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

All $n$-digit words from the alphabet $\{0, 1\}$ considered. These $2^n$ words should be in a sequence $w_0, w_1, w_2, ..., w_{2^-1}$ be arranged that $w_m$ from $w_{m-1}$ by changing of a single ornament ($m = 1, 2, 3, ..., 2n-1$). Prove that the following algorithm achievesthis : a) Start with $w_0 = 000... 00$. b) Let $w_{m-1} = a_1a_2a_3 ... a_n$ with $a_i \in \{0; 1\}$, $i = 1, 2, 3, ..., n$. Determine the exponent $e(m)$ of the highest power of two dividing $m$ and set $j = e(m)+1$. In $w_{m-1}$ replace the ornament $a_j$ with $1-aj$. this is now $w_m$.

There are 365 cards with 365 different numbers. Each step, we can choose 3 cards $a_{i},a_{j},a_{k}$ and we know the order of them (examble: $a_{i}<a_{j}<a_{k}$). With 2000 steps, can we order 365 cards from smallest to biggest??

Let $P(x),Q(x)$ be monic polynomials with integer coeeficients. Let $a_n=n!+n$ for all natural numbers $n$. Show that if $\frac{P(a_n)}{Q(a_n)}$ is an integer for all positive integer $n$ then $\frac{P(n)}{Q(n)}$ is an integer for every integer $n\neq0$. \\ [i]Hint (given in question): Try applying division algorithm for polynomials [/i]

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 $1$.

There are $256$ lumps of metal that have different weights in pairs. With the help of a beam balance , one may now compare every two lumps. Find the smallest number $m$ such that you can be sure to find the heaviest as well as the lightest lump with the weighing process.

For integers $n \ge k \ge 0$ we define the [i]bibinomial coefficient[/i] $\left( \binom{n}{k} \right)$ by \[ \left( \binom{n}{k} \right) = \frac{n!!}{k!!(n-k)!!} .\] Determine all pairs $(n,k)$ of integers with $n \ge k \ge 0$ such that the corresponding bibinomial coefficient is an integer. [i]Remark: The double factorial $n!!$ is defined to be the product of all even positive integers up to $n$ if $n$ is even and the product of all odd positive integers up to $n$ if $n$ is odd. So e.g. $0!! = 1$, $4!! = 2 \cdot 4 = 8$, and $7!! = 1 \cdot 3 \cdot 5 \cdot 7 = 105$.[/i]

We are given $ 3^{2k}$ apparently identical coins,one of which is fake,being lighter than the others. We also dispose of three apparently identical balances without weights, one of which is broken (and yields outcomes unrelated to the actual situations). How can we find the fake coin in $ 3k\plus{}1$ weighings?

The image shown below is a cross with length 2. If length of a cross of length $k$ it is called a $k$-cross. (Each $k$-cross ahs $6k+1$ squares.) [img]http://aycu08.webshots.com/image/4127/2003057947601864020_th.jpg[/img] a) Prove that space can be tiled with $1$-crosses. b) Prove that space can be tiled with $2$-crosses. c) Prove that for $k\geq5$ space can not be tiled with $k$-crosses.

On a $n \times n$ grid, each edge are written with $=$ or $\neq$. We need to filled every cells with color black or white. Find the largest constant $k$, such that for every $n>777771449$ and any layout of $=$ and $\neq$, we can always find a way to colored every cells, such that at least $k \cdot 2n(n-1)$ neighboring cells, there colors conform to the symbols on the edge. (Namely, two cells are filled with the same color if $=$ was written on their edge; two cells are filled with different colors if $\neq$ was written on their edge) [i]Proposed by chengbilly & sn6dh[/i]

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 $1$.

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 $AFBT$, $\lim_{k \to \infty}f(\text{x}_k)=f^*$

Alice and the Cheshire Cat play a game. At each step, Alice either (1) gives the cat a penny, which causes the cat to change the number of (magic) beans that Alice has from $ n$ to $ 5n$ or (2) gives the cat a nickel, which causes the cat to give Alice another bean. Alice wins (and the cat disappears) as soon as the number of beans Alice has is greater than $ 2008$ and has last two digits $ 42$. What is the minimum number of cents Alice can spend to win the game, assuming she starts with 0 beans?

Suppose that $n$ people $A_1$, $A_2$, $\ldots$, $A_n$, ($n \geq 3$) are seated in a circle and that $A_i$ has $a_i$ objects such that \[ a_1 + a_2 + \cdots + a_n = nN \] where $N$ is a positive integer. In order that each person has the same number of objects, each person $A_i$ is to give or to receive a certain number of objects to or from its two neighbours $A_{i-1}$ and $A_{i+1}$. (Here $A_{n+1}$ means $A_1$ and $A_n$ means $A_0$.) How should this redistribution be performed so that the total number of objects transferred is minimum?

There is $n$ black points in the plane.We do the following algorithm: Start from any point from those $n$ points and colour it red. Then connect this point to the nearest black point available and colour this new point red. Then do the same with this point but at any step , but you are never allowed to draw a line which intersects on of the current drawn segments. If you reach an intersection , the algorithm is over. Is it true that for any $n$ and at any initial position , we can start from a point st in the algorithm , we reach all the points?

A set of three nonnegative integers $\{x, y, z \}$ with $x<y<z$ is called historic if $\{z-y, y-x\}=\{1776,2001\}$. Show that the set of all nonnegative integers can be written as the union of pairwise disjoint historic sets.

