Answer: 277

Solution:

Let $v$ be Patrick's speed in miles per hour and $T$ his travel time in hours. Then Tanya travels for $T - 1$ hours at speed $v + 2,$ and José travels for $T - 2$ hours at speed $v + 9$ (which is $7$ more than Tanya's speed). Since all three cover the same distance, $$vT = (v+2)(T-1) = (v+9)(T-2).$$

Expanding the first equality gives $0 = 2T - v - 2,$ so $v = 2T - 2.$ Expanding the second gives $0 = 9T - 2v - 18,$ so $2v = 9T - 18.$ Substituting, $4T - 4 = 9T - 18,$ hence $T = \frac{14}{5}$ and $v = \frac{18}{5}.$

The distance is $vT = \frac{18}{5} \cdot \frac{14}{5} = \frac{252}{25},$ which is in lowest terms, so $m + n = 252 + 25 = 277.$

Answer: 62

Solution:

A palindrome with an even number of digits has each digit appearing in a mirrored pair, so its digit sum is even. Since $13$ is odd, the palindrome has an odd number of digits, and if $m$ is the middle digit, the rest of the digit sum $13 - m$ is split evenly between the two halves, so $m$ is odd. A one-digit palindrome would need $m = 13,$ which is impossible.

The palindrome is determined by its middle digit $m$ and the block of digits to the left of center: a nonempty string of nonzero digits with sum $s = \frac{13 - m}{2}.$ For $m = 1, 3, 5, 7, 9$ we get $s = 6, 5, 4, 3, 2.$ Since $s \le 6,$ every digit of such a string is automatically at most $9,$ so the number of strings is the number of compositions of $s,$ which is $2^{s-1}$ (each of the $s - 1$ gaps between units is either a break or not).

The total is $2^{5} + 2^{4} + 2^{3} + 2^{2} + 2^{1} = 32 + 16 + 8 + 4 + 2 = 62.$

Answer: 79

Solution:

A sphere of radius $42$ resting on the disk at $P$ has its center $42$ directly above $P.$ It lies inside the hemisphere of radius $200$ exactly when its center is within $200 - 42 = 158$ of the common center $O.$ If $d$ is the distance from $O$ to $P,$ the center of the sphere is at distance $\sqrt{d^2 + 42^2}$ from $O,$ so the condition is $d^2 + 42^2 \le 158^2.$

By difference of squares, $d^2 \le 158^2 - 42^2 = 116 \cdot 200 = 23200.$ Thus $\mathcal{T}$ is a disk of radius $\sqrt{23200},$ and the ratio of areas is $$\frac{23200}{200^2} = \frac{23200}{40000} = \frac{29}{50}.$$ Therefore $p + q = 29 + 50 = 79.$

Answer: 70

Solution:

Since $a + b + ab = (a+1)(b+1) - 1,$ an integer $n$ is representable exactly when $n + 1 = xy$ for distinct integers $x = a + 1$ and $y = b + 1$ that are each at least $2.$ So we count integers $n + 1$ in $\{2, 3, \ldots, 101\}$ that admit such a factorization.

A prime has no factorization into two factors that are both at least $2,$ and the square of a prime $p^2$ factors that way only as $p \cdot p,$ which is not allowed. Every other composite $M$ works: if $p$ is its smallest prime factor, then $M = p \cdot \frac{M}{p}$ with $\frac{M}{p} \gt p$ since $M \gt p^2.$ In $\{2, \ldots, 101\}$ there are $26$ primes (the $25$ primes below $100,$ together with $101$) and $4$ prime squares ($4,$ $9,$ $25,$ $49$).

The count is $100 - 26 - 4 = 70.$

Answer: 65

Solution:

Work in the complex plane with $B = 0$ and $A = 1.$ Rotating $z$ about $P$ through angle $\varphi$ counterclockwise gives $P + e^{i\varphi}(z - P).$ So $A' = e^{i\theta},$ and rotating $B$ clockwise through $\theta$ about $A'$ gives $$B' = A' + e^{-i\theta}(0 - A') = e^{i\theta} - e^{-i\theta}e^{i\theta} = e^{i\theta} - 1.$$

