c:$ $3 \cdot 201 = 603.$ A triple counted more than once has two variables equal to $100,$ which forces the third to be $100$ as well; the triple $(100, 100, 100)$ is counted three times, so the total is $603 - 2 = 601.$

Answer: 23

Small Hint:

The segment from $(\cos\theta, 0)$ to $(0, \sin\theta)$ lies on $\frac{x}{\cos\theta} + \frac{y}{\sin\theta} = 1;$ $\overline{AB}$ is the member $\theta = 60^\circ$

Big Hint:

Setting the $\theta$-derivative of $\frac{x}{\cos\theta} + \frac{y}{\sin\theta}$ to zero at $\theta = 60^\circ$ gives $y = 3\sqrt{3}\,x;$ intersect that with $\overline{AB}$

Solution:

The members of $\mathcal{F}$ are the segments from $(\cos\theta, 0)$ to $(0, \sin\theta)$ for $0 \lt \theta \lt 90^\circ,$ lying on the lines $\frac{x}{\cos\theta} + \frac{y}{\sin\theta} = 1;$ the segment $\overline{AB}$ is the member with $\theta = 60^\circ.$ For a point $(x, y)$ of $\overline{AB}$ with $x, y \gt 0,$ let $$g(\theta) = \frac{x}{\cos\theta} + \frac{y}{\sin\theta} - 1,$$ so the point lies on the member for angle $\theta$ exactly when $g(\theta) = 0.$ Note $g \to +\infty$ at both endpoints of $(0^\circ, 90^\circ)$ and $g(60^\circ) = 0.$ If $g'(60^\circ) \neq 0,$ then $g$ is negative on one side of $60^\circ,$ and the intermediate value theorem produces another zero on that side — the point is covered by another segment. So $C$ must satisfy $g'(60^\circ) = 0.$ Because both $\sec\theta$ and $\csc\theta$ are strictly convex on this interval, $g$ is strictly convex; thus for that point $60^\circ$ is the strict global minimum of $g,$ so no other segment contains it.

Now $g'(\theta) = \frac{x \sin\theta}{\cos^2\theta} - \frac{y \cos\theta}{\sin^2\theta},$ and $g'(60^\circ) = 0$ gives $x \sin^3 60^\circ = y \cos^3 60^\circ,$ i.e. $y = 3\sqrt{3}\,x.$ Intersecting with $\overline{AB}:$ $y = \frac{\sqrt{3}}{2} - \sqrt{3}\,x$ gives $3x = \frac{1}{2} - x,$ so $x = \frac{1}{8}$ and $y = \frac{3\sqrt{3}}{8},$ an interior point of $\overline{AB}.$

Therefore $OC^2 = \frac{1}{64} + \frac{27}{64} = \frac{28}{64} = \frac{7}{16},$ and $p + q = 7 + 16 = 23.$

Answer: 321

Small Hint:

Factor $2 - 2x + x^2$ $= (x - (1+\mathrm{i}))(x - (1-\mathrm{i})),$ and use $\prod_k (x - \omega^k) = x^{13} - 1$ over all $13$th roots of unity

Big Hint:

$(1+\mathrm{i})^2 = 2\mathrm{i}$ makes $(1+\mathrm{i})^{13}$ easy to compute; the whole product becomes $\left(1 - (1+\mathrm{i})^{13}\right)\left(1 - (1-\mathrm{i})^{13}\right)$

Solution:

Since $2 - 2x + x^2 = (x - 1)^2 + 1$ $= (x - (1+\mathrm{i}))(x - (1-\mathrm{i})),$ each factor of the product splits, and as $k$ runs from $0$ to $12,$ $\omega^k$ runs over all $13$th roots of unity. Because $\prod_k (x - \omega^k) = x^{13} - 1,$ for any $\alpha$ we get $\prod_k (\omega^k - \alpha) = (-1)^{13}(\alpha^{13} - 1)$ $= 1 - \alpha^{13}.$ Hence the product equals $$\left(1 - (1+\mathrm{i})^{13}\right)\left(1 - (1-\mathrm{i})^{13}\right).$$

Since $(1+\mathrm{i})^2 = 2\mathrm{i},$ we get $(1+\mathrm{i})^{13} = (1+\mathrm{i})(2\mathrm{i})^6$ $= -64(1 + \mathrm{i}) = -64 - 64\mathrm{i},$ and by conjugation $(1-\mathrm{i})^{13} = -64 + 64\mathrm{i}.$ So the product is $$\begin{gathered} (65 + 64\mathrm{i})(65 - 64\mathrm{i}) \\ = 65^2 + 64^2 = 4225 + 4096 \\ = 8321, \end{gathered}$$ whose remainder upon division by $1000$ is $321.$