In a country, mathematicians chose an $\alpha> 2$ and issued coins in denominations of 1 ruble, as well as $\alpha ^k$ rubles for each positive integer k. $\alpha$ was chosen so that the value of each coins, except the smallest, was irrational. Is it possible that any natural number of rubles can be formed with at most 6 of each denomination of coins?

Last year, Master Cheung is famous for multi-rotation. This year, he comes to DAMO to make noodles for sweeping monk. One day, software engineer Xiao Li talks with Master Cheung about his job. Xiao Li mainly researches and designs the algorithm to adjust the paramter of different kinds of products. These paramters can normally be obtainly by minimising loss function $f$ on $\mathbb{R}^n$. In the recent project of Xiao Li, this loss function is obtained by other topics. For safety consideration and technique reasons, this topic makes Xiao Li difficult to find the interal details of the function. They only provide a port to calculate the value of $f(\text x)$ for any $\text x\in\mathbb{R}^n$. Therefore, Xiao Li must only use the value of the function to minimise $f$. Also, every times calculating the value of $f$ will use a lot of calculating resources. It is good to know that the dimension $n$ is not very high (around $10$). Also, colleague who provides the function tells Xiao Li to assume $f$ is smooth first. This problem reminds Master Cheung of his antique radio. If you want to hear a programme from the radio, you need to turn the knob of the radio carefully. At the same time, you need to pay attention to the quality of the radio received, until the quality is the best. In this process, no one knows the relationship between the angle of turning the knob and the quality of the radio received. Master Cheung and Xiao Li realizes that minimising $f$ is same as adjusting the machine with multiple knobs: Assume every weight of $\text x$ is controlled by a knob. $f(\text x)$ is a certain performance of the machine. We only need to adjust every knobs again and again and observes the value of $f$ in the same time. Maybe there is hope to find the best $\text x$. As a result, two people suggest an iteration algorithm (named Automated Forward/Backward Tuning, $\text{AFBT}$, to minimise $f$. In $k$-th iteration, the algorithm adjusts the individual weight of $\text{x}_k$ to $2n$ points $\{\text x_k\pm t_k\text e^i:i=1,...,n\}$, where $t_k$ is the step size; then, make $y_k$ be the smallest one among the value of the function of thosse points. Then check if $\text y_k$ sufficiently makes $f$ decrease; then, take $\text x_{k+1}=\text y_k$, then make the step size doubled. Otherwise, make $\text x_{k+1}=\text x_k$ and makes the step size decrease in half. In the algorithm, $\text e^i$ is the $i$-th coordinate vector in $\mathbb{R}^n$. The weight of $i$-th is $1$. Others are $0$; $\mathbf{1}(\cdot)$ is indicator function. If $f(\text x_k)-f(\text y_k)$ is at least the square of $t_k$, then take the value of $\mathbf{1}(f(\text k)-f(y_k)\ge t^2_k)$ as $1$. Otherwise, take it as $0$. $\text{AFBT}$ algorithm Input $\text{x}_0\in \mathbb{R}^n$, $t_0>0$. For $k=0, 1, 2, ...$, perform the following loop: 1: #Calculate loss function. 2: $s_k:=\mathbb{1}[f(\text{x}_k)-f(\text{y}_k)\ge t^2_k]$ #Is it sufficiently decreasing? Yes: $s_k=1$; No: $s_k=0$. 3: $\text{x}_{k+1}:=(1-s_k)\text{x}_k+s_k\text{y}_k$ #Update the point of iteration. 4: $t_{k+1}:=2^{2S_k-1}t_k$ #Update step size. $s_k=1$: Step size doubles; $s_k=0$: Step size decreases by half. Now, we made assumption to the loss function $f:\mathbb{R}^n\to \mathbb{R}$. Assumption 1. Let $f$ be a convex function. For any $\text{x}, \text{y}\in \mathbb{R}^n$ and $\alpha \in [0, 1]$, we have $f((1-\alpha)\text{x}+\text{y})\le (1-\alpha)f(\text{x})+\alpha f(\text{y})$. Assumption 2. $f$ is differentiable on $\mathbb{R}^n$ and $\nabla f$ is L-Lipschitz continuous on $\mathbb{R}^n$. Assumption 3. The level set of $f$ is bounded. For any $\lambda\in\mathbb{R}$, set $\{\text x\in \mathbb{R}^n:f(\text x)\le \lambda\}$ is all bounded. Based on assumption 1 and 2, we can prove that $\left\langle \nabla f(\text x),\text y-\text x \right\rangle \le f(\text y)-f(\text x)\le \left\langle \nabla f(\text x),\text y-\text x\right\rangle+\frac{L}{2}||\text x-\text y||^2$ You can refer to any convex analysis textbook for more properties of convex function. Prove that under the assumption 1-3, for $AFBT$, $\lim_{k \to \infty}f(\text{x}_k)=f^*$

