di-zhang-fdu/OlympiadExtended huggingface dataset:
Subfield: Modern Physics


Question: 2.  After one day, the clock reading on the Earth surface and the satellite will differ due to both special and general relativistic effects. Calculate the difference due to each effect for one day. Calculate the total difference for one day. Which clock is faster, a clock on the Earth's surface or the satellite's clock?

Solution:
The general relativity effect is

$$
\begin{aligned}
\frac{d \tau_{g}}{d t} & =1+\frac{\Delta U}{m c^{2}} \\
\frac{d \tau_{g}}{d t} & =1+\frac{g R^{2}}{c^{2}} \frac{R-r}{R r}
\end{aligned}
$$

After one day, the difference is

$$
\begin{aligned}
\Delta \tau_{g} & =\frac{g R^{2}}{c^{2}} \frac{R-r}{R r} \Delta T \\
& =4.55 \times 10^{-5} \mathrm{~s} .
\end{aligned}
$$

The special relativity effect is

$$
\begin{aligned}
\frac{d \tau_{s}}{d t} & =\sqrt{1-\frac{v^{2}}{c^{2}}} \\
& =\sqrt{1-\left(\left(\frac{2 \pi g R^{2}}{T}\right)^{\frac{2}{3}}\right) \frac{1}{c^{2}}} \\
& \approx 1-\frac{1}{2}\left(\left(\frac{2 \pi g R^{2}}{T}\right)^{\frac{2}{3}}\right) \frac{1}{c^{2}} .
\end{aligned}
$$

After one day, the difference is

$$
\begin{aligned}
\Delta \tau_{s} & =-\frac{1}{2}\left(\left(\frac{2 \pi g R^{2}}{T}\right)^{\frac{2}{3}}\right) \frac{1}{c^{2}} \Delta T \\
& =-7.18 \times 10^{-6} \mathrm{~s} .
\end{aligned}
$$

The satelite's clock is faster with total $\Delta \tau=\Delta \tau_{g}+\Delta \tau_{s}=3.83 \times 10^{-5} \mathrm{~s}$.

Final answer: by general relativity effect, after one day, the difference is $4.55 \times 10^{-5} \mathrm{~s}$
by special relativity effect, after one day, the difference is $-7.18 \times 10^{-6} \mathrm{~s}$
The satelite's clock is faster with total $\Delta \tau=\Delta \tau_{g}+\Delta \tau_{s}=3.83 \times 10^{-5} \mathrm{~s}$
Subfield: Modern Physics

All matters in the universe have fundamental properties called spin, besides their mass and charge. Spin is an intrinsic form of angular momentum carried by particles. Despite the fact that quantum mechanics is needed for a full treatment of spin, we can still study the physics of spin using the usual classical formalism. In this problem, we are investigating the influence of magnetic field on spin using its classical analogue.

The classical torque equation of spin is given by

$$
\boldsymbol{\tau}=\frac{d \boldsymbol{L}}{d t}=\boldsymbol{\mu} \times \boldsymbol{B}
$$

In this case, the angular momentum $\boldsymbol{L}$ represents the "intrinsic" spin of the particles, $\boldsymbol{\mu}$ is the magnetic moment of the particles, and $\boldsymbol{B}$ is magnetic field. The spin of a particle is associated with a magnetic moment via the equation

$$
\boldsymbol{\mu}=-\gamma \boldsymbol{L}
$$

where $\gamma$ is the gyromagnetic ratio.

In this problem, the term "frequency" means angular frequency (rad/s), which is a scalar quantity. All bold letters represent vectors; otherwise they represent scalars.

Part A. Larmor precession

Question: 1.  Prove that the magnitude of magnetic moment $\mu$ is always constant under the influence of a magnetic field $\boldsymbol{B}$. For a special case of stationary (constant) magnetic field, also show that the angle between $\boldsymbol{\mu}$ and $\boldsymbol{B}$ is constant.

(Hint: You can use properties of vector products.)

