Small Hint:

Factor $a^2 - b^2 = (a - b)(a + b)$ and note the two factors always have the same parity

Big Hint:

Odd numbers and multiples of $4$ are all achievable; numbers that are $2$ more than a multiple of $4$ are not

Solution:

Write $n = a^2 - b^2 = (a - b)(a + b).$ The factors $a - b$ and $a + b$ differ by the even number $2b,$ so they have the same parity. If both are odd, $n$ is odd; if both are even, $n$ is divisible by $4.$ Hence no integer $n \equiv 2 \pmod 4$ is a difference of two squares.

Conversely, every odd number $2k + 1$ equals $(k+1)^2 - k^2,$ and every multiple of $4,$ say $4k,$ equals $(k+1)^2 - (k-1)^2$ (with $k - 1 \ge 0$ since $k \ge 1$).

Between $1$ and $1000$ there are $500$ odd numbers and $250$ multiples of $4,$ for a total of $500 + 250 = 750.$

Small Hint:

A rectangle is determined by choosing $2$ of the $9$ horizontal lines and $2$ of the $9$ vertical lines

Big Hint:

There are $(9 - k)^2$ squares of size $k \times k;$ sum over $k = 1$ to $8$

Solution:

A rectangle is determined by choosing two of the nine horizontal lines and two of the nine vertical lines, so $r = \binom{9}{2}^2 = 36^2 = 1296.$

A $k \times k$ square can be placed in $(9 - k)^2$ positions, so $$\begin{aligned} s &= \sum_{k=1}^{8} (9 - k)^2 \\ &= 8^2 + 7^2 + \cdots + 1^2 \\ &= \frac{8 \cdot 9 \cdot 17}{6} = 204. \end{aligned}$$

Then $\frac{s}{r} = \frac{204}{1296} = \frac{17}{108},$ which is in lowest terms, so $m + n = 17 + 108 = 125.$

Small Hint:

If the two-digit number is $a$ and the three-digit number is $b,$ the five-digit number is $1000a + b$

Big Hint:

Rewrite $1000a + b = 9ab$ as $b(9a - 1) = 1000a;$ since $\gcd(9a - 1, a) = 1,$ the factor $9a - 1$ must divide $1000$

Solution:

Let $a$ be the two-digit number and $b$ the three-digit number. The condition is $1000a + b = 9ab,$ which rearranges to $b(9a - 1) = 1000a.$ Since $\gcd(9a - 1, a) = 1,$ the number $9a - 1$ must divide $1000.$

For a two-digit $a,$ $9a - 1$ runs from $89$ to $890,$ and $9a - 1 \equiv 8 \pmod 9.$ The only divisor of $1000$ in that range congruent to $8$ modulo $9$ is $125,$ giving $a = 14$ and $b = \frac{1000 \cdot 14}{125} = 112,$ which is indeed a three-digit number. Check: $14112 = 9 \cdot 14 \cdot 112.$

The requested sum is $14 + 112 = 126.$

Small Hint:

The two radius-$5$ centers are $10$ apart, and the other two centers both lie on the perpendicular bisector of that segment

Big Hint:

Right triangles give midpoint distances $12$ and $\sqrt{r^2 + 10r};$ rule out opposite sides before subtracting them

Solution:

Let the radius-$5$ circles have centers $P_1$ and $P_2,$ so $P_1P_2 = 5 + 5 = 10,$ and let $M$ be the midpoint. The radius-$8$ circle’s center $Q$ satisfies $QP_1 = QP_2 = 13,$ so $Q$ lies on the perpendicular bisector of $\overline{P_1P_2}$ at distance $\sqrt{13^2 - 5^2} = 12$ from $M.$ Likewise the fourth circle, of radius $r,$ has its center $R$ on the same perpendicular bisector with $RP_1 = 5 + r,$ so $RM = \sqrt{(5+r)^2 - 25}$ $= \sqrt{r^2 + 10r}.$

The centers $Q$ and $R$ cannot lie on opposite sides of $M:$ that would give $12 + RM = 8 + r,$ hence $RM = r - 4,$ but squaring would yield $18r = 16,$ contrary to $r \ge 4.$ Nor can $R$ lie beyond $Q,$ since then $RM - 12 = 8 + r$ would force $RM = r + 20,$ which is incompatible with $RM^2 = r^2 + 10r.$ Thus $R$ lies between $M$ and $Q,$ and external tangency to the radius-$8$ circle gives $$12 - \sqrt{r^2 + 10r} = 8 + r.$$ Then $\sqrt{r^2 + 10r} = 4 - r,$ and squaring yields $r^2 + 10r = 16 - 8r + r^2,$ so $18r = 16$ and $r = \frac{8}{9}.$