2010 MOPpers are assigned numbers 1 through 2010. Each one is given a red slip and a blue slip of paper. Two positive integers, A and B, each less than or equal to 2010 are chosen. On the red slip of paper, each MOPper writes the remainder when the product of A and his or her number is divided by 2011. On the blue slip of paper, he or she writes the remainder when the product of B and his or her number is divided by 2011. The MOPpers may then perform either of the following two operations: [list] [*] Each MOPper gives his or her red slip to the MOPper whose number is written on his or her blue slip. [*] Each MOPper gives his or her blue slip to the MOPper whose number is written on his or her red slip.[/list] Show that it is always possible to perform some number of these operations such that each MOPper is holding a red slip with his or her number written on it. [i]Brian Hamrick.[/i]

Let $n_1,n_2, \cdots, n_{26}$ be pairwise distinct positive integers satisfying (1) for each $n_i$, its digits belong to the set $\{1,2\}$; (2) for each $i,j$, $n_i$ can't be obtained from $n_j$ by adding some digits on the right. Find the smallest possible value of $\sum_{i=1}^{26} S(n_i)$, where $S(m)$ denotes the sum of all digits of a positive integer $m$.

Let $n$  be a positive integer. A row of $n+ 1$ squares is written from left to right, numbered $0, 1, 2, \cdots, n$ Two frogs, named Alphonse and Beryl, begin a race starting at square 0. For each second that passes, Alphonse and Beryl make a jump to the right according to the following rules: if there are at least eight squares to the right of Alphonse, then Alphonse jumps eight squares to the right. Otherwise, Alphonse jumps one square to the right. If there are at least seven squares to the right of Beryl, then Beryl jumps seven squares to the right. Otherwise, Beryl jumps one square to the right. Let A(n) and B(n) respectively denote the number of seconds for Alphonse and Beryl to reach square n. For example, A(40) = 5 and B(40) = 10. (a) Determine an integer n>200 for which $B(n) <A(n)$. (b) Determine the largest integer n for which$ B(n) \le A(n)$.

a)Does there exist for any rational number $\frac{a}{b}$ some rational numbers $x_1,x_2,....x_n$ such that $x_1*x_2*....*x_n=1$ and $x_1+x_2+....+x_n=\frac{a}{b}$ a)Does there exist for any rational number $\frac{a}{b}$ some rational numbers $x_1,x_2,....x_n$ such that $x_1*x_2*....*x_n=\frac{a}{b}$ and $x_1+x_2+....+x_n=1$

Let $n > 1$ be an integer. In a circular arrangement of $n$ lamps $L_0, \ldots, L_{n-1},$ each of of which can either ON or OFF, we start with the situation where all lamps are ON, and then carry out a sequence of steps, $Step_0, Step_1, \ldots .$ If $L_{j-1}$ ($j$ is taken mod $n$) is ON then $Step_j$ changes the state of $L_j$ (it goes from ON to OFF or from OFF to ON) but does not change the state of any of the other lamps. If $L_{j-1}$ is OFF then $Step_j$ does not change anything at all. Show that: (i) There is a positive integer $M(n)$ such that after $M(n)$ steps all lamps are ON again, (ii) If $n$ has the form $2^k$ then all the lamps are ON after $n^2-1$ steps, (iii) If $n$ has the form $2^k + 1$ then all lamps are ON after $n^2 - n + 1$ steps.

Let $ k \equal{} \sqrt[3]{3}$. a, Find all polynomials $ p(x)$ with rationl coefficients whose degree are as least as possible such that $ p(k \plus{} k^2) \equal{} 3 \plus{} k$. b, Does there exist a polynomial $ p(x)$ with integer coefficients satisfying $ p(k \plus{} k^2) \equal{} 3 \plus{} k$

If $a_1, a_2, \cdots, a_7$ are not necessarily distinct real numbers such that $1 < a_i < 13$ for all $i$, then show that we can choose three of them such that they are the lengths of the sides of a triangle.

Given a set $X$ and a function $f: X \rightarrow X$, for each $x \in X$ we define $f^1(x)=f(x)$ and, for each $j \ge 1$, $f^{j+1}(x)=f(f^j(x))$. We say that $a \in X$ is a fixed point of $f$ if $f(a)=a$. For each $x \in \mathbb{R}$, let $\pi (x)$ be the quantity of positive primes lesser or equal to $x$. Given an positive integer $n$, we say that $f: \{1,2, \dots, n\} \rightarrow \{1,2, \dots, n\}$ is [i]catracha[/i] if $f^{f(k)}(k)=k$, for every $k=1, 2, \dots n$. Prove that: (a) If $f$ is catracha, $f$ has at least $\pi (n) -\pi (\sqrt{n}) +1$ fixed points. (b) If $n \ge 36$, there exists a catracha function $f$ with exactly $ \pi (n) -\pi (\sqrt{n}) + 1$ fixed points.

In the nation of Onewaynia, certain pairs of cities are connected by roads. Every road connects exactly two cities (roads are allowed to cross each other, e.g., via bridges). Some roads have a traffic capacity of 1 unit and other roads have a traffic capacity of 2 units. However, on every road, traffic is only allowed to travel in one direction. It is known that for every city, the sum of the capacities of the roads connected to it is always odd. The transportation minister needs to assign a direction to every road. Prove that he can do it in such a way that for every city, the difference between the sum of the capacities of roads entering the city and the sum of the capacities of roads leaving the city is always exactly one. [i]Proposed by Zuming Feng and Yufei Zhao[/i]