Solution:
From the two equations given in the text, we obtain the relation

$$
\frac{d \boldsymbol{\mu}}{d t}=-\gamma \boldsymbol{\mu} \times \mathbf{B}
\tag{1}
$$

Taking the dot product of eq (1). with $\boldsymbol{\mu}$, we can prove that

$$
\begin{aligned}
\boldsymbol{\mu} \cdot \frac{d \boldsymbol{\mu}}{d t} & =-\gamma \boldsymbol{\mu} \cdot(\boldsymbol{\mu} \times \mathbf{B}) \\
\frac{d|\boldsymbol{\mu}|^{2}}{d t} & =0 \\
\mu=|\boldsymbol{\mu}| & =\text { const. }
\end{aligned}
\tag{2}
$$

Taking the dot product of eq. (1) with $\mathbf{B}$, we also prove that

$$
\begin{aligned}
\mathbf{B} \cdot \frac{d \boldsymbol{\mu}}{d t} & =-\gamma \mathbf{B} \cdot(\boldsymbol{\mu} \times \mathbf{B}), \\
\mathbf{B} \cdot \frac{d \boldsymbol{\mu}}{d t} & =0, \\
\mathbf{B} \cdot \boldsymbol{\mu} & =\text { const. }
\end{aligned}
\tag{3}
$$

An acute reader will notice that our master equation in (1) is identical to the equation of motion for a charged particle in a magnetic field

$$
\frac{d \mathbf{v}}{d t}=\frac{q}{m} \mathbf{v} \times \mathbf{B}
\tag{4}
$$

Hence, the same argument for a charged particle in magnetic field can be applied in this case.
Subfield: Modern Physics

All matters in the universe have fundamental properties called spin, besides their mass and charge. Spin is an intrinsic form of angular momentum carried by particles. Despite the fact that quantum mechanics is needed for a full treatment of spin, we can still study the physics of spin using the usual classical formalism. In this problem, we are investigating the influence of magnetic field on spin using its classical analogue.

The classical torque equation of spin is given by

$$
\boldsymbol{\tau}=\frac{d \boldsymbol{L}}{d t}=\boldsymbol{\mu} \times \boldsymbol{B}
$$

In this case, the angular momentum $\boldsymbol{L}$ represents the "intrinsic" spin of the particles, $\boldsymbol{\mu}$ is the magnetic moment of the particles, and $\boldsymbol{B}$ is magnetic field. The spin of a particle is associated with a magnetic moment via the equation

$$
\boldsymbol{\mu}=-\gamma \boldsymbol{L}
$$

where $\gamma$ is the gyromagnetic ratio.

In this problem, the term "frequency" means angular frequency (rad/s), which is a scalar quantity. All bold letters represent vectors; otherwise they represent scalars.

Part A. Larmor precession
Context question:
1.  Prove that the magnitude of magnetic moment $\mu$ is always constant under the influence of a magnetic field $\boldsymbol{B}$. For a special case of stationary (constant) magnetic field, also show that the angle between $\boldsymbol{\mu}$ and $\boldsymbol{B}$ is constant.

(Hint: You can use properties of vector products.)
Context answer:
\boxed{证明题}


Question: 2.  A uniform magnetic field $\boldsymbol{B}$ exists and it makes an angle $\phi$ with a particle's magnetic moment $\boldsymbol{\mu}$. Due to the torque by the magnetic field, the magnetic moment $\boldsymbol{\mu}$ rotates around the field $\boldsymbol{B}$, which is also known as Larmor precession. Determine the Larmor precession frequency $\omega_{0}$ of the magnetic moment with respect to $\boldsymbol{B}=B_{0} \boldsymbol{k}$.