Let $n$ be an positive integer. Find the smallest integer $k$ with the following property; Given any real numbers $a_1 , \cdots , a_d $ such that $a_1 + a_2 + \cdots + a_d = n$ and $0 \le a_i \le 1$ for $i=1,2,\cdots ,d$, it is possible to partition these numbers into $k$ groups (some of which may be empty) such that the sum of the numbers in each group is at most $1$.

Let us consider a first-order language $L$ with a ternary predicate $\operatorname{Plus}$. Hence (well-formed) formulas of $L$ are built of symbols for variables, logical connectives, quantifiers, brackets, and the predicate symbol $\operatorname{Plus}$. $$(\exists x_1)(\forall x_2):\operatorname{Plus}(x_2,x_1,x_2)\wedge(\forall x_3):\neg\operatorname{Plus}(x_1,x_3,x_3)$$ is an example of such a formula. Recall that a formula is [i]closed[/i] iff each variable symbol occurs within the scope of a quantifier. Show that there exists an algorithm which decides whether or not a given closed formula of $L$ is true for the set $\mathbb N$ of natural numbers ($\{0,1,2,\ldots\}$) where $\operatorname{Plus}(x,y,z)$ is interpreted as $x+y=z$.

Find a method by which one can compute the coefficients of $P(x) = x^6 + a_1x^5 + \cdots+ a_6$ from the roots of $P(x) = 0$ by performing not more than $15$ additions and $15$ multiplications.

For a polynomial $ P(x)$ with integer coefficients, $ r(2i \minus{} 1)$ (for $ i \equal{} 1,2,3,\ldots,512$) is the remainder obtained when $ P(2i \minus{} 1)$ is divided by $ 1024$. The sequence \[ (r(1),r(3),\ldots,r(1023)) \] is called the [i]remainder sequence[/i] of $ P(x)$. A remainder sequence is called [i]complete[/i] if it is a permutation of $ (1,3,5,\ldots,1023)$. Prove that there are no more than $ 2^{35}$ different complete remainder sequences.

Prove that there exist two infinite sequences $ \{a_n\}_{n\ge 1}$ and $ \{b_n\}_{n\ge 1}$ of positive integers such that the following conditions hold simultaneously: $ (i)$ $ 0 < a_1 < a_2 < a_3 < \cdots$; $ (ii)$ $ a_n < b_n < a_n^2$, for all $ n\ge 1$; $ (iii)$ $ a_n \minus{} 1$ divides $ b_n \minus{} 1$, for all $ n\ge 1$ $ (iv)$ $ a_n^2 \minus{} 1$ divides $ b_n^2 \minus{} 1$, for all $ n\ge 1$ [19 points out of 100 for the 6 problems]

Let $ x \in \mathbb{Q}^+$. Prove that there exits $\alpha_1,\alpha_2,...,\alpha_k \in \mathbb{N}$ and pairwe distinct such that \[x= \sum_{i=1}^{k} \frac{1}{\alpha_i}\]

Let $n$ be a positive integer. The numbers $\{1, 2, ..., n^2\}$ are placed in an $n \times n$ grid, each exactly once. The grid is said to be [i]Muirhead-able[/i] if the sum of the entries in each column is the same, but for every $1 \le i,k \le n-1$, the sum of the first $k$ entries in column $i$ is at least the sum of the first $k$ entries in column $i+1$. For which $n$ can one construct a Muirhead-able array such that the entries in each column are decreasing? [i]Proposed by Evan Chen[/i]

We are given a bag of sugar, a two-pan balance, and a weight of $1$ gram. How do we obtain $1$ kilogram of sugar in the smallest possible number of weighings?

We have $n$ countries. Each country have $m$ persons who live in that country ($n>m>1$). We divide $m \cdot n$ persons into $n$ groups each with $m$ members such that there don't exist two persons in any groups who come from one country. Prove that one can choose $n$ people into one class such that they come from different groups and different countries.

Given a positive integer $n$, let $a_1,a_2,..,a_n$ be a sequence of nonnegative integers. A sequence of one or more consecutive terms of $a_1,a_2,..,a_n$ is called $dragon$ if their aritmetic mean is larger than 1. If a sequence is a $dragon$, then its first term is the $head$ and the last term is the $tail$. Suppose $a_1,a_2,..,a_n$ is the $head$ or/and $tail$ of some $dragon$ sequence; determine the minimum value of $a_1+a_2+\cdots +a_n$ in terms of $n$.

