Solution:

Let the common difference be $d.$ Since the first term is $4$ and both $24$ and $34$ appear, $d$ divides $24 - 4 = 20$ and $34 - 4 = 30,$ so $d$ divides $\gcd(20, 30) = 10.$ The difference must be positive to reach $24$ and $34$ from $4,$ so $d \in \{1, 2, 5, 10\}$ (and each of these works, since $d \mid 20$ and $d \mid 30$ put both targets in the sequence).

The $10$th term is $4 + 9d,$ so the requested sum is $$\sum_{d \in \{1,2,5,10\}} (4 + 9d) = 4 \cdot 4 + 9(1 + 2 + 5 + 10) = 16 + 162 = 178.$$

Solution:

Put $A = (0, 0)$ and $B = (10, 1).$ Every horizontal and diagonal move goes rightward, so the bug's $x$-coordinate never decreases, and it crosses each of the $10$ vertical strips exactly once, using exactly one of that square's three rightward segments: the bottom edge, the top edge, or the diagonal.

These ten choices determine the whole path. Each crossing arrives at a definite height (bottom edge: low; top edge or diagonal: high) and departs at a definite height (bottom edge or diagonal: low; top edge: high), so at each vertical line the bug traverses the vertical segment exactly when the arrival and departure heights differ — and each vertical segment is needed at most once, so no segment repeats. The same applies at the ends: the bug starts low at $A$ and finishes high at $B,$ using the end verticals if necessary. Conversely, every sequence of choices yields a valid path.

Therefore $N = 3^{10},$ and $\sqrt{N} = 3^5 = 243.$

Solution:

Place $A = (0, 0)$ and $E = (20, 0)$ with the pentagon above line $AE,$ and write $h = AB.$ The right angles at $A$ and $E$ make $AB$ and $ED$ vertical: $B = (0, h)$ and $D = (20, k)$ with $k = DE.$ At $B$ the side $BC = 14\sqrt{2}$ makes a $45^\circ$ angle with the downward ray $BA,$ heading into the pentagon, so $C = (14, h - 14).$ Similarly at $D,$ the side $DC$ makes a $45^\circ$ angle with the downward ray $DE,$ so $C = (20 - s, k - s)$ where $s = \frac{DC}{\sqrt{2}}.$ Matching coordinates gives $s = 6$ and $k = h - 8.$ The interior angle at $C$ is then the reflex angle $270^\circ$ (angle sum $90 + 45 + 270 + 45 + 90 = 540$), and $DE = h - 8 \lt AB$ automatically.

The shoelace formula on $A(0,0),$ $B(0,h),$ $C(14, h-14),$ $D(20, h-8),$ $E(20, 0)$ gives area $$\frac{1}{2}\left|{-14h} + (-6h + 168) + (-20h + 160)\right| = 20h - 164.$$ The condition $16 \mid 20h - 164$ reduces to $4h \equiv 4 \pmod{16},$ that is, $h \equiv 1 \pmod 4.$ For $C$ to lie strictly on the same side of line $AE$ as $B$ and $D,$ we need $h \gt 14.$

So $h$ runs over $17, 21, 25, \ldots, 2025,$ which is $\frac{2025 - 17}{4} + 1 = 503$ values.

Solution:

If $n$ has a single digit $d,$ then the numeral $d$ has value $d$ in every base, so $f(n) = n:$ all $9$ one-digit numbers work. If $n$ has digits $d_{k-1} \ldots d_1 d_0$ with $k \ge 2,$ then $b \le 10$ always, and if $b \lt 10$ then $f(n) = \sum d_i b^i \lt \sum d_i 10^i = n$ because the leading digit satisfies $d_{k-1} b^{k-1} \lt d_{k-1} 10^{k-1}.$ So a multi-digit $n$ satisfies $f(n) = n$ exactly when $b = 10,$ that is, when some digit of $n$ equals $9.$

Two-digit numbers containing a $9:$ the numbers $90$ through $99$ plus $19, 29, \ldots, 89,$ for $10 + 8 = 18.$ Three-digit numbers containing a $9:$ $900 - 8 \cdot 9 \cdot 9 = 252,$ subtracting the numbers with no $9$ (leading digit $1$–$8,$ others $0$–$8$).

