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For finite sets $E_1,\dots ,E_s$, the inclusion-exclusion principle statesIt counts each element once: an element in exactly $q$ of the sets contributes ${\sum }_{j=1}^q(-1{)}^{j+1}(\genfrac{}{}{0ex}{}{q}{j})=1$, by the binomial theorem.

For the cyclic digit run problem, let $E_j$ mean that the three digits starting at position $j$ form a run, for $j=1,2,3,4$. Every run of length at least three contains one of these triples, so we need $|E_1\cup E_2\cup E_3\cup E_4|$. Introduce the five differencesA starting digit and these five residues determine a unique string, including runs that cross $9$ and $0$. In these coordinates $E_j$ requires $(d_j,d_{j+1})=(1,1)$ or $(-1,-1)$.

Each single event fixes two differences, with two choices of sign; the initial digit and the other three differences are free. Thus $|E_j|=10\cdot 2\cdot 10^3=20000$. For intersections of two events, adjacent starting positions force three consecutive differences to have the same sign, giving $10\cdot 2\cdot 10^2=2000$. There are three such pairs. The other three pairs have disjoint constrained difference positions, with independent signs and one free difference, giving $10\cdot 4\cdot 10=400$ each. Consequently the sum of pair-intersection sizes is $3\cdot 2000+3\cdot 400=7200$.

For triple intersections, $E_1\cap E_2\cap E_3$ and $E_2\cap E_3\cap E_4$ each force four consecutive differences to have the same sign, leaving one free difference, and each has size $10\cdot 2\cdot 10=200$. The other two triples, $E_1\cap E_2\cap E_4$ and $E_1\cap E_3\cap E_4$, divide the five differences into two blocks with independent signs; each has size $10\cdot 4=40$. The triple-intersection sum is therefore $480$. Finally, all four events force all five differences to have the same sign, giving $20$ strings.

The inclusion-exclusion principle now gives the number of stringsUsing differences makes the overlap conditions explicit and prevents double-counting long cyclic digit runs.

Multiplying the desired representation by $b_1{b}_2$ turns it into the integer equationSince $b_1$ and $b_2$ are coprime integers, the necessary modular multiplicative inverses exist. Choose the unique representativesThen $a-a_1{b}_2-a_2{b}_1$ is divisible by both $b_1$ and $b_2$, hence by their product, because they are coprime. Definingtherefore gives an integer and proves existence. Any other representation, reduced modulo $b_1$ and $b_2$, must have the same $a_1$ and $a_2$, since their allowed ranges contain exactly one representative of each residue. The integer equation then forces the same $n$. This proves uniqueness.

For the prime-power fraction decomposition, put $d_i=p_i^{n_i}$ and $B_i=b/d_i$. The distinct prime-power factors $d_i$ are pairwise coprime, and $B_i$ has a modular multiplicative inverse modulo $d_i$. DefineEvery term $q_j{B}_j$ with $j\ne i$ is divisible by $d_i$, while $q_i{B}_i\equiv a(\mathrm{mod}{d}_i)$. Thus $a-{\sum }_i{q}_i{B}_i$ is divisible by all the pairwise coprime $d_i$ and therefore by $b$. With $n=(a-{\sum }_i{q}_i{B}_i)/b$, division by $b$ yields the required representation. Reducing any such representation modulo each $d_i$ proves uniqueness exactly as above. This is the residue-solving mechanism of the Chinese remainder theorem.

For $315=9\cdot 5\cdot 7$, the complementary factors are $35,63,45$. The required modular inverses givebecause $35\cdot 8\equiv 1(\mathrm{mod}9)$, $63\cdot 2\equiv 1(\mathrm{mod}5)$, and $45\cdot 5\equiv 1(\mathrm{mod}7)$. Their numerator sum is $280+126+225=631=1+2\cdot 315$. Thus the unique decomposition is

In a circular orbit, $r=a$ is constant, so the radial acceleration vanishes. The radial equation requiresFor every $a>0$, this equation permits a real nonzero specific angular momentum, $h=\pm \sqrt{k{a}^{n+3}}$, and hence a circular orbit. This chooses $h$ for the desired radius; it does not claim that one prescribed value of $h$ permits every radius.

