Solution:

Since $24 = 2^3 \cdot 3,$ a square $N^2$ is a multiple of $24$ exactly when $N$ is a multiple of $12:$ the factor $2^3$ forces $N$ to contain $2^2,$ the factor $3$ forces $N$ to contain $3,$ and conversely $(12m)^2 = 144m^2$ is always a multiple of $24.$ Also $N^2 \lt 10^6$ exactly when $N \lt 1000.$

The multiples of $12$ less than $1000$ are $12, 24, \ldots, 996,$ and there are $\frac{996}{12} = 83$ of them.

Solution:

Measure time $t$ in seconds from when Al steps on. Al stands on the walkway, so he is at $6t;$ Bob moves at $6 + 4 = 10$ feet per second, so he is at $10(t - 2);$ Cy walks beside the walkway at $8$ feet per second, so he is at $8(t - 4).$ All three are moving once $t \ge 4.$

The middle person's position doubled must equal the sum of the other two. If Bob were in the middle, $20(t-2) = 6t + 8(t-4)$ gives $t = \frac{4}{3} \lt 4,$ impossible. If Cy were in the middle, $16(t-4) = 6t + 10(t-2)$ reduces to $-64 = -20,$ with no solution. If Al is in the middle, $12t = 10(t-2) + 8(t-4) = 18t - 52,$ so $t = \frac{26}{3}.$

At that moment Al is at $6 \cdot \frac{26}{3} = 52$ feet, while Bob and Cy are at $\frac{200}{3}$ and $\frac{112}{3},$ whose average is indeed $52.$ The middle person is $52$ feet from the start.

Solution:

By the binomial theorem, $z^2 = (81 - b^2) + 18bi$ and $z^3 = (729 - 27b^2) + (243b - b^3)i.$ Setting the imaginary parts equal gives $18b = 243b - b^3.$

Since $b$ is positive we may divide by $b,$ leaving $b^2 = 243 - 18 = 225,$ so $b = 15.$

Solution:

All four bodies lie on one line exactly when every pair of planets is collinear with the star, i.e. when each pair's angular positions differ by a multiple of $180^\circ$ — half a revolution. In $n$ years the planets complete $\frac{n}{60},$ $\frac{n}{84},$ and $\frac{n}{140}$ revolutions, so the pairwise differences are $$\frac{n}{60} - \frac{n}{84} = \frac{n}{210}, \qquad \frac{n}{84} - \frac{n}{140} = \frac{n}{210}, \qquad \frac{n}{60} - \frac{n}{140} = \frac{n}{105}.$$

We need $\frac{n}{210}$ and $\frac{n}{105}$ to be multiples of $\frac{1}{2}.$ The first requires $n$ to be a multiple of $105,$ and any such $n$ makes $\frac{n}{105}$ an integer. The smallest positive choice is $n = 105.$

Solution:

Adding $9$ to $F$ adds exactly $5$ to $\frac{5}{9}(F - 32),$ hence $5$ to the rounded Celsius value, hence $9$ to the final Fahrenheit value. So $T$ returns to itself if and only if $T + 9$ does, and it suffices to check nine consecutive temperatures. Checking $32$ through $40:$ the final values are $32, 34, 34, 36, 36, 37, 37, 39, 39,$ so exactly the five temperatures $32, 34, 36, 37, 39$ survive.

The range from $32$ through $994$ contains $963 = 107 \cdot 9$ integers, contributing $107 \cdot 5 = 535$ survivors. The remaining $995, \ldots, 1000$ behave like $32, \ldots, 37,$ of which $32, 34, 36, 37$ survive, adding $4$ more.

The total is $535 + 4 = 539.$

Solution:

Split the journey at the landmarks $13$ and $26.$ From $0$ the frog climbs the multiples of $3$ and may jump to $13$ from any of $0, 3, 6, 9, 12,$ giving $5$ routes from $0$ to $13;$ likewise there are $5$ routes from $13$ to $26$ (jump to $26$ from $13, 15, 18, 21, 24$) and $5$ from $26$ to $39.$ To skip $13$ entirely the frog must take the multiple-of-$3$ option every time through $12 \to 15,$ then jump to $26$ from one of $15, 18, 21, 24:$ $4$ routes from $0$ to $26$ avoiding $13.$ Similarly there are $4$ routes from $13$ to $39$ avoiding $26,$ and $4$ from $0$ to $39$ avoiding both.