The total is $9 + 18 + 252 = 279.$

Solution:

Say there are $r$ red and $b$ blue marbles, $r, b \ge 7.$ The condition is $\binom{r}{4}\binom{b}{3} = \binom{r}{5}\binom{b}{2}.$ Since $\binom{r}{5} = \binom{r}{4}\frac{r-4}{5}$ and $\binom{b}{3} = \binom{b}{2}\frac{b-2}{3},$ cancelling gives $$\frac{b - 2}{3} = \frac{r - 4}{5}, \qquad \text{that is,} \qquad b = \frac{3r - 2}{5}.$$

So $r \equiv 4 \pmod 5,$ and $b \ge 7$ requires $3r - 2 \ge 35,$ so $r \ge 14.$ The five smallest choices are $r = 14, 19, 24, 29, 34$ with $b = 8, 11, 14, 17, 20,$ giving $n = 22, 30, 38, 46, 54.$

The sum is $22 + 30 + 38 + 46 + 54 = 190.$

Solution:

Completing the square, $2y = (x - 4)^2 - 4,$ so with $u = x - 4$ the parabola is the set of points $\left(4 + u,\ \frac{u^2}{2} - 2\right)$ and the center $(4, 39)$ lies on its axis. The circle is tangent to the parabola at a point exactly when the two curves share a tangent line there, i.e. when the radius to that point is normal to the parabola — which happens exactly at critical points of the squared distance $$D(u) = u^2 + \left(\frac{u^2}{2} - 41\right)^2, \qquad D'(u) = 2u + u\left(u^2 - 82\right) = u\left(u^2 - 80\right).$$

At $u = \pm\sqrt{80}:$ $D = 80 + (40 - 41)^2 = 81,$ so $r = 9$ (the circle touches the parabola at two symmetric points). At $u = 0,$ the point is the vertex $(4, -2)$ at distance $41,$ where the parabola and the circle of radius $41$ both have horizontal tangent lines, so $r = 41$ also works.

The sum is $9 + 41 = 50.$

Solution:

Each roll is an Alice roll, a Bob roll, or a Carol roll, each with probability $\frac{1}{3}.$ The event succeeds exactly when the rolls before the first Carol roll include at least two Alice rolls and at least two Bob rolls. The first Carol roll is roll $k + 1$ with probability $\left(\frac{2}{3}\right)^k \frac{1}{3},$ and given this, the first $k$ rolls form one of $2^k$ equally likely Alice/Bob strings. For $k \ge 3$ the bad strings — at most one Alice, or at most one Bob — number $(k + 1) + (k + 1) = 2k + 2,$ and no string is bad in both ways. Hence $$P = \sum_{k \ge 4} \frac{1}{3}\left(\frac{2}{3}\right)^k \left(1 - \frac{2k + 2}{2^k}\right).$$

The first piece is $\sum_{k \ge 4} \left(\frac{2}{3}\right)^k = \frac{16}{27}.$ For the second, $\sum_{k \ge 0} \frac{k + 1}{3^k} = \frac{1}{(1 - 1/3)^2} = \frac{9}{4},$ so $$\sum_{k \ge 4} \frac{2k + 2}{3^k} = 2\left(\frac{9}{4} - 1 - \frac{2}{3} - \frac{1}{3} - \frac{4}{27}\right) = 2 \cdot \frac{11}{108} = \frac{11}{54}.$$

Therefore $P = \frac{1}{3}\left(\frac{16}{27} - \frac{11}{54}\right) = \frac{1}{3} \cdot \frac{21}{54} = \frac{7}{54},$ and $100m + n = 700 + 54 = 754.$

Solution:

Let $a = AB = BC$ and $b = AC,$ so $s = a + \frac{b}{2}.$ The incircle touches $AC$ at its midpoint (tangent length from $A$ is $s - a = \frac{b}{2}$), so $AI^2 = CI^2 = r^2 + \frac{b^2}{4}.$ By Heron's formula, $r^2 = \frac{(s-a)^2(s-b)}{s} = \frac{b^2}{4} \cdot \frac{2a - b}{2a + b},$ and therefore $$AI^2 = \frac{b^2}{4}\left(\frac{2a - b}{2a + b} + 1\right) = \frac{ab^2}{2a + b}.$$ The perimeter condition is $125\,(2\,AI + b) = 6\,(2a + b).$