To test circular-orbit stability for a power-law central force, keep that chosen specific angular momentum fixed and write $r=a+\rho$, with $|\rho |\ll a$. The radial equation becomesExpanding to first order and using $h^2=k{a}^{n+3}$ givesFor $n>-3$, the linearized equation is the harmonic oscillator equation, withThe radial disturbance undergoes bounded small oscillations, rather than growing. Equivalently, the effective potential has a strict local minimum: its second derivative at $a$ is $(n+3)k{a}^{n-1}>0$. By the effective potential stability criterion, sufficiently small radial disturbances remain small, proving stability for $n>-3$. For comparison, $n<-3$ gives exponential growth. At $n=-3$, the balancing value $h^2=k$ makes the radial acceleration identically zero, so a small radial velocity produces drift rather than a restoring oscillation.

For a nonempty set $S\subseteq \mathbb{R}$, an upper bound in a partially ordered set is a real number $u$ such that $s\le u$ for every $s\in S$. Its supremum, or least upper bound, is an upper bound $L$ with $L\le u$ for every upper bound $u$. The least-upper-bound property asserts that every nonempty set of real numbers bounded above has a real supremum.

Let $(x_n)$ be a bounded, increasing sequence; here increasing allows $x_{n+1}=x_n$. Apply the least-upper-bound property to its set of terms and put $L=\sup \{x_n:n\ge 1\}$. Given $\epsilon >0$, the number $L-\epsilon$ cannot be an upper bound, so some $x_N$ satisfies $x_N>L-\epsilon$. The monotone sequence property then givesHence $|x_n-L|<\epsilon$ for all $n\ge N$. This proves the monotone bounded sequence result, with limit $\boxed{x_n⟶L}$.

For the two series, define their partial sums $A_N={\sum }_{n=1}^N{a}_n$ and $B_N={\sum }_{n=1}^N{b}_n$. Positivity makes both partial-sum sequences increasing, and termwise comparison gives $0<B_N\le A_N$. Since $(A_N)$ is a convergent sequence, it is bounded above, and therefore so is $(B_N)$. Applying the monotone bounded sequence result already proved shows that $(B_N)$ converges. By the definition of a convergent series, ${\sum }_{n=1}^{\infty }{b}_n$ converges. This is the comparison test for series for positive terms.

The extended Euclidean algorithm givesBack-substituting in these equations produces the Bezout identityThus one integer solution is $\boxed{x=12,y=-7}$. In fact, subtracting this Bezout identity from any other solution and using that $17$ and $29$ are coprime integers gives all solutions: $x=12+29t$, $y=-7-17t$, where $t\in \mathbb{Z}$.

For the modular arithmetic calculation, $29$ is a prime number and $17$ is coprime to it, so Fermat little theorem gives $17^{28}\equiv 1(\mathrm{mod}29)$. Since $138=5\cdot 28-2$, and the preceding Bezout identity identifies $12$ as the modular multiplicative inverse of $17$, we obtainConsequently the least positive representative is $\boxed{28}$.

Use an Earth-centred rotating reference frame, with Earth's constant angular velocity vector $\mathbf{\omega }$. The vectors $\mathbf{r}$, $\mathbf{v}$, and $\mathbf{a}$ are the particle's position, velocity, and acceleration measured in that rotating frame. The vector $\mathbf{g}$ is the gravitational acceleration due to the actual gravitational force, expressed in the same axes; it excludes the separate centrifugal acceleration. The remaining term $-2\mathbf{\omega }\times \mathbf{v}$ is the Coriolis acceleration. A constant rotation axis and an unaccelerated Earth-centred origin account for the absence of an Euler term or a translational-origin term.

For the size estimates, Earth's angular speed isWith drop height $H=20\mathrm{m}$, unperturbed free fall gives the fall timeand final speed $gT=20{\mathrm{m}\mathrm{s}}^{-1}$. Thus the gravitational term has size $10{\mathrm{m}\mathrm{s}}^{-2}$, the Coriolis acceleration is at most about $2\omega gT\approx 2.8\times 10^{-3}{\mathrm{m}\mathrm{s}}^{-2}$, and the centrifugal acceleration is at most ${\omega }^2{R}_0\approx 2.9\times 10^{-2}{\mathrm{m}\mathrm{s}}^{-2}$. These are characteristic sizes $10^1$, $10^{-3}$, and $10^{-2}$ in acceleration units; geometric factors can reduce them, and the Coriolis term is zero at release.