Combining the segments: through both landmarks, $5 \cdot 5 \cdot 5 = 125;$ through $13$ only, $5 \cdot 4 = 20;$ through $26$ only, $4 \cdot 5 = 20;$ through neither, $4.$ The total is $125 + 20 + 20 + 4 = 169.$

Solution:

The difference $\lceil x \rceil - \lfloor x \rfloor$ equals $1$ when $x$ is not an integer and $0$ when it is. Now $\log_{\sqrt{2}} k$ is an integer exactly when $k = (\sqrt{2})^j$ for some integer $j,$ and for $k$ to be an integer, $j$ must be even — that is, $k$ must be a power of $2.$ The powers at most $1000$ are $2^0, 2^1, \ldots, 2^9 = 512.$

Therefore $$N = \sum_{k=1}^{1000} k - \sum_{j=0}^{9} 2^j = \frac{1000 \cdot 1001}{2} - 1023 = 500500 - 1023 = 499477,$$ and the remainder upon division by $1000$ is $477.$

Solution:

If $Q_1$ and $Q_2$ had no common root, their product — of degree $4$ — would divide the cubic $P(x),$ which is impossible. So they share a root $r,$ and $2Q_1(r) - Q_2(r) = 0.$ Computing, $2Q_1(x) - Q_2(x) = -15x - 3k,$ so $r = -\frac{k}{5}.$

Substituting into $Q_1(r) = 0$ gives $\frac{k^2}{25} - (k - 29)\frac{k}{5} - k = 0;$ multiplying by $25$ and simplifying yields $-4k^2 + 120k = 0,$ so $k = 0$ or $k = 30.$

For $k = 30,$ $Q_1(x) = x^2 + x - 30 = (x + 6)(x - 5)$ and $Q_2(x) = 2x^2 + 17x + 30 = (x + 6)(2x + 5),$ and both divide $P(x) = (x + 6)(x - 5)(2x + 5).$ The largest value is $30.$

Solution:

The hypotenuse is $AB = \sqrt{30^2 + 16^2} = 34.$ Let $T_1$ and $T_2$ be the points where the circles touch $AB.$ Both centers lie at distance $r$ from line $AB$ on the side away from the triangle, so $\overline{O_1 O_2}$ is parallel to $AB$ and $T_1 T_2 = O_1 O_2 = 2r,$ since the circles are externally tangent. Thus $AB = AT_1 + 2r + T_2 B.$

Circle $O_1$ is inscribed in the angle at $A$ between ray $AB$ and the extension of $\overline{CA}$ beyond $A,$ which measures $180^\circ - \angle A.$ Its tangent length from $A$ is therefore $AT_1 = r\big/\tan\left(90^\circ - \tfrac{A}{2}\right) = r \tan\frac{A}{2}.$ With $\sin A = \frac{16}{34}$ and $\cos A = \frac{30}{34},$ the half-angle formula gives $\tan\frac{A}{2} = \frac{\sin A}{1 + \cos A} = \frac{16}{64} = \frac{1}{4},$ and similarly $\tan\frac{B}{2} = \frac{30}{34 + 16} = \frac{3}{5}.$

So $34 = \frac{r}{4} + 2r + \frac{3r}{5} = \frac{57r}{20},$ giving $r = \frac{680}{57}.$ Since $680 = 2^3 \cdot 5 \cdot 17$ and $57 = 3 \cdot 19$ share no common factor, $p + q = 680 + 57 = 737.$

Solution:

Shade three of the six rows in column $1:$ $\binom{6}{3} = 20$ ways. Let $k$ be the number of rows shaded in both columns $1$ and $2;$ column $2$ can then be chosen in $\binom{3}{k}\binom{3}{3-k}$ ways. After these two columns, $k$ rows are complete with two shaded squares, $6 - 2k$ rows have one, and $k$ rows have none.