Solution:
For a magnetic moment making an angle of $\phi$ with $\mathbf{B}$,

$$
\begin{aligned}
\frac{d \boldsymbol{\mu}}{d t} & =-\gamma \boldsymbol{\mu} \times \mathbf{B} \\
|\boldsymbol{\mu}| \sin \phi \frac{d \theta}{d t} & =\gamma|\boldsymbol{\mu}| B_{0} \sin \phi \\
\omega_{0}=\frac{d \theta}{d t} & =\gamma B_{0}
\end{aligned}
\tag{5}
$$

Final answer: $\omega_{0}=\gamma B_{0}$
Subfield: Modern Physics


Question: 1.  Let us consider an oven source of silver atoms, which has a small opening. The atoms stream out of the opening along $-y$ direction (see Figure below) and experience a spatial varying field $\boldsymbol{B}_{1}$. The field $\boldsymbol{B}_{1}$ has strong bias field component in the $z$ direction, where the atoms with different magnetic moment $\mu_{z}= \pm \gamma \hbar$ are split in the $z$ direction. At a distance $D$ from the oven source, a screen $S C_{1}$ is put to allow only spin up atoms to pass (blocking spin down atoms). Thus, at the instant after passing the screen, the atoms are prepared in spin up states. After the screen, the atoms enter a region of nonhomogenous field $\boldsymbol{B}_{2}$ where the atoms feel a force

$$
F_{x}=\mu_{x} C
$$

The field $\boldsymbol{B}_{2}$ has strong bias field component in the $x$ direction, where the atoms have magnetic moment $\mu_{x}= \pm \gamma \hbar$.



![](https://cdn.mathpix.com/cropped/2023_12_21_a0eec930b34aa88d323eg-1.jpg?height=437&width=597&top_left_y=347&top_left_x=798)

In order to determine $\mu_{x}$ by observing the splitting in $x$ direction, show that the following condition must be fulfilled:

$$
\frac{1}{\hbar}\left|\mu_{x}\right| \Delta x C t \gg 1
$$

where $t$ is the duration after leaving the screen $S C_{1}$ and $\Delta x$ is the opening width on $S C_{1}$.

Solution:
In the $x$ direction, the uncertainty in position due to the screen opening is $\Delta x$. According to the uncertainty principle, the atom momentum uncertainty $\Delta p_{x}$ is given by

$$
\Delta p_{x} \approx \frac{\hbar}{\Delta x},
$$

which translates into an uncertainty in the $x$ velocity of the atom,

$$
v_{x} \approx \frac{\hbar}{m \Delta x}
$$

Consequently, during the time of flight $t$ of the atoms through the device, the uncertainty in the width of the beam will grow by an amount $\delta x$ given by

$$
\delta x=\Delta v_{x} t \approx \frac{\hbar}{m \Delta x} t
$$



So, the width of the beams is growing linearly in time. Meanwhile, the two beams are separating at a rate determined by the force $F_{x}$ and the separation between the beams after a time $t$ becomes

$$
d_{x}=2 \times \frac{1}{2} \frac{F_{x}}{m} t^{2}=\frac{1}{m}\left|\mu_{x}\right| C t^{2}
$$

In order to be able to distinguish which beam a particle belongs to, the separation of the two beams must be greater than the widths of the beams; otherwise the two beams will overlap and it will be impossible to know what the $x$ component of the atom spin is. Thus, the condition must be satisfied is

$$
\begin{aligned}
d_{x} & \gg \delta x \\
\frac{1}{m}\left|\mu_{x}\right| C t^{2} & \gg \frac{\hbar}{m \Delta x} t, \\
\frac{1}{\hbar}\left|\mu_{x}\right| \Delta x C t & \gg 1 .
\end{aligned}
\tag{15}
$$

Final answer: 证明题
Subfield: Geometry


Question: Let $P$ and $P^{\prime}$ be two convex quadrilateral regions in the plane (regions contain their boundary). Let them intersect, with $O$ a point in the intersection. Suppose that for every line $\ell$ through $O$ the segment $\ell \cap P$ is strictly longer than the segment $\ell \cap P^{\prime}$. Is it possible that the ratio of the area of $P^{\prime}$ to the area of $P$ is greater than $1.9 ?$

Solution:
The answer is in the affirmative: Given a positive $\epsilon<2$, the ratio in question may indeed be greater than $2-\epsilon$.



