In a small town, there are $n \times n$ houses indexed by $(i, j)$ for $1 \leq i, j \leq n$ with $(1, 1)$ being the house at the top left corner, where $i$ and $j$ are the row and column indices, respectively. At time 0, a fire breaks out at the house indexed by $(1, c)$, where $c \leq \frac{n}{2}$. During each subsequent time interval $[t, t+1]$, the fire fighters defend a house which is not yet on fire while the fire spreads to all undefended [i]neighbors[/i] of each house which was on fire at time t. Once a house is defended, it remains so all the time. The process ends when the fire can no longer spread. At most how many houses can be saved by the fire fighters? A house indexed by $(i, j)$ is a [i]neighbor[/i] of a house indexed by $(k, l)$ if $|i - k| + |j - l|=1$.

If the Highest Common Divisor of $ 6432$ and $ 132$ is diminished by $ 8$, it will equal: $ \textbf{(A)}\ \minus{}6 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ \minus{}2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$

Let $S$ be a set of $n$ points in the coordinate plane. Say that a pair of points is [i]aligned[/i] if the two points have the same $x$-coordinate or $y$-coordinate. Prove that $S$ can be partitioned into disjoint subsets such that (a) each of these subsets is a collinear set of points, and (b) at most $n^{3/2}$ unordered pairs of distinct points in $S$ are aligned but not in the same subset.

Each cell of an $m\times n$ board is filled with some nonnegative integer. Two numbers in the filling are said to be [i]adjacent[/i] if their cells share a common side. (Note that two numbers in cells that share only a corner are not adjacent). The filling is called a [i]garden[/i] if it satisfies the following two conditions: (i) The difference between any two adjacent numbers is either $0$ or $1$. (ii) If a number is less than or equal to all of its adjacent numbers, then it is equal to $0$. Determine the number of distinct gardens in terms of $m$ and $n$.

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 $n_1,n_2, \cdots, n_{26}$ 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 $m$.

Prove that in a plane, arbitrary $ n$ points can be overlapped by discs that the sum of all the diameters is less than $ n$, and the distances between arbitrary two are greater than $ 1$. (where the distances between two discs that have no common points are defined as that the distances between its centers subtract the sum of its radii; the distances between two discs that have common points are zero)

Larry and Rob are two robots travelling in one car from Argovia to Zillis. Both robots have control over the steering and steer according to the following algorithm: Larry makes a 90 degrees left turn after every $ \ell$ kilometer driving from start, Rob makes a 90 degrees right turn after every $ r$ kilometer driving from start, where $ \ell$ and $ r$ are relatively prime positive integers. In the event of both turns occurring simultaneously, the car will keep going without changing direction. Assume that the ground is flat and the car can move in any direction. Let the car start from Argovia facing towards Zillis. For which choices of the pair ($ \ell$, $ r$) is the car guaranteed to reach Zillis, regardless of how far it is from Argovia?

Let $ P_1$ be a regular $ r$-gon and $ P_2$ be a regular $ s$-gon $ (r\geq s\geq 3)$ such that each interior angle of $ P_1$ is $ \frac {59}{58}$ as large as each interior angle of $ P_2$. What's the largest possible value of $ s$?

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

What is the rightmost nonzero digit of $1000000!$?

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.

Consider an odd prime $p$ and a positive integer $N < 50p$. Let $a_1, a_2, \ldots , a_N$ be a list of positive integers less than $p$ such that any specific value occurs at most $\frac{51}{100}N$ times and $a_1 + a_2 + \cdots· + a_N$ is not divisible by $p$. Prove that there exists a permutation $b_1, b_2, \ldots , b_N$ of the $a_i$ such that, for all $k = 1, 2, \ldots , N$, the sum $b_1 + b_2 + \cdots + b_k$ is not divisible by $p$. [i]Will Steinberg, United Kingdom[/i]

Let $G$ be a simple graph with $3n^2$ vertices ($n\geq 2$). It is known that the degree of each vertex of $G$ is not greater than $4n$, there exists at least a vertex of degree one, and between any two vertices, there is a path of length $\leq 3$. Prove that the minimum number of edges that $G$ might have is equal to $\frac{(7n^2- 3n)}{2}$.

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]