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.