The empty rows must be shaded in both columns $3$ and $4.$ Column $3$ takes those $k$ rows plus $3 - k$ of the $6 - 2k$ singly-shaded rows, in $\binom{6-2k}{3-k}$ ways, and column $4$ is then forced: it must cover the empty rows and exactly the singly-shaded rows skipped by column $3.$

Summing, $$N = 20\sum_{k=0}^{3} \binom{3}{k}\binom{3}{3-k}\binom{6-2k}{3-k} = 20(20 + 54 + 18 + 1) = 1860,$$ so the remainder is $860.$

Solution:

For a positive integer $k,$ the condition $|k - \sqrt{p}| \lt \frac{1}{2}$ means $\left(k - \frac{1}{2}\right)^2 \lt p \lt \left(k + \frac{1}{2}\right)^2,$ which for integers $p$ is exactly $k^2 - k + 1 \le p \le k^2 + k.$ So $b(p) = k$ for precisely $2k$ values of $p.$

Since $44^2 + 44 = 1980,$ the blocks $k = 1, \ldots, 44$ exactly cover $p \le 1980$ and contribute $$\sum_{k=1}^{44} k \cdot 2k = 2 \cdot \frac{44 \cdot 45 \cdot 89}{6} = 58740.$$ The remaining $27$ values $p = 1981, \ldots, 2007$ each have $b(p) = 45,$ adding $27 \cdot 45 = 1215.$

Thus $S = 58740 + 1215 = 59955,$ and the remainder is $955.$

Solution:

Since $AC$ makes a $75^\circ$ angle with the positive $x$-axis, the rotation is by $15^\circ.$ Let $B'$ and $C'$ be the images of $B$ and $C.$ Because $\angle B'AB = 15^\circ$ and $\angle ABC = 75^\circ,$ segment $AB'$ is perpendicular to $BC;$ let $D$ be their intersection, and let $E = BC \cap B'C'$ and $F = AC \cap B'C'.$ The common region is the quadrilateral $ADEF,$ whose area is $[AB'F] - [EB'D].$

In triangle $AB'F,$ $\angle FAB' = 75^\circ - 15^\circ = 60^\circ$ and $\angle AB'F = 75^\circ,$ so $\angle AFB' = 45^\circ,$ and the law of sines gives $B'F = 20\sin 60^\circ/\sin 45^\circ = 10\sqrt{6}.$ With $\sin 75^\circ = \frac{\sqrt{6} + \sqrt{2}}{4},$ $$[AB'F] = \tfrac{1}{2} \cdot 20 \cdot 10\sqrt{6}\,\sin 75^\circ = 50(3 + \sqrt{3}).$$

In right triangle $ABD,$ $AD = 20\cos 15^\circ$ and $BD = 20\sin 15^\circ,$ so $[ABD] = 200\sin 15^\circ\cos 15^\circ = 100\sin 30^\circ = 50,$ and $B'D = 20(1 - \cos 15^\circ).$ Triangles $EB'D$ and $ABD$ are similar (right angles at $D,$ and $\angle EB'D = \angle ABD = 75^\circ$), so, using $\cos 15^\circ = \frac{\sqrt{6} + \sqrt{2}}{4},$ $$[EB'D] = 50\left(\frac{1 - \cos 15^\circ}{\sin 15^\circ}\right)^2 = 50\left(15 + 8\sqrt{3} - 6\sqrt{6} - 10\sqrt{2}\right).$$ Therefore $[ADEF] = 50(3 + \sqrt{3}) - 50(15 + 8\sqrt{3} - 6\sqrt{6} - 10\sqrt{2}) = 500\sqrt{2} - 350\sqrt{3} + 300\sqrt{6} - 600,$ so $(p, q, r, s) = (500, -350, 300, -600)$ and $\frac{p - q + r - s}{2} = \frac{1750}{2} = 875.$

Solution:

Place the base at $A = (0,0,0),$ $B = (4,0,0),$ $C = (4,4,0),$ $D = (0,4,0);$ the apex is then $E = (2, 2, 2\sqrt{2}),$ since $2^2 + 2^2 + 8 = 16.$ The given midpoints are $R = (1, 1, \sqrt{2}),$ $S = (4, 2, 0),$ and $T = (2, 4, 0),$ and all three satisfy $x + y + 2\sqrt{2}\,z = 6,$ the equation of the cutting plane.