Then $$AB'^2 = \left|e^{i\theta} - 2\right|^2 = (\cos\theta - 2)^2 + \sin^2\theta = 5 - 4\cos\theta.$$ Setting this equal to $\left(\frac{4}{3}\right)^2 = \frac{16}{9}$ gives $4\cos\theta = 5 - \frac{16}{9} = \frac{29}{9},$ so $\cos\theta = \frac{29}{36}$ (indeed positive, consistent with $\theta$ acute). Thus $m + n = 29 + 36 = 65.$

Answer: 441

Solution:

Let $t = \log_{2026} x.$ Taking $\log_{2026}$ of both sides of $x^{(\log_{2026} x)/20} = 26x$ gives $$\frac{t^2}{20} = \log_{2026} 26 + t, \qquad \text{that is} \qquad t^2 - 20t - 20\log_{2026} 26 = 0.$$ The discriminant $400 + 80\log_{2026} 26$ is positive, so there are two real roots $t_1, t_2,$ each giving a valid positive solution $x = 2026^{t}.$

By Vieta's formulas $t_1 + t_2 = 20,$ so the product of the solutions is $2026^{t_1} \cdot 2026^{t_2} = 2026^{20}.$ Since $2026 = 2 \cdot 1013$ and $1013$ is prime, $P = 2^{20} \cdot 1013^{20}$ has $21 \cdot 21 = 441$ positive divisors.

Answer: 396

Solution:

A function from a finite set onto itself is a bijection, so $\pi$ is a permutation of six elements, and the condition says $\pi^6$ is the identity. A permutation satisfies $\pi^6 = \mathrm{id}$ exactly when every cycle in its cycle decomposition has length dividing $6.$ Among the possible lengths $1$ through $6,$ only $4$ and $5$ fail to divide $6.$

We subtract the permutations containing a $4$-cycle or a $5$-cycle from $6! = 720.$ Cycle type $4+1+1$ gives $\frac{6!}{4 \cdot 2!} = 90,$ type $4+2$ gives $\frac{6!}{4 \cdot 2} = 90,$ and type $5+1$ gives $\frac{6!}{5} = 144,$ for $90 + 90 + 144 = 324$ excluded permutations.

The count is $720 - 324 = 396.$

Answer: 244

Solution:

Since $17017 = 7 \cdot 11 \cdot 13 \cdot 17,$ the divisors of $17017^{17}$ are $7^a 11^b 13^c 17^d$ with each exponent between $0$ and $17.$ Modulo $12$ we have $13 \equiv 1,$ and $7^2 \equiv 11^2 \equiv 17^2 \equiv 1$ (as $17 \equiv 5$), so the residue of a divisor is $7^{\alpha} \, 11^{\beta} \, 5^{\delta} \pmod{12},$ where $\alpha, \beta, \delta$ are the parities of $a, b, d.$

The four possible values $1, 5, 7, 11$ multiply like the group $\{1, 5, 7, 11\}$ mod $12,$ in which $7 \cdot 11 \equiv 5.$ Checking the eight parity patterns, the residue is $5$ exactly when $(\alpha, \beta, \delta) = (0, 0, 1)$ or $(1, 1, 0).$ Each parity condition is satisfied by $9$ of the $18$ choices of that exponent, while $c$ is free with $18$ choices.

Therefore $N = 2 \cdot 9 \cdot 9 \cdot 9 \cdot 18 = 26244,$ and the remainder mod $1000$ is $244.$

Answer: 29

Solution:

Let $f_1, \ldots, f_6$ be the top faces rolled, independent and uniform over the six faces. Sticker $i$ goes on face $f_i$ and ends up visible exactly when $f_j \ne f_i$ for all $j \gt i$ (sticker $6$ is always visible). So the conditioning event is $f_3, f_4, f_5, f_6 \ne f_2$ and $f_5, f_6 \ne f_4.$ Counting choices in the order $f_1, f_2, f_3, f_4, f_5, f_6$ gives $6 \cdot 6 \cdot 5 \cdot 5 \cdot 4 \cdot 4 = 14400$ sequences out of $6^6.$

A face is blank exactly when it never appears among $f_1, \ldots, f_6,$ so exactly one blank face means the sequence takes exactly $5$ distinct values, i.e. there is exactly one coincidence $f_i = f_j$ with $i \lt j$ and all other values distinct. The coincidence must not violate the conditioning: pairs $(2, j)$ and $(4, 5),$ $(4, 6)$ are forbidden, leaving the $9$ pairs $(1,2),$ $(1,3),$ $(1,4),$ $(1,5),$ $(1,6),$ $(3,4),$ $(3,5),$ $(3,6),$ $(5,6).$ For each allowed pair, the five distinct values can be assigned in $6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 = 720$ ways, and every constraint holds automatically because the only repeated value occupies an allowed pair. That gives $9 \cdot 720 = 6480$ sequences.

