The polynomial factors into integer linear factors exactly when its roots are integers, that is, when the discriminant $a^2 - 4b$ equals $c^2$ for some integer $c \ge 0.$ Given $a,$ such a $b \ge 0$ exists exactly when $4b = a^2 - c^2 = (a-c)(a+c)$ for some $c$ with $0 \le c \le a$ and $c \equiv a \pmod 2,$ and distinct such $c$ give distinct values $b = \frac{a^2 - c^2}{4}.$

For odd $a$ the valid $c$ are $1, 3, \ldots, a,$ which is $\frac{a+1}{2}$ choices; for even $a$ they are $0, 2, \ldots, a,$ which is $\frac{a}{2} + 1$ choices.

Summing over $a = 1, \ldots, 100:$ the odd $a$ contribute $1 + 2 + \cdots + 50 = 1275,$ and the even $a$ contribute $2 + 3 + \cdots + 51 = 1325.$ Thus $S = 2600,$ and the remainder is $600.$

Writing out the place values, $n = 196a + 14b + c$ $= 225a + 15c + b$ $= 222a + 37c,$ where $a$ and $c$ are base-$6$ digits with $1 \le a \le 5$ and $0 \le c \le 5,$ and $0 \le b \le 13.$

Equating the last two expressions gives $b = 22c - 3a.$ Substituting into $196a + 14b + c = 225a + 15c + b$ (which says $13b = 29a + 14c$) yields $13(22c - 3a) = 29a + 14c,$ so $272c = 68a,$ that is $a = 4c.$ The digit bounds force $c = 1,$ $a = 4,$ and then $b = 22 - 12 = 10,$ which is a valid digit in bases $14$ and $15.$

Therefore $n = 222 \cdot 4 + 37 \cdot 1 = 925.$ Indeed $925 = 196 \cdot 4 + 14 \cdot 10 + 1,$ confirming the base-$14$ form. The answer is $925.$

There are $10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 = 30240$ equally likely ordered layouts of $5$ of the $10$ distinct cards. Kathy is happy exactly when the color pattern consists of one block of reds and one block of greens: the patterns are RRRRR, GGGGG, and the eight mixed patterns $\text{R}^r\text{G}^{5-r}$ and $\text{G}^{5-r}\text{R}^r$ for $r = 1, 2, 3, 4.$

A pattern using $r$ red and $5 - r$ green positions can be filled in $\frac{5!}{(5-r)!} \cdot \frac{5!}{r!}$ ways (ordered choices of which red cards and which green cards appear). For $r = 5, 4, 3, 2, 1, 0$ these counts are $120,$ $600,$ $1200,$ $1200,$ $600,$ $120.$ The happy layouts number $$\begin{aligned} &120 + 120 \\ &{}+ 2\,(600 + 1200 + 1200 + 600) \\ &{}= 7440. \end{aligned}$$

The probability is $\frac{7440}{30240} = \frac{31}{126},$ so $m + n = 31 + 126 = 157.$

By the law of cosines in $\triangle ABC,$ $$\begin{aligned} \cos A &= \frac{10^2 + 10^2 - 12^2}{2 \cdot 10 \cdot 10} \\ &= \frac{56}{200} = \frac{7}{25}. \end{aligned}$$

Let $x = AD = DE = EC,$ so $AE = 10 - x.$ The law of cosines in $\triangle ADE$ gives $$\begin{aligned} &x^2 = x^2 + (10 - x)^2 \\ &\quad {}- 2x(10 - x)\cdot\frac{7}{25}, \end{aligned}$$ so $(10 - x)^2 = \frac{14}{25}\,x(10 - x).$ Since $x \lt 10,$ we may divide by $10 - x$ to get $10 - x = \frac{14x}{25},$ hence $250 = 39x$ and $x = \frac{250}{39}.$

As $\gcd(250, 39) = 1,$ the answer is $250 + 39 = 289.$

Because $\log_4 u = \log_2 \sqrt{u},$ the first equation is equivalent to $(2x + y)^2 = x^2 + xy + 7y^2$ together with $2x + y \gt 0.$ Expanding gives $3x^2 + 3xy - 6y^2 = 0,$ which factors as $3(x - y)(x + 2y) = 0.$ So $x = y$ or $x = -2y,$ with $(x, y) \ne (0, 0).$

Similarly the second equation says $(3x + y)^2 = 3x^2 + 4xy + Ky^2,$ that is $6x^2 + 2xy + y^2 = Ky^2.$ If $x = y$ (taking $x \gt 0$ so both logarithms are defined), then $K = 6 + 2 + 1 = 9.$ If $x = -2y$ (taking $y \lt 0,$ so $2x + y = -3y \gt 0$ and $3x + y = -5y \gt 0$), then $24y^2 - 4y^2 + y^2 = Ky^2,$ so $K = 21.$

Both cases occur, so the product of all possible values is $9 \cdot 21 = 189.$

Write $z = e^{i\theta}$ with $\theta \in [0, 2\pi).$ Then $z^{720} - z^{120}$ is real exactly when $\sin 720\theta = \sin 120\theta,$ which happens when the angles are equal or supplementary modulo $2\pi:$ either $720\theta = 120\theta + 2\pi k,$ giving $\theta = \frac{\pi k}{300},$ or $720\theta = \pi - 120\theta + 2\pi k,$ giving $\theta = \frac{(2k+1)\pi}{840}.$

The first family has $600$ values in $[0, 2\pi)$ and the second has $840.$ They cannot coincide: $\frac{\pi k}{300} = \frac{(2j+1)\pi}{840}$ would give $14k = 5(2j + 1),$ equating an even number with an odd one.