Since $AI$ is rational, write $\sqrt{\frac{a}{2a + b}} = \frac{p}{q}$ in lowest terms. Then $aq^2 = p^2(2a + b)$ forces $p^2 \mid a;$ writing $a = mp^2$ gives $b = m(q^2 - 2p^2),$ $2a + b = mq^2,$ and $AI = \frac{mp(q^2 - 2p^2)}{q}.$ The perimeter condition then loses $m$ entirely: $$125\,(q^2 - 2p^2)(2p + q) = 6q^3.$$ Since $\gcd(125, 6) = 1$ we get $5 \mid q;$ and $q$ must be even, since for odd $q$ both factors on the left are odd while the right side is even. Writing $q = 10w$ and simplifying, $(50w^2 - p^2)(p + 5w) = 12w^3.$ Both factors on the left are coprime to $w$ (as $\gcd(p, q) = 1$), so $w = 1,$ and $(50 - p^2)(p + 5) = 12$ has the unique solution $p = 7.$

So $a = 49m,$ $b = 2m,$ and $AI = \frac{7m}{5},$ which is an integer exactly when $5 \mid m.$ Taking $m = 5$ gives $\triangle ABC$ with sides $245, 245, 10$ and $\triangle AIC$ with sides $7, 7, 10,$ whose perimeters $500$ and $24$ are indeed in ratio $125 : 6.$ The minimum possible $AB$ is $245.$

Solution:

Each term is $\frac{1}{10^k - 1} = \sum_{j \ge 1} 10^{-kj},$ so summing over $k$ and collecting the exponent $n = kj,$ $$S = \sum_{n \ge 1} \frac{d(n)}{10^n},$$ where $d(n)$ is the number of divisors of $n.$ Hence $10^{100} S = \sum_{n = 1}^{100} d(n)\,10^{100 - n} + T$ with $T = \sum_{m \ge 1} d(100 + m)\,10^{-m}.$

From $d(101) = 2,$ $d(102) = 8,$ $d(103) = 2,$ $d(104) = 8,$ the tail starts $0.2 + 0.08 + 0.002 + 0.0008 = 0.2828,$ and since $d(N) \lt 2\sqrt{N},$ the remaining terms contribute less than $\sum_{m \ge 5} \frac{2\sqrt{100 + m}}{10^m} \lt 0.001.$ So $0 \lt T \lt 1$ and $$\left\lfloor 10^{100} S \right\rfloor = \sum_{n = 1}^{100} d(n)\,10^{100 - n}.$$

Modulo $1000,$ every term with $n \le 97$ is a multiple of $1000,$ leaving $d(98) \cdot 100 + d(99) \cdot 10 + d(100).$ Since $d(98) = 6,$ $d(99) = 6,$ and $d(100) = 9,$ the remainder is $600 + 60 + 9 = 669.$

Solution:

Since $\omega$ is tangent to $BC$ at $D,$ the power of $B$ gives $BD^2 = BE \cdot BA$ and the power of $C$ gives $CD^2 = CF \cdot CA.$ The angle bisector gives $\frac{BD}{DC} = \frac{AB}{AC} = \frac{8}{9},$ so $BD = 8t$ and $CD = 9t,$ where $t = CD - BD$ is a positive integer. Then $$BE = \frac{64t^2}{200} = \frac{8t^2}{25}, \qquad CF = \frac{81t^2}{225} = \frac{9t^2}{25},$$ so $AE = 200 - \frac{8t^2}{25}$ and $AF = 225 - \frac{9t^2}{25}.$

For $AE$ and $AF$ to be integers we need $25 \mid t^2,$ that is, $t = 5s.$ Then $AE = 200 - 8s^2 \gt 0$ forces $s \le 4,$ and $BC = 17t = 85s.$ At $s = 4:$ $BC = 340,$ with $BD = 160,$ $CD = 180,$ $AE = 72,$ $AF = 81$ all positive integers, and the sides $200, 225, 340$ form a valid triangle since $200 + 225 \gt 340.$

The greatest possible value of $BC$ is $340.$

Solution:

The roots of the cubic are $\alpha^2, \beta^2, \gamma^2.$ Fix square roots $s_1, s_2, s_3$ of them; then $\alpha + \beta + \gamma$ ranges over the eight expressions $\pm s_1 \pm s_2 \pm s_3,$ which come in four pairs $\pm v.$ Generically all eight are distinct. A coincidence $v(\varepsilon) = v(\varepsilon')$ between choices that are not opposite forces $s_i = \pm s_j$ for some $i \ne j,$ which collapses the eight values to at most six. So exactly seven values occur precisely when one choice satisfies $\pm s_1 \pm s_2 \pm s_3 = 0$ — its opposite is then the same value $0$ — and no further degeneracies occur.