Therefore $p = \frac{6480}{14400} = \frac{9}{20},$ and $m + n = 9 + 20 = 29.$

Answer: 156

Solution:

Place $B = (0,0),$ $C = (14,0),$ $A = (5,12).$ The circumcenter lies on $x = 7,$ and equating distances to $B$ and $A$ gives $O = \left(7, \frac{33}{8}\right).$ The direction of $\overline{AC}$ is $C - A = (9, -12),$ parallel to $(3, -4).$ A rotation through $\varphi$ makes $\overline{A'C'}$ vertical exactly when it sends $(3,-4)$ to $(0, \pm 5),$ so $(\cos\varphi, \sin\varphi) = \left(\frac{4}{5}, -\frac{3}{5}\right)$ or $\left(-\frac{4}{5}, \frac{3}{5}\right).$ Rotating each vertex about $O$ and checking the line $B'C'$ shows that $A'$ and $B$ are on opposite sides only for $\cos\varphi = \frac{4}{5},$ $\sin\varphi = -\frac{3}{5}.$

With this rotation, $P' = O + R(P - O)$ gives $$A' = \left(\tfrac{81}{8}, \tfrac{93}{8}\right), \qquad B' = \left(-\tfrac{43}{40}, \tfrac{201}{40}\right), \qquad C' = \left(\tfrac{81}{8}, -\tfrac{27}{8}\right).$$ For example, $A - O = \left(-2, \tfrac{63}{8}\right)$ rotates to $\left(\tfrac{25}{8}, \tfrac{15}{2}\right),$ giving $A' = \left(\tfrac{81}{8}, \tfrac{93}{8}\right).$

The hexagon $A A' C C' B B'$ is simple with these vertices in order, so the shoelace formula on $(5,12),$ $\left(\tfrac{81}{8}, \tfrac{93}{8}\right),$ $(14,0),$ $\left(\tfrac{81}{8}, -\tfrac{27}{8}\right),$ $(0,0),$ $\left(-\tfrac{43}{40}, \tfrac{201}{40}\right)$ gives area $\frac{1557}{10} = 155.7.$ The closest integer is $156.$

Answer: 896

Solution:

View the grid as a graph whose $112$ edges join adjacent cells. Each edge contributes its larger endpoint value positively and its smaller one negatively, so $M = \sum_v c_v a_v,$ where $a_v$ is the entry in cell $v$ and $c_v$ is the number of neighbors of $v$ with smaller entries minus the number with larger entries. Then $|c_v| \le \deg(v),$ which is $4$ for the $36$ interior cells, $3$ for the $24$ edge cells, and $2$ for the $4$ corners, and $\sum_v c_v = 0$ since each edge contributes $+1$ and $-1.$

Because $\sum_v c_v = 0,$ $$M = \sum_v c_v \left(a_v - \tfrac{65}{2}\right) \le \sum_v \deg(v)\left|a_v - \tfrac{65}{2}\right|.$$ By the rearrangement inequality this is maximized by pairing the $36$ values farthest from $\frac{65}{2}$ (namely $1$–$18$ and $47$–$64,$ whose deviations total $828$) with the interior cells, the next $24$ values ($19$–$30$ and $35$–$46,$ totaling $192$) with the edge cells, and $31$–$34$ (totaling $4$) with the corners. Hence $M \le 4 \cdot 828 + 3 \cdot 192 + 2 \cdot 4 = 3896.$

Equality requires every cell holding a value at most $32$ to be smaller than all its neighbors and every value at least $33$ to be larger, which a checkerboard achieves: put $1$–$32$ on the black cells ($1$–$18$ on interior blacks, $19$–$30$ on edge blacks, $31$–$32$ on black corners) and $33$–$64$ on the white cells ($33$–$34$ on corners, $35$–$46$ on edges, $47$–$64$ in the interior). Every neighbor pair then compares white over black, so $M = 3896,$ and the answer is $3896 \bmod 1000 = 896.$

Answer: 161

Solution:

A sphere of radius $\rho$ tangent to $\mathcal{P}$ at $P$ has center $P + \rho k,$ where $k$ is the upward unit normal of $\mathcal{P}.$ Write $\mathcal{T}$ as $\{x : n \cdot x = c\}$ with unit normal $n = (n_1, n_2, h)$ in coordinates where $\mathcal{P}$ is the $xy$-plane. Tangency with all spheres on the same side means $n \cdot (P + \rho k) - c = \rho$ for each sphere, that is $n_1 x_P + n_2 y_P - c = (1 - h)\rho.$ Here $h \ne 1,$ since otherwise the left side would be constant while the radii differ. So $\rho = g(P)$ for the affine function $g(x,y) = \frac{n_1 x + n_2 y - c}{1 - h}.$

Take $A = (0,0),$ $B = (6,0),$ $C = (0,4).$ The affine function with $g(A) = 1,$ $g(B) = 2,$ $g(C) = 3$ is $g(x,y) = 1 + \frac{x}{6} + \frac{y}{2}.$ The centroid is $G = \left(2, \frac{4}{3}\right),$ and line $BC$ is $2x + 3y = 12.$ Since $2 \cdot 2 + 3 \cdot \frac{4}{3} - 12 = -4,$ reflecting gives $$D = G + \frac{8}{13}\,(2, 3) = \left(\frac{42}{13},\ \frac{124}{39}\right).$$

Therefore $r = g(D) = 1 + \frac{1}{6} \cdot \frac{42}{13} + \frac{1}{2} \cdot \frac{124}{39} = 1 + \frac{21}{39} + \frac{62}{39} = \frac{122}{39}.$ (Such a plane exists: the normal condition $(1-h)^2\left|\nabla g\right|^2 = 1 - h^2$ with $\left|\nabla g\right|^2 = \frac{5}{18}$ gives $h = -\frac{13}{23}.$) Since $\gcd(122, 39) = 1,$ the answer is $122 + 39 = 161.$

Answer: 39

Solution:

Work in the ring $\mathbb{F}_{503}[x]/(x^{502} - 1).$ Reducing $(1 + x)^{10000} = \sum_k \binom{10000}{k} x^k$ replaces each exponent $k$ by $k \bmod 502,$ so $$(1+x)^{10000} \equiv \sum_{r=0}^{501} S_r \, x^r \pmod{503,\ x^{502} - 1}.$$

Since $503$ is prime, $(1+x)^{503} \equiv 1 + x^{503} \pmod{503},$ and $x^{503} = x \cdot x^{502} \equiv x,$ so $(1+x)^{503} \equiv 1 + x$ in this ring. Writing $10000 = 19 \cdot 503 + 443,$ $$(1+x)^{10000} = \left((1+x)^{503}\right)^{19} (1+x)^{443} \equiv (1+x)^{19} (1+x)^{443} = (1+x)^{462}.$$ As $462 \lt 502,$ no exponents fold, so $S_r \equiv \binom{462}{r} \pmod{503}$ for $0 \le r \le 501,$ where $\binom{462}{r} = 0$ for $r \gt 462.$

For $0 \le r \le 462$ the binomial coefficient $\binom{462}{r}$ is not divisible by $503:$ both $462$ and $r$ are single digits in base $503,$ so Lucas' theorem gives a nonzero value (indeed $\binom{462}{r} = \frac{462!}{r!\,(462-r)!}$ involves no factor of $503$). Hence $S_r \equiv 0 \pmod{503}$ exactly for $r = 463, 464, \ldots, 501,$ which is $501 - 463 + 1 = 39$ values.

Answer: 681

Solution:

In an equiangular pentagon each side direction turns by the exterior angle $72^\circ,$ so the sides are the vectors $s_k u_k$ for $k = 1, \ldots, 5,$ where $u_k = (\cos 72k^\circ, \sin 72k^\circ)$ and $\sum_k s_k u_k = 0.$ Write $Q = \sum s_k^2 = 308,$ $P_1 = \sum_{k} s_k s_{k+1},$ and $P_2 = \sum_k s_k s_{k+2}$ (indices cyclic). Each diagonal is a sum of two consecutive side vectors, so its square is $s_{k+1}^2 + s_{k+2}^2 + 2 s_{k+1} s_{k+2} \cos 72^\circ,$ and summing all five gives $$800 = 2Q + 2\cos 72^\circ \, P_1, \qquad \text{so} \qquad 2\cos 72^\circ \, P_1 = 800 - 616 = 184.$$