To show this, consider a square $A B C D$ centred at $O$, and let $A^{\prime}, B^{\prime}$, and $C^{\prime}$ be the reflections of $O$ in $A, B$, and $C$, respectively. Notice that, if $\ell$ is a line through $O$, then the segments $\ell \cap A B C D$ and $\ell \cap A^{\prime} B^{\prime} C^{\prime}$ have equal lengths, unless $\ell$ is the line $A C$.



Next, consider the points $M$ and $N$ on the segments $B^{\prime} A^{\prime}$ and $B^{\prime} C^{\prime}$, respectively, such that $B^{\prime} M / B^{\prime} A^{\prime}=B^{\prime} N / B^{\prime} C^{\prime}=(1-\epsilon / 4)^{1 / 2}$. Finally, let $P^{\prime}$ be the image of the convex quadrangle $B^{\prime} M O N$ under the homothety of ratio $(1-\epsilon / 4)^{1 / 4}$ centred at $O$. Clearly, the quadrangles $P \equiv A B C D$ and $P^{\prime}$ satisfy the conditions in the statement, and the ratio of the area of $P^{\prime}$ to the area of $P$ is exactly $2-\epsilon / 2$.



<img_3582>
Subfield: Combinatorics


Question: $2 n$ distinct tokens are placed at the vertices of a regular $2 n$-gon, with one token placed at each vertex. A move consists of choosing an edge of the $2 n$-gon and interchanging the two tokens at the endpoints of that edge. Suppose that after a finite number of moves, every pair of tokens have been interchanged exactly once. Prove that some edge has never been chosen.

Solution:
Step 1. Enumerate all the tokens in the initial arrangement in clockwise circular order; also enumerate the vertices of the $2 n$-gon accordingly. Consider any three tokens $i<j<k$. At each moment, their cyclic order may be either $i, j, k$ or $i, k, j$, counted clockwise. This order changes exactly when two of these three tokens have been switched. Hence the order has been reversed thrice, and in the final arrangement token $k$ stands on the arc passing clockwise from token $i$ to token $j$. Thus, at the end, token $i+1$ is a counter-clockwise neighbor of token $i$ for all $i=1,2, \ldots, 2 n-1$, so the tokens in the final arrangement are numbered successively in counter-clockwise circular order.



This means that the final arrangement of tokens can be obtained from the initial one by reflection in some line $\ell$.



Step 2. Notice that each token was involved into $2 n-1$ switchings, so its initial and final vertices have different parity. Hence $\ell$ passes through the midpoints of two opposite sides of a $2 n$-gon; we may assume that these are the sides $a$ and $b$ connecting $2 n$ with 1 and $n$ with $n+1$, respectively.



During the process, each token $x$ has crossed $\ell$ at least once; thus one of its switchings has been made at edge $a$ or at edge $b$. Assume that some two its switchings were performed at $a$ and at $b$; we may (and will) assume that the one at $a$ was earlier, and $x \leq n$. Then the total movement of token $x$ consisted at least of: (i) moving from vertex $x$ to $a$ and crossing $\ell$ along $a$; (ii) moving from $a$ to $b$ and crossing $\ell$ along $b$; (iii) coming to vertex $2 n+1-x$. This tales at least $x+n+(n-x)=2 n$ switchings, which is impossible.



Thus, each token had a switching at exactly one of the edges $a$ and $b$.



Step 3. Finally, let us show that either each token has been switched at $a$, or each token has been switched at $b$ (then the other edge has never been used, as desired). To the contrary, assume that there were switchings at both $a$ and at $b$. Consider the first such switchings, and let $x$ and $y$ be the tokens which were moved clockwise during these switchings and crossed $\ell$ at $a$ and $b$, respectively. By Step 2, $x \neq y$. Then tokens $x$ and $y$ initially were on opposite sides of $\ell$.