Thus $m + n = 8 + 9 = 17.$

Small Hint:

Find the fractions with numerator $1$ or $2$ closest to $\frac{2}{7}$ on each side

Big Hint:

$r$ must be closer to $\frac{2}{7}$ than to $\frac{1}{4}$ or $\frac{1}{3};$ count four-place decimals between $\frac{15}{56}$ and $\frac{13}{42}$

Solution:

Among fractions with numerator $1$ or $2,$ the closest neighbors of $\frac{2}{7} \approx 0.2857$ are $\frac{1}{4} = 0.25$ below (note $\frac{2}{8} = \frac{1}{4}$) and $\frac{1}{3} \approx 0.3333$ above (note $\frac{2}{6} = \frac{1}{3}$); no other candidate lies between them. So $\frac{2}{7}$ is the unique closest fraction to $r$ exactly when $r$ is closer to $\frac{2}{7}$ than to both $\frac{1}{4}$ and $\frac{1}{3},$ i.e. when $r$ lies strictly between the midpoints $$\begin{aligned} \frac{1}{2}\left(\frac{1}{4} + \frac{2}{7}\right) &= \frac{15}{56} \\ &= 0.26785\ldots \end{aligned}$$ and $$\begin{aligned} \frac{1}{2}\left(\frac{2}{7} + \frac{1}{3}\right) &= \frac{13}{42} \\ &= 0.30952\ldots. \end{aligned}$$

The four-place decimals in that interval are $0.2679, 0.2680, \ldots, 0.3095,$ and there are $3095 - 2679 + 1 = 417$ of them.

Small Hint:

The $n$-gon’s interior angle, the $60^\circ$ angle, and $\angle BA_1A_n$ together fill $360^\circ$ at $A_1$

Big Hint:

Setting $\angle BA_1A_n$ equal to the interior angle of a regular $m$-gon gives $\frac{1}{m} = \frac{1}{6} - \frac{1}{n},$ so $n - 6$ must divide $36$

Solution:

Since $B$ is outside the $n$-gon, the angles at $A_1$ — the interior angle $\angle A_nA_1A_2 = \frac{(n-2)180^\circ}{n},$ the equilateral angle $\angle A_2A_1B = 60^\circ,$ and $\angle BA_1A_n$ — fill a full revolution, so $$\begin{aligned} \angle BA_1A_n &= 300^\circ - \frac{(n-2)180^\circ}{n} \\ &= 120^\circ + \frac{360^\circ}{n}. \end{aligned}$$ Also $A_nA_1 = A_1B,$ since both equal the side of the $n$-gon.

For $A_n,$ $A_1,$ $B$ to be consecutive vertices of a regular $m$-gon, this angle must be the $m$-gon’s interior angle: $120^\circ + \frac{360^\circ}{n} = 180^\circ - \frac{360^\circ}{m},$ which simplifies to $\frac{1}{m} = \frac{1}{6} - \frac{1}{n},$ so $m = \frac{6n}{n - 6} = 6 + \frac{36}{n - 6}.$

Thus $n - 6$ must divide $36,$ and the largest choice is $n - 6 = 36,$ i.e. $n = 42$ (with $m = 7$).

Small Hint:

Set up coordinates: the car is at $\left(\frac{2t}{3}, 0\right)$ and the storm center at $\left(\frac{t}{2}, 110 - \frac{t}{2}\right)$

Big Hint:

Entering and leaving happen when the squared distance equals $51^2;$ that is a quadratic in $t,$ so use Vieta’s formulas for $t_1 + t_2$

Solution:

Put the car at the origin at $t = 0,$ with east as the positive $x$-direction and north as the positive $y$-direction. At time $t$ the car is at $\left(\frac{2t}{3}, 0\right),$ and the storm center, moving southeast at speed $\frac{\sqrt{2}}{2}$ (components $\frac{1}{2}$ east and $\frac{1}{2}$ south), is at $\left(\frac{t}{2}, 110 - \frac{t}{2}\right).$

The car is on the storm boundary when the squared distance is $51^2:$ $$\begin{aligned} &\left(\frac{2t}{3} - \frac{t}{2}\right)^2 \\ &\quad {}+ \left(110 - \frac{t}{2}\right)^2 = 51^2, \end{aligned}$$ that is $\frac{t^2}{36} + \frac{t^2}{4}$ $- 110t + 12100 - 2601 = 0,$ or $\frac{5}{18}t^2 - 110t + 9499 = 0.$