For a suitable approximation to the rotational deflection of a falling body, take $\mathbf{g}$ constant over the short fall, start at rest relative to Earth, and use the zeroth-order velocity ${\mathbf{v}}^{(0)}(t)=\mathbf{g}t$ inside the small Coriolis term. In the centrifugal term replace $\mathbf{r}$ by ${\mathbf{R}}_0$, because $H/R_0\ll 1$. Integrating the resulting acceleration twice, with initial position ${\mathbf{r}}_{\mathrm{i}}$, givesSince ${\mathbf{r}}_{\mathrm{i}}+\mathbf{g}{T}^2/2={\mathbf{R}}_0$, evaluating at the unperturbed fall time yields the printed first-iterate estimate

The approximation keeps the leading Coriolis contribution and the leading centrifugal contribution enhanced by Earth's large radius. Feeding the Coriolis velocity correction back into the equation produces terms of size ${\omega }^{2}g{T}^4$, smaller than the retained centrifugal displacement by a factor of order $H/R_0$. Replacing $\mathbf{r}$ by ${\mathbf{R}}_0$ in that term has the same relative error. Here $\omega T\approx 1.4\times 10^{-4}$ and $H/R_0\approx 3.3\times 10^{-6}$, so both approximations are well controlled. Strictly, the displayed vector is the position at time $T$: its small upward component is removed by the slightly longer actual fall time. That arrival-time correction changes the retained horizontal displacements only at higher order. Thus the horizontal components of the printed estimate give the landing point on the locally flat surface; its vertical component should not be interpreted as a landing above the ground.

Let ${\hat{\mathbf{e}}}_E,{\hat{\mathbf{e}}}_N,{\hat{\mathbf{e}}}_U$ point east, north, and vertically upwards. They form a right-handed basis, with ${\hat{\mathbf{e}}}_N\times {\hat{\mathbf{e}}}_U={\hat{\mathbf{e}}}_E$. At latitude $\lambda$, in the spherical-Earth approximation,The required cross products areAt $\lambda =45^{\circ }$, the northerly and easterly displacements are thereforeThe negative northerly component means a southward displacement. With the supplied values, these are approximately $2.9\mathrm{cm}$ south and $1.3\mathrm{mm}$ east. The eastward effect comes from Coriolis acceleration, while the leading southward effect here comes from centrifugal acceleration.

For unit mass, the inverse-square force has potential energy $-k/r$, with zero potential at infinity. Conservation of angular momentum gives $r^2\dot{\theta }=h$. Differentiate the polar equation of the Kepler orbit:Consequently the total mechanical energy iswhere $k=h^2/\ell$. This also explains the sign distinction between an elliptic orbit, a parabolic Kepler orbit, and a hyperbolic Kepler orbit.

At the original periapsis, the parabolic Kepler orbit has $e=1$, $r=d=\ell /2$, and purely tangential speed $V$. ThusThe outward radial impulse has zero moment about the Sun, so it preserves angular momentum and hence $\ell$. Its additional velocity is perpendicular to the original velocity, so the new mechanical energy is $E^{\prime }={\alpha }^2{V}^2/2$. Applying the energy formula givesFor $\alpha >0$, the new orbit is a hyperbolic Kepler orbit; $\alpha =0$ leaves the parabola unchanged.

To fix the orientation of the sketch, put the Sun at the origin, the impulse point at $(d,0)$, and take the original motion there upwards. The new velocity is $(\alpha V,V)$. The eccentricity vector isIt points towards the new periapsis, which is rotated clockwise through $\beta =\arctan (2\alpha )$ from the original periapsis direction. Thus, measured using the original polar angle $\theta$, the new orbit isIn particular, the impulse point is already on the outgoing part of the new orbit and is not its new periapsis. The old parabola satisfies $x=d-y^2/(4d)$. Both orbits pass through $(d,0)$ and have the same focus, but the hyperbola's new periapsis axis is tilted.