Now consider the switching of tokens $x$ and $y$; there was exactly one such switching, and we assume that it has been made on the same side of $\ell$ as vertex $y$. Then this switching has been made after token $x$ had traced $a$. From this point on, token $x$ is on the clockwise arc from token $y$ to $b$, and it has no way to leave out from this arc. But this is impossible, since token $y$ should trace $b$ after that moment. A contradiction.
Subfield: Combinatorics


Question: Xenia and Sergey play the following game. Xenia thinks of a positive integer $N$ not exceeding 5000. Then she fixes 20 distinct positive integers $a_{1}, a_{2}, \ldots, a_{20}$ such that, for each $k=1,2, \ldots, 20$, the numbers $N$ and $a_{k}$ are congruent modulo $k$. By a move, Sergey tells Xenia a set $S$ of positive integers not exceeding 20 , and she tells him back the set $\left\{a_{k}: k \in S\right\}$ without spelling out which number corresponds to which index. How many moves does Sergey need to determine for sure the number Xenia thought of?

Solution:
Sergey can determine Xenia's number in 2 but not fewer moves.



We first show that 2 moves are sufficient. Let Sergey provide the set $\{17,18\}$ on his first move, and the set $\{18,19\}$ on the second move. In Xenia's two responses, exactly one number occurs twice, namely, $a_{18}$. Thus, Sergey is able to identify $a_{17}, a_{18}$, and $a_{19}$, and thence the residue of $N$ modulo $17 \cdot 18 \cdot 19=5814>5000$, by the Chinese Remainder Theorem. This means that the given range contains a single number satisfying all congruences, and Sergey achieves his goal.



To show that 1 move is not sufficient, let $M=\operatorname{lcm}(1,2, \ldots, 10)=2^{3} \cdot 3^{2} \cdot 5 \cdot 7=2520$. Notice that $M$ is divisible by the greatest common divisor of every pair of distinct positive integers not exceeding 20. Let Sergey provide the set $S=\left\{s_{1}, s_{2}, \ldots, s_{k}\right\}$. We show that there exist pairwise distinct positive integers $b_{1}, b_{2}, \ldots, b_{k}$ such that $1 \equiv b_{i}\left(\bmod s_{i}\right)$ and $M+1 \equiv b_{i-1}\left(\bmod s_{i}\right)$ (indices are reduced modulo $k$ ). Thus, if in response Xenia provides the set $\left\{b_{1}, b_{2}, \ldots, b_{k}\right\}$, then Sergey will be unable to distinguish 1 from $M+1$, as desired.



To this end, notice that, for each $i$, the numbers of the form $1+m s_{i}, m \in \mathbb{Z}$, cover all residues modulo $s_{i+1}$ which are congruent to $1(\equiv M+1)$ modulo $\operatorname{gcd}\left(s_{i}, s_{i+1}\right) \mid M$. Xenia can therefore choose a positive integer $b_{i}$ such that $b_{i} \equiv 1\left(\bmod s_{i}\right)$ and $b_{i} \equiv M+1\left(\bmod s_{i+1}\right)$. Clearly, such choices can be performed so as to make the $b_{i}$ pairwise distinct, as required.

Final answer: 2
Subfield: Number Theory


Question: A number of 17 workers stand in a row. Every contiguous group of at least 2 workers is a brigade. The chief wants to assign each brigade a leader (which is a member of the brigade) so that each worker's number of assignments is divisible by 4 . Prove that the number of such ways to assign the leaders is divisible by 17 .

Solution:
Assume that every single worker also forms a brigade (with a unique possible leader). In this modified setting, we are interested in the number $N$ of ways to assign leadership so that each worker's number of assignments is congruent to 1 modulo 4.



Consider the variables $x_{1}, x_{2}, \ldots, x_{17}$ corresponding to the workers. Assign each brigade (from the $i$-th through the $j$-th worker) the polynomial $f_{i j}=x_{i}+x_{i+1}+\cdots+x_{j}$, and form the product $f=\prod_{1 \leq i \leq j \leq 17} f_{i j}$. The number $N$ is the sum $\Sigma(f)$ of the coefficients of all monomials $x_{1}^{\alpha_{1}} x_{2}^{\alpha_{2}} \ldots x_{17}^{\alpha_{17}}$ in the expansion of $f$, where the $\alpha_{i}$ are all congruent to 1 modulo 4 . For any polynomial $P$, let $\Sigma(P)$ denote the corresponding sum. From now on, all polynomials are considered with coefficients in the finite field $\mathbb{F}_{17}$.