Hence $N = 600 + 840 = 1440,$ and the remainder is $440.$

The chords of a unit regular hexagon have lengths $1,$ $\sqrt{3},$ and $2.$ Among the $\binom{6}{3} = 20$ triangles in one hexagon, $6$ have sides $1, 1, \sqrt{3}$ and $2$ are equilateral with side $\sqrt{3};$ the other $12,$ with sides $1, \sqrt{3}, 2,$ are scalene. So each base contributes $8$ isosceles triangles, for $16$ in all.

Otherwise two vertices lie on one base ($2$ choices of that base) and one on the other. A vertex of the top base at horizontal distance $d$ from a bottom vertex is at distance $\sqrt{d^2 + 4} \ge 2$ from it. If the bottom pair is adjacent (chord $1$): the perpendicular bisector of a hexagon edge passes through no vertices, and no slant side can equal $1,$ so there are no isosceles triangles. If the pair has one vertex between them (chord $\sqrt{3},$ $6$ pairs): the top vertices above that middle vertex and above the opposite vertex are equidistant from the pair, giving $6 \cdot 2 = 12.$ If the pair is diametrically opposite (chord $2,$ $3$ pairs): no vertex lies above the perpendicular bisector, but the top vertex directly above either endpoint gives a slant side $\sqrt{0 + 4} = 2$ equal to the chord, giving $3 \cdot 2 = 6.$

The total is $16 + 2\,(12 + 6) = 52.$

All interior angles are $120^\circ,$ so opposite sides are parallel. Attaching equilateral triangles to two opposite sides produces a parallelogram, which forces $AB + BC = DE + EF$ and $FA + AB = CD + DE.$ Hence $EF = 2$ and $FA = 16.$

Walking from one side to the opposite side along the two connecting sides shows that the distance between a pair of opposite sides is $\frac{\sqrt{3}}{2}$ times the sum of those two connecting sides: the strips have widths $\frac{\sqrt{3}}{2}(BC + CD) = 9\sqrt{3}$ between $AB$ and $DE,$ $\frac{\sqrt{3}}{2}(CD + DE) = 11\sqrt{3}$ between $BC$ and $EF,$ and $\frac{\sqrt{3}}{2}(DE + EF) = 7\sqrt{3}$ between $CD$ and $FA.$ Any circle inside the hexagon fits in the narrowest strip, so $d \le 7\sqrt{3}.$

A circle of diameter $7\sqrt{3}$ tangent to lines $CD$ and $FA$ can be centered so that it also touches $DE$ exactly and has distances $6\sqrt{3},$ $5\sqrt{3},$ and $\frac{11\sqrt{3}}{2}$ from lines $EF,$ $BC,$ and $AB,$ all more than its radius $\frac{7\sqrt{3}}{2},$ so it fits inside the hexagon. Therefore $d = 7\sqrt{3}$ and $d^2 = 147.$

The pairs of distinct elements summing to $16$ are $\{1,15\}, \{2,14\}, \ldots, \{7,9\}$ (seven pairs), and those summing to $24$ are $\{4,20\}, \{5,19\}, \ldots, \{11,13\}$ (eight pairs). First count subsets containing a $16$-pair and a $24$-pair that are disjoint. Of the $7 \cdot 8 = 56$ combinations, the ones sharing an element $x$ require $16 - x,$ $x,$ and $24 - x$ all to be valid, which happens for the $10$ values $x \in \{4, \ldots, 15\}$ other than $8$ and $12.$ No four-element set arises from two different disjoint combinations (a second decomposition would force a $16$-pair to coincide with a $24$-pair), so this case gives $56 - 10 = 46$ subsets.

In the remaining subsets every $16$-pair meets every $24$-pair, so some center $a$ has both $b = 16 - a$ and $c = 24 - a$ in the subset. There are $10$ possible centers ($a \in \{4, \ldots, 15\}$ with $a \ne 8, 12$), and the fourth element can be any of the $17$ remaining numbers, giving $170$ center–subset counts. Exactly $6$ subsets admit two centers and are counted twice: $\{1,7,9,15\},$ $\{2,6,10,14\},$ $\{3,5,11,13\},$ $\{5,11,13,19\},$ $\{6,10,14,18\},$ and $\{7,9,15,17\}.$ This case gives $170 - 6 = 164$ subsets, none of which contain disjoint pairs.

The total is $46 + 164 = 210.$

From any inner point the bug has exactly two moves, counterclockwise along the inner circle or outward along a spoke; from any outer point it has exactly two, clockwise along the outer circle or inward along a spoke. Call a move $X$ if it is counterclockwise or inward and $Y$ if it is clockwise or outward. Then every string in $\{X, Y\}^{15}$ describes exactly one $15$-step path from $A.$

A step arrives on the inner circle exactly when it is an $X,$ so the path ends on the inner circle exactly when its last move is an $X;$ in that case the numbers of inward and outward moves are equal. Measuring angular position in fifths of a turn (counterclockwise $+1,$ clockwise $-1,$ spokes $0$), the path returns to $A$ exactly when it ends on the inner circle and the net rotation is a multiple of $5,$ that is, when the last move is $X$ and $\#X - \#Y \equiv 0 \pmod 5.$ With $15$ moves this means the number of $X$s is $5,$ $10,$ or $15.$

Fixing the last move as $X,$ the first $14$ moves contain $4,$ $9,$ or $14$ $X$s, so $$\begin{aligned} n &= \binom{14}{4} + \binom{14}{9} + \binom{14}{14} \\ &= 1001 + 2002 + 1 = 3004, \end{aligned}$$ and the remainder is $4.$