Set $S_n = \sum_{p=1}^n (p^5+p^7)$. Determine the greatest common divisor of $S_n$ and $S_{3n}.$

Find all positive integers $n$ for which there exist two distinct numbers of $n$ digits, $\overline{a_1a_2\ldots a_n}$ and $\overline{b_1b_2\ldots b_n}$, such that the number of $2n$ digits $\overline{a_1a_2\ldots a_nb_1b_2\ldots b_n}$ is divisible by $\overline{b_1b_2\ldots b_na_1a_2\ldots a_n}$.

Prove that every bijective function $ f: \mathbb{Z}\rightarrow\mathbb{Z}$ can be written in the way $ f\equal{}u\plus{}v$ where $ u,v: \mathbb{Z}\rightarrow\mathbb{Z}$ are bijective functions.

Let $X$ be a set with $n$ elements. Find the largest number of subsets of $X$, each with $3$ elements, so that no two of them are disjoint.

Find all functions $f:\mathbb{N}\rightarrow \mathbb{N}$ such that for all $m,n\in\mathbb{N}$, \[(2^m+1)f(n)f(2^mn)=2^mf(n)^2+f(2^mn)^2+(2^m-1)^2n. \]

Define real number $y$ as the fractional part of real number $x$ such that $0\leq y<1$ and $x-y$ is integer. Denote this by $<x>$. For real number $a$, define an infinite sequence $\{a_n\}\ (n=1,\ 2,\ 3,\ \cdots)$ inductively as follows. (i) $a_1=<a>$ (ii) If $a\n\neq 0$, then $a_{n+1}=\left<\frac{1}{a_n}\right>$, if $a_n=0$, then $a_{n+1}=0$. (1) For $a=\sqrt{2}$, find $a_n$. (2) For any natural number $n$, find real number $a\geq \frac 13$ such that $a_n=a$. (3) Let $a$ be a rational number. When we express $a=\frac{p}{q}$ with integer $p$, natural number $q$, prove that $a_n=0$ for any natural number $n\geq q$. [i]2011 Tokyo University entrance exam/Science, Problem 2[/i]

Prove that any positive integer can be represented as a sum of Fibonacci numbers, no two of which are consecutive.

Find all prime numbers $ p,q,r$, such that $ \frac{p}{q}\minus{}\frac{4}{r\plus{}1}\equal{}1$

Two players play alternately on a $ 5 \times 5$ board. The first player always enters a $ 1$ into an empty square and the second player always enters a $ 0$ into an empty square. When the board is full, the sum of the numbers in each of the nine $ 3 \times 3$ squares is calculated and the first player's score is the largest such sum. What is the largest score the first player can make, regardless of the responses of the second player?

Determine whether it's possible to cover a $K_{2012}$ with a) 1000 $K_{1006}$'s; b) 1000 $K_{1006,1006}$'s. [i]David Yang.[/i]

Let $\alpha_1,\ldots,\alpha_d,\beta_1,\ldots,\beta_e\in\mathbb{C}$ be such that the polynomials $f_1(x) =\prod_{i=1}^d(x-\alpha_i)$ and $f_2(x) =\prod_{i=1}^e(x-\beta_i)$ have integer coefficients. Suppose that there exist polynomials $g_1, g_2 \in\mathbb{Z}[x]$ such that $f_1g_1 +f_2g_2 = 1$. Prove that $\left|\prod_{i=1}^d \prod_{j=1}^e (\alpha_i - \beta_j)\right|=1$

A number of robots are placed on the squares of a finite, rectangular grid of squares. A square can hold any number of robots. Every edge of each square of the grid is classified as either passable or impassable. All edges on the boundary of the grid are impassable. You can give any of the commands up, down, left, or right. All of the robots then simultaneously try to move in the specified direction. If the edge adjacent to a robot in that direction is passable, the robot moves across the edge and into the next square. Otherwise, the robot remains on its current square. You can then give another command of up, down, left, or right, then another, for as long as you want. Suppose that for any individual robot, and any square on the grid, there is a finite sequence of commands that will move that robot to that square. Prove that you can also give a finite sequence of commands such that all of the robots end up on the same square at the same time.

There are $ 63$ points arbitrarily on the circle $ \mathcal{C}$ with its diameter being $ 20$. Let $ S$ denote the number of triangles whose vertices are three of the $ 63$ points and the length of its sides is no less than $ 9$. Fine the maximum of $ S$.