Recall that for any positive integer $n$, and any integers $a_{1}, a_{2}, \ldots, a_{n}$, there exist indices $i \leq j$ such that $a_{i}+a_{i+1}+\cdots+a_{j}$ is divisible by $n$. Consequently, $f\left(a_{1}, a_{2}, \ldots, a_{17}\right)=0$ for all $a_{1}, a_{2}, \ldots, a_{17}$ in $\mathbb{F}_{17}$.



Now, if some monomial in the expansion of $f$ is divisible by $x_{i}^{17}$, replace that $x_{i}^{17}$ by $x_{i}$; this does not alter the above overall vanishing property (by Fermat's Little Theorem), and preserves $\Sigma(f)$. After several such changes, $f$ transforms into a polynomial $g$ whose degree in each variable does not exceed 16 , and $g\left(a_{1}, a_{2}, \ldots, a_{17}\right)=0$ for all $a_{1}, a_{2}, \ldots, a_{17}$ in $\mathbb{F}_{17}$. For such a polynomial, an easy induction on the number of variables shows that it is identically zero. Consequently, $\Sigma(g)=0$, so $\Sigma(f)=0$ as well, as desired.
Subfield: Geometry


Question: Let $A B C$ be a triangle, let $D$ be the touchpoint of the side $B C$ and the incircle of the triangle $A B C$, and let $J_{b}$ and $J_{c}$ be the incentres of the triangles $A B D$ and $A C D$, respectively. Prove that the circumcentre of the triangle $A J_{b} J_{c}$ lies on the bisectrix of the angle $B A C$.

Solution:
Let the incircle of the triangle $A B C$ meet $C A$ and $A B$ at points $E$ and $F$, respectively. Let the incircles of the triangles $A B D$ and $A C D$ meet $A D$ at points $X$ and $Y$, respectively. Then $2 D X=D A+D B-A B=D A+D B-B F-A F=D A-A F$; similarly, $2 D Y=D A-A E=2 D X$. Hence the points $X$ and $Y$ coincide, so $J_{b} J_{c} \perp A D$.



Now let $O$ be the circumcentre of the triangle $A J_{b} J_{c}$. Then $\angle J_{b} A O=\pi / 2-\angle A O J_{b} / 2=$ $\pi / 2-\angle A J_{c} J_{b}=\angle X A J_{c}=\frac{1}{2} \angle D A C$. Therefore, $\angle B A O=\angle B A J_{b}+\angle J_{b} A O=\frac{1}{2} \angle B A D+$ $\frac{1}{2} \angle D A C=\frac{1}{2} \angle B A C$, and the conclusion follows.



<img_3637>



Fig. 1
Subfield: Number Theory


Question: Let $p \geq 5$ be a prime number. For a positive integer $k$ we denote by $R(k)$ the remainder of $k$ when divided by $p$. Determine all positive integers $a<p$ such that



$$

m+R(m a)>a

$$



for every $m=1,2, \ldots, p-1$.

Solution:
The required integers are $p-1$ along with all the numbers of the form $\lfloor p / q\rfloor, q=$ $2, \ldots, p-1$. In other words, these are $p-1$, along with the numbers $1,2, \ldots,\lfloor\sqrt{p}\rfloor$, and also the (distinct) numbers $\lfloor p / q\rfloor, q=2, \ldots,\left\lfloor\sqrt{p}-\frac{1}{2}\right\rfloor$.



We begin by showing that these numbers satisfy the conditions in the statement. It is readily checked that $p-1$ satisfies the required inequalities, since $m+R(m(p-1))=m+(p-m)=$ $p>p-1$ for all $m=1, \ldots, p-1$.