That condition is the vanishing of $$(s_1 + s_2 + s_3)(-s_1 + s_2 + s_3)(s_1 - s_2 + s_3)(s_1 + s_2 - s_3) = 2\sum_{i \lt j} r_i r_j - \sum_i r_i^2 = 4e_2 - e_1^2,$$ where $r_i = s_i^2$ are the roots and $e_1, e_2$ their elementary symmetric functions. By Vieta's formulas $e_1 = \frac{n}{6}$ and $e_2 = n - 11,$ so $\frac{n^2}{36} = 4(n - 11),$ i.e. $n^2 - 144n + 1584 = 0,$ with roots $n = 12$ and $n = 132.$

For $n = 132$ the cubic's roots are distinct and nonzero (the constant term is $400 \ne 0$), so the only coincidence is the value $0$ and exactly seven sums occur. The greatest such integer is $132.$

Solution:

The four faces have side multiset $\{5\sqrt{10} \times 8,\ 10 \times 2,\ 18 \times 2\},$ and each edge lies on two faces, so the tetrahedron $ABCD$ has $AB = 10$ and $CD = 18$ as opposite edges and the other four edges equal to $5\sqrt{10}.$ Place $$A = (-5, 0, 12), \quad B = (5, 0, 12), \quad C = (0, -9, 0), \quad D = (0, 9, 0),$$ which is consistent since $AC^2 = 25 + 81 + 144 = 250 = (5\sqrt{10})^2.$ The configuration is symmetric under $x \to -x$ and under $y \to -y,$ so both centers lie on the $z$-axis.

For $S = (0, 0, s),$ equating distances to $A$ and $C$ gives $25 + (12 - s)^2 = 81 + s^2,$ so $s = \frac{11}{3}.$ For $R = (0, 0, t),$ face $ABC$ has plane $4y - 3z + 36 = 0$ and face $ACD$ has plane $12x + 5z = 0,$ so equal distances require $$\frac{36 - 3t}{5} = \frac{5t}{13} \implies t = \frac{117}{16},$$ and by the two mirror symmetries this point is equidistant (at distance $\frac{45}{16}$) from all four faces.

Therefore $RS = \frac{117}{16} - \frac{11}{3} = \frac{351 - 176}{48} = \frac{175}{48},$ which is in lowest terms, so $m + n = 175 + 48 = 223.$

Solution:

A cousin $T$ is the image of an injection sending each $x \in S$ to $x - 1$ or $x + 1,$ landing outside $S.$ If $x - 1, x, x + 1 \in S$ then $x$ has nowhere to go, so every maximal block of consecutive elements of $S$ has size $1$ or $2.$ A double block $\{a, a+1\}$ is forced to map to $\{a - 1, a + 2\},$ while a singleton $\{a\}$ chooses $a - 1$ or $a + 1.$ Two blocks can fight over a value only when exactly one integer separates them, so group blocks into chains: consecutive blocks with gaps of exactly one. Within a chain the only consistent patterns are "the first $i$ blocks shift left and the rest shift right," since a block choosing right and its successor choosing left would collide; a double block acts as both left and right, forcing the switch to happen exactly at it. Hence a chain of $k$ singletons produces $k + 1$ distinct images, a chain containing one double produces exactly $1,$ and a chain with two doubles produces $0.$ Distinct patterns give distinct sets $T,$ and choices in different chains are independent, so the number of cousins is the product of $(k_i + 1)$ over the all-singleton chains.

We need $\prod (k_i + 1) = 4040 = 2^3 \cdot 5 \cdot 101$ while minimizing the element count $\sum k_i$ (chains with doubles only waste elements). Replacing a composite factor $f = gh$ with the two factors $g, h \ge 2$ strictly lowers the cost, because $(g - 1) + (h - 1) \lt gh - 1.$ So the optimum uses the prime factorization: $$\sum (f - 1) = 1 + 1 + 1 + 4 + 100 = 107,$$ realized by five chains of $1, 1, 1, 4, 100$ singletons — runs of every-other integer — placed far apart.