The roots are $t_1$ and $t_2,$ so by Vieta’s formulas $t_1 + t_2 = \frac{110 \cdot 18}{5} = 396,$ and $\frac{1}{2}(t_1 + t_2) = 198.$

Small Hint:

Record each row by the pair of columns holding its $1$’s, and split into cases by how row $2$ overlaps row $1$

Big Hint:

When rows $1$ and $2$ share $2,$ $1,$ or $0$ columns, count how many ways rows $3$ and $4$ can finish the columns

Solution:

Each row must contain two $1$’s and two $-1$’s, so identify each row with the pair of columns holding its $1$’s; each column must end up chosen by exactly two rows. There are $\binom{4}{2} = 6$ choices for row $1.$ Classify by how row $2$ overlaps row $1.$

If row $2$ uses the same pair ($1$ way), those two columns are full, so rows $3$ and $4$ must both use the complementary pair: $1$ completion. If row $2$ uses the complementary pair ($1$ way), every column has one $1$ so far, so rows $3$ and $4$ need only be a complementary pair themselves: $6$ choices for row $3,$ row $4$ forced, giving $6$ completions. If row $2$ shares exactly one column with row $1$ ($2 \cdot 2 = 4$ ways), one column is full, two have one $1,$ and one is empty; rows $3$ and $4$ must each take the empty column together with one of the two half-filled columns, so there are $2$ completions.

The total is $6\,(1 \cdot 1 + 1 \cdot 6 + 4 \cdot 2) = 6 \cdot 15$ $= 90.$

Small Hint:

Since $2 \lt a^2 \lt 3$ and $0 \lt a^{-1} \lt 1,$ the condition says $a^{-1} = a^2 - 2$

Big Hint:

The cubic $a^3 - 2a - 1 = 0$ factors with root $a = \frac{1 + \sqrt{5}}{2};$ use $a^2 = a + 1$ repeatedly to reduce $a^{12}$

Solution:

From $2 \lt a^2 \lt 3$ we get $\sqrt{2} \lt a \lt \sqrt{3},$ so $0 \lt a^{-1} \lt 1$ and $\langle a^{-1}\rangle = a^{-1},$ while $\langle a^2\rangle = a^2 - 2.$ The condition becomes $a^{-1} = a^2 - 2,$ i.e. $a^3 - 2a - 1 = 0,$ which factors as $$(a + 1)(a^2 - a - 1) = 0.$$ Since $a \gt 0,$ we get $a = \frac{1 + \sqrt{5}}{2},$ the golden ratio, and indeed $a^2 = a + 1 \approx 2.618$ lies in $(2, 3).$

Using $a^2 = a + 1$ repeatedly: $a^4 = (a+1)^2 = 3a + 2,$ $a^8 = (3a+2)^2$ $= 9(a+1) + 12a + 4$ $= 21a + 13,$ and $a^{12} = a^8 a^4$ $= (21a + 13)(3a + 2)$ $= 63(a+1) + 81a + 26$ $= 144a + 89.$ Also $a^{-1} = a - 1$ from $a^2 = a + 1.$

Therefore $a^{12} - 144a^{-1}$ $= 144a + 89 - 144(a - 1)$ $= 89 + 144 = 233.$

Small Hint:

Check that any two distinct cards are completed to a complementary set by exactly one third card

Big Hint:

Count pairs of cards, then figure out how many different pairs give rise to the same three-card set

Solution:

Given any two distinct cards, there is exactly one card completing them to a complementary set: in each attribute, if the two cards agree, the third card must share that value, and if they differ, the third must take the one remaining value. The completing card is distinct from both (the two given cards differ somewhere, and in that attribute the third card differs from each).

