The class of $(1,1)$ isso its size is Euler's totient functionThe unit group acts freely on every pair in $Y$: if $\lambda a\equiv a$ and $\lambda b\equiv b$, multiplying the Bezout identity by $\lambda -1$ gives $\lambda \equiv 1(\mathrm{mod}n)$. Hence every equivalence class has size $\phi (n)$. Dividing the answer from part (a) by this size gives

Assume $ad\equiv bc(\mathrm{mod}n)$ and choose $r,s$ as in part (b). Put $\lambda =rc+sd$. Multiplying $ra+sb\equiv 1$ by $c$ and using $ad\equiv bc$ givesSimilarly, multiplication by $d$ gives $\lambda b\equiv d(\mathrm{mod}n)$.

Moreover, $\lambda$ is a unit modulo n: if a prime divided both $\lambda$ and $n$, the two congruences would make it divide $c,d,n$, contrary to $(c,d)\in Y$. Thus $R$ means precisely that two pairs differ by multiplication by a unit modulo $n$. Reflexivity, symmetry using ${\lambda }^{-1}$, and transitivity using products of units now show that $R$ is an equivalence relation.

The condition $\gcd (a,b,n)=1$ says that the ideal of $\mathbb{Z}$ generated by $a,b,n$ is all of $\mathbb{Z}$. The multi-integer Bezout identity therefore supplies integers $r,s,t$ such thatEquivalently, first apply Bézout's identity to $a,b$ and then to $\gcd (a,b),n$.

The inclusion-exclusion principle states that for finite sets $A_1,\dots ,A_k$,For each prime $p\mid n$, exclude pairs whose two coordinates are divisible by $p$. For a set of distinct prime divisors $p_1,\dots ,p_j$, exactly $n^2/(p_1\cdots p_j{)}^2$ pairs have both coordinates divisible by their product. Inclusion-exclusion therefore gives

We prove both identities by mathematical induction. They hold at $n=1$. Assume they hold at $n$. Using $F_{n-1}=F_{n+1}-F_n$,Adding this to $F_{2n+1}=F_n^2+F_{n+1}^2$ and using $F_{n+2}=F_n+F_{n+1}$ givesThese are precisely the two assertions with $n$ replaced by $n+1$, completing the induction.

Fermat's little theorem states that for a prime $p$ and integer $a$,if $p\nmid a$, equivalently $a^{p-1}\equiv 1(\mathrm{mod}p)$. For the latter form, multiplication by $a$ permutes the nonzero residue classes modulo $p$. Multiplying the congruences $a,2a,\dots ,(p-1)a$ and cancelling $(p-1)!$ gives $a^{p-1}\equiv 1$. The first form also covers $p\mid a$.

Since $803=50\cdot 16+3$, the theorem gives

Apply part (a) first to the position $\mathbf{r}$:Applying it again to both terms on the right giveswhere $(\dot{\mathbf{\omega }}{)}_S=(\dot{\mathbf{\omega }}{)}_{S^{\prime }}$ because $\mathbf{\omega }\times \mathbf{\omega }=0$. Substituting the first formula into the last term produces

In the Earth-fixed rotating reference frame, the sideways apparent force is the Coriolis acceleration multiplied by the ski-doo's mass:At the South Pole, $\mathbf{\omega }$ is vertical and antiparallel to the local outward normal, while $\mathbf{v}$ is horizontal, soThe force points to the left of the direction of travel, which is westward when the ski-doo is moving north after crossing the pole.

Let $\{{\mathbf{e}}_i(t)\}$ be an orthonormal basis fixed in $S^{\prime }$ and write $\mathbf{a}=a_i{\mathbf{e}}_i$. A basis vector rotating with angular velocity $\mathbf{\omega }$ satisfiesDifferentiating the components and the basis vectors in $S$ therefore gives the rotating-frame derivative formula

At time $t_0$, the accelerated proton has momentum $p=qE{t}_0$ and, by the relativistic energy-momentum relation, energyThe incident cosmic microwave background photon adds energy $E_{\gamma }$. Since the final state contains only the Delta-plus baryon, conservation of relativistic energy gives its laboratory-frame Lorentz factorIts rest-frame decay time $t_{\Delta }$ is a proper time, so time dilation gives the laboratory decay time

Write $E=|\mathbf{E}|$ and $\hat{\mathbf{E}}=\mathbf{E}/E$. The constant electric force gives momentumCombining $\mathbf{p}=\gamma m\dot{\mathbf{x}}$ with the relativistic energy-momentum relation givesThe dimensional analysis from part (a) confirms that $qEt$ has the dimensions of momentum, so $qEt/(mc)$ is dimensionless and the expression has the dimensions of velocity.

The speed increases monotonically from zero and approaches the speed of light asymptotically:Integrating in the direction of the field givesAt small times, the Taylor expansion of the square root gives the nonrelativistic limitwhich is motion from rest with the constant acceleration $q\mathbf{E}/m$.

The Lorentz force and Newton's second law giveWriting $\Omega =qB/m$, the component equations areThe initial velocity therefore givesand integration with the initial position givesThis is helical motion in a uniform magnetic field: a helix of radius $v/\Omega$ around the line $x=1+v/\Omega$, $y=0$, with its axis parallel to the magnetic field.

A three-cycle is an even permutation, as is a product of two disjoint three-cycles. Thus every element counted in part (c) lies in $A_7$, and every order-three element of $A_7$ was already counted there. The answer remains $\boxed{350}$.

An element of order three in $S_7$ has cycle type $31^4$ or $3^{2}1$. There aresingle three-cycles. For two disjoint three-cycles, choose the two unordered three-element supports and an orientation on each:Hence $S_7$ contains $\boxed{350}$ elements of order three.