Now, consider any number $a$ of the form $a=\lfloor p / q\rfloor$, where $q$ is an integer greater than 1 but less than $p$; then $p=a q+r$ with $0<r<q$. Choose any integer $m \in(0, p)$ and write $m=x q+y$ with $x, y \in \mathbb{Z}, 0<y \leq q$ (notice that $x$ is nonnegative). Then



$$

R(m a)=R(a y+x a q)=R(a y+x p-x r)=R(a y-x r) .

$$



Since $a y-x r \leq a y \leq a q<p$, we obtain $R(a y-x r) \geq a y-x r$ and hence



$$

m+R(m a) \geq(x q+y)+(a y-x r)=x(q-r)+y(a+1) \geq a+1

$$



by $q>r$ and $y \geq 1$. Thus $a$ satisfies the required condition.



Finally, we show that if an integer $a \in(0, p-1)$ satisfies the required condition then $a$ is indeed of the form $a=\lfloor p / q\rfloor$ for some integer $q \in(0, p)$. This is clear for $a=1$, so we may (and will) assume that $a \geq 2$.



Write $p=a q+r$ with $q, r \in \mathbb{Z}$ and $0<r<a$; since $a \geq 2$ we have $q<p / 2$. Choose $m=q+1<p ;$ we have $R(m a)=R(a q+a)=R(p+(a-r))=a-r$, so



$$

a<m+R(m a)=q+1+a-r,

$$



which yields $r<q+1$. Moreover, if $r=q$, then $p=q(a+1)$ which is impossible by $1<a+1<p$. Thus $r<q$, and we have



$$

0 \leq \frac{p}{q}-a=\frac{r}{q}<1,

$$



which proves $a=\lfloor p / q\rfloor$.

Final answer: the required integers are $p-1$ along with all the numbers of the form $\lfloor p / q\rfloor, q=$ $2, \ldots, p-1$
Subfield: Number Theory


Question: 
Prove that every positive integer $n$ can be written uniquely in the form



$$

n=\sum_{j=1}^{2 k+1}(-1)^{j-1} 2^{m_{j}}

$$



where $k \geq 0$ and $0 \leq m_{1}<m_{2}<\cdots<m_{2 k+1}$ are integers.



This number $k$ is called the weight of $n$.

Solution:
We show by induction on the integer $M \geq 0$ that every integer $n$ in the range $-2^{M}+1$ through $2^{M}$ can uniquely be written in the form $n=\sum_{j=1}^{\ell}(-1)^{j-1} 2^{m_{j}}$ for some integers $\ell \geq 0$ and $0 \leq m_{1}<m_{2}<\cdots<m_{\ell} \leq M$ (empty sums are 0 ); moreover, in this unique representation $\ell$ is odd if $n>0$, and even if $n \leq 0$. The integer $w(n)=\lfloor\ell / 2\rfloor$ is called the weight of $n$.



Existence once proved, uniqueness follows from the fact that there are as many such representations as integers in the range $-2^{M}+1$ through $2^{M}$, namely, $2^{M+1}$.



To prove existence, notice that the base case $M=0$ is clear, so let $M \geq 1$ and let $n$ be an integer in the range $-2^{M}+1$ through $2^{M}$.



If $-2^{M}+1 \leq n \leq-2^{M-1}$, then $1 \leq n+2^{M} \leq 2^{M-1}$, so $n+2^{M}=\sum_{j=1}^{2 k+1}(-1)^{j-1} 2^{m_{j}}$ for some integers $k \geq 0$ and $0 \leq m_{1}<\cdots<m_{2 k+1} \leq M-1$ by the induction hypothesis, and $n=\sum_{j=1}^{2 k+2}(-1)^{j-1} 2^{m_{j}}$, where $m_{2 k+2}=M$.



The case $-2^{M-1}+1 \leq n \leq 2^{M-1}$ is covered by the induction hypothesis.



Finally, if $2^{M-1}+1 \leq n \leq 2^{M}$, then $-2^{M-1}+1 \leq n-2^{M} \leq 0$, so $n-2^{M}=\sum_{j=1}^{2 k}(-1)^{j-1} 2^{m_{j}}$ for some integers $k \geq 0$ and $0 \leq m_{1}<\cdots<m_{2 k} \leq M-1$ by the induction hypothesis, and $n=\sum_{j=1}^{2 k+1}(-1)^{j-1} 2^{m_{j}}$, where $m_{2 k+1}=M$.