The least possible number of elements is $107.$

Solution:

Since $a - b \equiv a + b \pmod 2,$ the running value after $k$ steps has the same parity as $a_1 + \cdots + a_k.$ So term $a_k$ is subtracted exactly when $a_k$ is even and the prefix sum $a_1 + \cdots + a_{k-1}$ is odd, and the final value is $12$ minus twice the total of the subtracted terms. We must count compositions of $12$ in which the even terms sitting where the prefix sum is odd total exactly $6.$ The prefix parity flips exactly at odd terms, so the odd terms come in $2m$ (the total is even), and the subtracted terms are precisely the even terms lying between the $(2i-1)$st and $2i$th odd terms; these $m$ "odd stretches" must hold even terms totaling $6,$ while the other $m + 1$ stretches hold even terms totaling $6 - A,$ where $A$ is the sum of the odd terms.

Let $f_r(t)$ be the number of ways to fill $r$ ordered stretches with sequences of even terms totaling $2t.$ One stretch is a composition of $2t$ into even parts, i.e. of $t:$ $f_1(t) = 2^{t-1}$ for $t \ge 1$ and $f_1(0) = 1;$ convolving gives the values needed below: $f_r(0) = 1,$ $f_r(1) = r,$ $f_2(2) = 5,$ and $f_1(3), f_2(3), f_3(3) = 4, 12, 25.$ Compositions of $A$ into $2m$ odd parts number $\binom{(A - 2m)/2 + 2m - 1}{2m - 1}.$

Casework on $m$ and $A:$ for $m = 1:$ $A = 2, 4, 6$ give $1 \cdot 4 \cdot f_2(2) = 20,$ $2 \cdot 4 \cdot f_2(1) = 16,$ and $3 \cdot 4 \cdot 1 = 12.$ For $m = 2:$ $A = 4, 6$ give $1 \cdot 12 \cdot f_3(1) = 36$ and $4 \cdot 12 \cdot 1 = 48.$ For $m = 3:$ $A = 6$ gives $1 \cdot 25 \cdot 1 = 25.$ The total is $20 + 16 + 12 + 36 + 48 + 25 = 157.$

Solution:

Work modulo $3:$ entries $3$ are $0$ and entries $1, 2$ are $\pm 1.$ Because the differences of $\{0, 1, 3\}$ hit every nonzero residue mod $7$ exactly once, the seven triples $\{i, i+1, i+3\}$ are the lines of a Fano plane on the positions: every pair of positions lies on exactly one line, and any two lines meet in exactly one point. Let $Z$ be the set of positions holding a $3$ and $k = |Z|.$ A product term survives exactly when its line avoids $Z,$ contributing $(-1)^{\#\{\text{2's on the line}\}},$ and the linear condition constrains the $7 - k$ values $\pm 1$ to sum to $0$ mod $3.$

Casework on $k.$ $k = 7:$ the all-$3$s tuple works: $1.$ $k = 6:$ a single $\pm 1$ can't sum to $0:$ none. $k = 5:$ no line survives; the two nonzero entries must be a $1$ and a $2:$ $\binom{7}{2} \cdot 2 = 42.$ $k = 4:$ three $\pm 1$s sum to $0$ only if all equal, and the three nonzero positions must not form a line, else its product is $\pm 1:$ $(35 - 7) \cdot 2 = 56.$ $k = 3:$ four $\pm 1$s must split two and two; exactly one line avoids a non-line $Z$ (spoiling the sum), while a line $Z$ is avoided by no line: $7 \cdot \binom{4}{2} = 42.$ $k = 2:$ five $\pm 1$s must go four and one; exactly two lines avoid $Z,$ meeting at a point $p$ and covering the five positions, and their products cancel exactly when the lone minority value avoids $p:$ $\binom{7}{2} \cdot 2 \cdot 4 = 168.$ $k = 1:$ six $\pm 1$s sum to $0$ if all equal or three of each; the four lines avoiding $Z$ pairwise meet in the six nonzero positions, and since the product of all four line-products is $+1,$ we need exactly two negative lines. All-equal gives $0$ or $4$ negative lines; for three $2$'s, viewing positions as edges of $K_4$ on the four lines, a line is negative exactly when it has odd degree in the chosen $3$-edge set, and exactly the $12$ three-edge paths (of the $\binom{6}{3} = 20$ subsets) give two odd degrees: $7 \cdot 12 = 84.$ $k = 0:$ seven $\pm 1$s need two or five $2$'s, which make $4$ or $3$ lines negative respectively, but $7 - 2t \equiv 0 \pmod 3$ needs $t \equiv 2 \pmod 3:$ none.

The total is $1 + 42 + 56 + 42 + 168 + 84 = 393.$