So the $\binom{27}{2} = 351$ pairs of cards each extend to one complementary set, and each complementary set is produced by $\binom{3}{2} = 3$ of these pairs. The number of sets is $\frac{351}{3} = 117.$

Small Hint:

Multiply numerator and denominator by $2\sin\frac{1}{2}^\circ$ and telescope using product-to-sum identities

Big Hint:

Sum-to-product turns the result into $\cot 22.5^\circ,$ which the half-angle formula evaluates exactly

Solution:

Multiply numerator and denominator by $2\sin\frac{1}{2}^\circ.$ Since $2\cos n^\circ \sin\frac{1}{2}^\circ$ $= \sin\left(n + \frac{1}{2}\right)^\circ$ $- \sin\left(n - \frac{1}{2}\right)^\circ$ and $2\sin n^\circ \sin\frac{1}{2}^\circ$ $= \cos\left(n - \frac{1}{2}\right)^\circ$ $- \cos\left(n + \frac{1}{2}\right)^\circ,$ both sums telescope: $$\begin{aligned} x &= \frac{\sin 44.5^\circ - \sin 0.5^\circ}{\cos 0.5^\circ - \cos 44.5^\circ} \\ &= \frac{2\cos 22.5^\circ \sin 22^\circ}{2\sin 22.5^\circ \sin 22^\circ} \\ &= \cot 22.5^\circ, \end{aligned}$$ using the sum-to-product identities in the last step.

By the half-angle formula, $\cot 22.5^\circ = \frac{1 + \cos 45^\circ}{\sin 45^\circ} = \sqrt{2} + 1.$ Since $1.41^2 \lt 2 \lt 1.42^2,$ we have $241 \lt 100\sqrt{2} + 100 \lt 242.$ Therefore the greatest integer not exceeding $100x$ is $241.$

Small Hint:

$f(f(x)) = x$ means $f$ is its own inverse; comparing coefficients forces $d = -a$

Big Hint:

$19$ and $97$ are the roots of $cx^2 + (d - a)x - b = 0,$ and the missing value is the horizontal asymptote $\frac{a}{c}$

Solution:

Composing, $f(f(x)) = \frac{(a^2 + bc)x + b(a + d)}{c(a + d)x + (bc + d^2)},$ and this equals $x$ identically only if $c(a + d) = 0.$ Since $c \ne 0,$ we get $d = -a,$ so $f(x) = \frac{ax + b}{cx - a}.$

A fixed point satisfies $cx^2 - ax = ax + b,$ i.e. $cx^2 - 2ax - b = 0,$ whose roots are $19$ and $97.$ By Vieta’s formulas, $19 + 97 = \frac{2a}{c},$ so $\frac{a}{c} = 58.$

Finally, $y$ is in the range exactly when $y = \frac{ax + b}{cx - a}$ has a solution, i.e. $x(cy - a) = ay + b.$ This solves for $x$ unless $cy - a = 0;$ and when $y = \frac{a}{c},$ the right side $\frac{a^2}{c} + b = \frac{a^2 + bc}{c}$ is nonzero (otherwise the fractional-linear expression would be degenerate, contrary to its two distinct fixed points). So the unique number not in the range is $\frac{a}{c} = 58.$

Small Hint:

For $0 \le x \le 2$ the expression $\bigl||x| - 2\bigr| - 1$ equals $1 - x,$ and for $2 \le x \le 4$ it equals $x - 3$

Big Hint:

The graph is a union of diamonds $|x - a| + |y - b| = 1;$ count the possible centers and multiply by one diamond’s perimeter

Solution:

Let $f(t) = \bigl|\,||t| - 2| - 1\,\bigr|,$ so the equation is $f(x) + f(y) = 1.$ The function $f$ is even, and for $t \ge 0:$ on $[0, 2],$ $||t| - 2| - 1 = (2 - t) - 1$ $= 1 - t,$ so $f(t) = |t - 1|;$ on $[2, 4],$ $f(t) = |t - 3|;$ and for $t \gt 4,$ $f(t) = t - 3 \gt 1,$ which is too large. So on the relevant range, $f(t) = |t - a|$ where $a \in \{-3, -1, 1, 3\}$ is the nearest of those four values to $t.$

Therefore $S$ is the union of the $16$ taxicab circles $$\begin{aligned} &|x - a| + |y - b| = 1, \\ &\qquad a, b \in \{-3, -1, 1, 3\}, \end{aligned}$$ which meet only at isolated points. Each is a square (diamond) with diagonal $2,$ hence side $\sqrt{2}$ and perimeter $4\sqrt{2}.$

The total length is $16 \cdot 4\sqrt{2} = 64\sqrt{2},$ so $a + b = 64 + 2 = 66.$

Small Hint:

Fix $v;$ if $w = v e^{\frac{2\pi i k}{1997}},$ then $|v + w| = 2\left|\cos\frac{\pi k}{1997}\right|$

Big Hint:

Since $\sqrt{2 + \sqrt{3}} = 2\cos 15^\circ,$ count $k$ with $\frac{k}{1997} \le \frac{1}{12}$ or $\frac{k}{1997} \ge \frac{11}{12}$