Parametrizing edges $\overline{BE}$ and $\overline{DE}$ shows the plane meets them at $U = \left(\frac{7}{2}, \frac{1}{2}, \frac{\sqrt{2}}{2}\right)$ and $V = \left(\frac{1}{2}, \frac{7}{2}, \frac{\sqrt{2}}{2}\right).$ The cross-section is the pentagon $RUSTV$ with $RU = RV = \sqrt{7},$ $US = VT = \sqrt{3},$ $ST = 2\sqrt{2},$ and diagonal $UV = 3\sqrt{2}.$

Split the pentagon along $\overline{UV}.$ Isosceles triangle $RUV$ has height $\sqrt{7 - \frac{9}{2}} = \sqrt{\frac{5}{2}}$ and area $\frac{1}{2} \cdot 3\sqrt{2} \cdot \sqrt{\frac{5}{2}} = \frac{3\sqrt{5}}{2}.$ Isosceles trapezoid $USTV$ has height $\sqrt{3 - \frac{1}{2}} = \sqrt{\frac{5}{2}}$ and area $\frac{1}{2}(3\sqrt{2} + 2\sqrt{2})\sqrt{\frac{5}{2}} = \frac{5\sqrt{5}}{2}.$ The total is $4\sqrt{5} = \sqrt{80},$ so $p = 80.$

Solution:

For $n \ge 3,$ both $a_{n+1}a_{n-1} = a_n^2 + 2007$ and $a_n a_{n-2} = a_{n-1}^2 + 2007$ hold. Subtracting and regrouping gives $a_{n-1}(a_{n+1} + a_{n-1}) = a_n(a_n + a_{n-2}),$ so $\frac{a_{n+1} + a_{n-1}}{a_n}$ has the same value for every $n \ge 2.$ Since $a_3 = \frac{3^2 + 2007}{3} = 672,$ that value is $\frac{672 + 3}{3} = 225,$ and the sequence satisfies $a_{n+1} = 225a_n - a_{n-1}.$

Multiplying $a_{n+1} + a_{n-1} = 225a_n$ by $a_{n+1}$ and substituting $a_{n+1}a_{n-1} = a_n^2 + 2007$ yields $a_{n+1}^2 + a_n^2 + 2007 = 225\,a_n a_{n+1},$ so $$\frac{a_{n+1}^2 + a_n^2}{a_{n+1}a_n} = 225 - \frac{2007}{a_n a_{n+1}}.$$

The sequence increases: $a_3 = 672 \gt a_2,$ and $a_{n+1} = 225a_n - a_{n-1} \gt a_n$ whenever $a_n \gt a_{n-1}.$ Hence $a_{2006}a_{2007} \gt 672^2 \gt 2007,$ so the fraction lies strictly between $224$ and $225,$ and the answer is $224.$

Solution:

Let $s = AB$ and $t = AE.$ Using the $60^\circ$ angles at $A,$ $B,$ $C$ and the area formula $\frac{1}{2}xy\sin 60^\circ:$ $[AEF] = \frac{\sqrt{3}}{4} \cdot 5t,$ $[BFD] = \frac{\sqrt{3}}{4}(s-5)(s-2),$ and $[CDE] = \frac{\sqrt{3}}{4} \cdot 2(s-t).$ Subtracting all three from $[ABC] = \frac{\sqrt{3}}{4}s^2$ and simplifying, $$[DEF] = \frac{\sqrt{3}}{4}\bigl(5(s - t) + 2t - 10\bigr) = 14\sqrt{3},$$ so $5(s - t) + 2t = 66.$

At $E,$ the angles $\angle AEF$ and $\angle CED$ sum to $180^\circ - 60^\circ = 120^\circ,$ while in triangle $AEF$ the angles $\angle AEF$ and $\angle AFE$ also sum to $120^\circ.$ Hence $\angle AFE = \angle CED,$ and since $\angle A = \angle C = 60^\circ,$ triangles $AEF$ and $CDE$ are similar. Then $\frac{AE}{AF} = \frac{CD}{CE}$ gives $\frac{t}{5} = \frac{2}{s - t},$ so $t(s - t) = 10.$

Substituting $s - t = \frac{10}{t}$ into $5(s - t) + 2t = 66$ gives $\frac{50}{t} + 2t = 66,$ or $t^2 - 33t + 25 = 0,$ so $t = \frac{33 \pm \sqrt{989}}{2}.$ From $\frac{25}{t} = 33 - t$ we get $\frac{10}{t} = \frac{2}{5}(33 - t),$ so $s = t + \frac{10}{t} = \frac{3t + 66}{5} = \frac{231 \pm 3\sqrt{989}}{10}.$ Both values yield valid configurations, so $r = 989.$