Expanding $\left|\sum_k s_k u_k\right|^2 = 0,$ the angle between $u_k$ and $u_{k+1}$ is $72^\circ$ and between $u_k$ and $u_{k+2}$ is $144^\circ:$ $$0 = Q + 2\cos 72^\circ \, P_1 + 2\cos 144^\circ \, P_2 = 308 + 184 - 2\cos 36^\circ \, P_2,$$ so $2\cos 36^\circ \, P_2 = 492.$ Using $\cos 72^\circ = \frac{\sqrt{5} - 1}{4}$ and $\cos 36^\circ = \frac{\sqrt{5} + 1}{4},$ we get $2P_1 = \frac{184}{\cos 72^\circ} = 184\left(\sqrt{5} + 1\right)$ and $2P_2 = \frac{492}{\cos 36^\circ} = 492\left(\sqrt{5} - 1\right).$

The square of the perimeter is $$\left(\sum s_k\right)^2 = Q + 2P_1 + 2P_2 = 308 + 184\sqrt{5} + 184 + 492\sqrt{5} - 492 = 676\sqrt{5}.$$ Therefore $m + n = 676 + 5 = 681.$

Answer: 83

Solution:

Since the five loops cover $\sum \left(2(a_i + b_i) - 4\right) = 100$ cells, $\sum (a_i + b_i) = 60.$ Every loop has an even number of cells, so no odd-by-odd rectangle can be exactly filled by loops; and filling a rectangle whose shortest even side is $e$ requires at least $\frac{e}{2}$ loops, since peeling off an outermost loop shrinks that side by exactly $2$ while splitting a rectangle into smaller ones only adds up such requirements. Now consider the outermost loops of a partition (those whose rectangles lie inside no other loop's rectangle): their rectangles tile the $10 \times 10$ square. If outermost rectangle $R_i$ has shortest even side $e_i,$ it uses $n_i \ge \frac{e_i}{2}$ loops and covers at most $10\,e_i$ cells. Summing over the tiling, $100 \le \sum 10\,e_i \le 20 \sum n_i = 100,$ so equality holds throughout: each $R_i$ spans the full $10$ in one direction, has even width $e_i,$ and is filled with exactly $\frac{e_i}{2}$ loops. Two full-length slabs in different directions would overlap, so the outermost rectangles are the whole square or parallel slabs, and the same equality argument repeats inside every loop's inner rectangle.

Let $s(w)$ be the number of ways to fill a full-height slab of even width $w$ with $\frac{w}{2}$ loops. A width-$2$ slab is a single loop: $s(2) = 1.$ A width-$4$ slab is a $10 \times 4$ loop around an $8 \times 2$ loop: $s(4) = 1.$ A width-$6$ slab is a $10 \times 6$ loop around an $8 \times 4$ region holding two loops — either nested ($8 \times 4$ around $6 \times 2$) or two $8 \times 2$ slabs — so $s(6) = 2.$ A width-$8$ slab surrounds an $8 \times 6$ region holding three loops: an $8 \times 6$ loop around a $6 \times 4$ region with two loops ($2$ ways as before), or full-height strips of widths $2 + 2 + 2$ ($1$ way), or widths $2 + 4$ in two orders ($2$ ways), so $s(8) = 5.$ The same recursion counts the full square: a $10 \times 10$ loop around an $8 \times 8$ region with four loops, where the $4 \times 4,$ $6 \times 6,$ and $8 \times 8$ regions admit $3,$ then $3 + 3 + 3 = 9,$ then $9 + 9 + 9 = 27$ fillings (single nested loop, vertical strips, or horizontal strips at each stage).

Finally, tally the outermost structures. The single $10 \times 10$ rectangle gives $27$ partitions. For parallel slabs, the widths form a composition of $10$ into even parts with at least two parts, and orientations (vertical or horizontal) double the count: $(2,2,2,2,2)$ gives $1;$ $(4,2,2,2)$ in $4$ orders gives $4;$ $(4,4,2)$ in $3$ orders gives $3;$ $(6,2,2)$ in $3$ orders gives $3 \cdot 2 = 6;$ $(6,4)$ in $2$ orders gives $2 \cdot 2 = 4;$ and $(8,2)$ in $2$ orders gives $2 \cdot 5 = 10,$ for $28$ per orientation. The total is $27 + 2 \cdot 28 = 83.$