### 数学竞赛代考|AIME代考美国数学邀请赛|Basic Counting Techniques

AIME资格认证的变化

AMC办公室将从2月下旬开始向学校邮寄2012年AMC 10和2012年AMC 12报告，并持续到3月初至3月中旬。 在该AMC 10和AMC 12报告中，将列出学校的AIME合格者名单。

AIME的目的是在AMC10或AMC12之外，为北美许多具有特殊数学能力的高中生提供进一步的挑战和认可。得分最高的美国公民和合法居住在美国和加拿大的学生（根据加权平均分，获得合格分数）被邀请参加美国数学竞赛。
AIME（美国数学邀请考试）是介于AMC10或AMC12和USAMO之间的考试。所有参加AMC 12的学生，如果在可能的150分中取得100分或以上的成绩，或在前5%的学生被邀请参加AIME考试。所有参加AMC 10的学生，在可能的150分中取得120分或以上，或进入前2.5%的学生也有资格参加AIME。本学年AIME I的日期为 ，AIME II的日期为 ， 。美国数学邀请考试没有额外的注册费，除非你选择参加第二次考试。额外的管理/运输费是要收取的，前10名学生的最低费用为，超过10名学生的最低费用为。这在AMC 10/12和AIME教师手册中有更详细的解释。

## 数学竞赛代考|AIME代考美国数学邀请赛|Casework and Complementary Counting

Two basic approaches for counting problems are casework and complementary counting. Unfortunately, “casework” means exactly what it means: subdivide a counting problem into manageable, organized cases. Complementary counting is a technique where we count the number of outcomes that we don’t want, then subtract that result from the total number of possible outcomes. Look for words such as “at least” or “not” – these often signal that complementary counting may be a viable method.

Example 1.3.1. (AIME II 2007 #1) A mathematical organization is producing a set of commemorative license plates. Each plate contains a sequence of five characters chosen from the four letters in AIME and the four digits in 2007. No character may appear in a sequence more times than it appears among the four letters in AIME or the four digits in 2007. A set of plates in which each possible sequence appears exactly once contains $N$ license plates. Find $\frac{N}{10}$.

Solution. Let’s break it down into cases based on how many letters there are. If there are 4 , there are 3 ways to choose the digit and 5! ways to arrange them, for a subtotal of 360 . If there are 3 , there are 4 ways to choose them and 3 ways to choose two different digits, plus one way to choose the same digits. This gives a subtotal of $4(3 \cdot 120+60)=1680$. If there are 2 , there are 6 ways to choose them and 1 way to choose three different digits and 2 ways otherwise. This gives a subtotal of $6(120+2 \cdot 60)=1440$. Finally, if there is only 1 letter, there are 4 ways to choose it and 60 ways to arrange it. This gives $N=360+1680+1440+60=3540 \Rightarrow 354$.

Example 1.3.2. (AIME I 2002 #1) Many states use a sequence of three letters followed by a sequence of three digits as their standard license-plate pattern. Given that each threeletter three-digit arrangement is equally likely, the probability that such a license plate will contain at least one palindrome (a three-letter arrangement or a three-digit arrangement that reads the same left-to-right as it does right-to-left) is $m / n$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution. We just need to calculate the probability that neither the digits nor the letters form a palindrome. In both cases, the third character just cannot equal the first. So the probability is $1-\left(\frac{9}{10}\right)\left(\frac{25}{26}\right)=\frac{7}{52} \Rightarrow 59$.

## 数学竞赛代考|AIME代考美国数学邀请赛|Examples

1. (HMMT Feb-2018-Combinatorics-1) Consider a $2 \times 3$ grid where each entry is one of 0,1 , and 2 . For how many such grids is the sum of the numbers in every row and in every column a multiple of 3 ? One valid grid is shown below.
$$\left[\begin{array}{lll} 1 & 2 & 0 \ 2 & 1 & 0 \end{array}\right]$$
2. (PUMaC-2012-Team-2.1.3) Suppose you draw 5 vertices of a convex pentagon (but not the sides!). Let $N$ be the number of ways you can draw at least 0 straight line segments between the vertices so that no two line segments intersect in the interior of the pentagon. What is $N-64$ ? (Note what the question is asking for! You have been warned!)
3. (AIME-2010-I-7) Define an ordered triple $(\mathcal{A}, \mathcal{B}, \mathcal{C})$ of sets to be minimally intersecting if $|\mathcal{A} \cap \mathcal{B}|=|\mathcal{B} \cap \mathcal{C}|=|\mathcal{C} \cap \mathcal{A}|=1$ and $\mathcal{A} \cap \mathcal{B} \cap \mathcal{C}=\emptyset$. For example, $({1,2},{2,3},{1,3,4})$ is a minimally intersecting triple. Let $N$ be the number of minimally intersecting ordered triples of sets for which each set is a subset of ${1,2,3,4,5,6,7}$. Find the remainder when $N$ is divided by 1000 .
Note: $|S|$ represents the number of elements in the set $S$.
4. (HMMT Feb-2016-Guts-6) Consider a $2 \times n$ grid of points and a path consisting of $2 n-1$ straight line segments connecting all these $2 n$ points, starting from the bottom left corner and ending at the upper right corner. Such a path is called efficient if each point is only passed through once and no two line segments intersect. How many efficient paths are there when $n=2016$ ?
5. (AIME-2016-I-8) For a permutation $p=\left(a_1, a_2, \ldots, a_9\right)$ of the digits $1,2, \ldots, 9$, let $s(p)$ denote the sum of the three 3 -digit numbers $a_1 a_2 a_3, a_4 a_5 a_6$, and $a_7 a_8 a_9$. Let $m$ be the minimum value of $s(p)$ subject to the condition that the units digit of $s(p)$ is 0 . Let $n$ denote the number of permutations $p$ with $s(p)=m$. Find $|m-n|$.
6. (HMMT Feb-2010-Combinatorics-4) Manya has a stack of $85=1+4+16+64$ blocks comprised of 4 layers (the $k$ th layer from the top has $4^{k-1}$ blocks; see the diagram below). Each block rests on 4 smaller blocks, each with dimensions half those of the larger block. Laura removes blocks one at a time from this stack, removing only block that currently have no blocks on top of them. Find the number of ways Laura can remove precisely 5 blocks from Manya’s stack (the order in which they are removed matters).

## 数学竞赛代考|AIME代考美国数学邀请赛|Examples

1. (HMMT Feb-2018-Combinatorics-1) 考虑一个2×3网格，其中每个条目都是 0,1 和 2 之一。有多少个这样的格子，每行每列的数字之和是3的倍数？一个有效的网格如下所示。
[120 210]
2. (PUMaC-2012-Team-2.1.3) 假设您绘制了一个凸五边形的 5 个顶点（但不是边！）。让否是您可以在顶点之间绘制至少 0 条直线段的方式的数量，以便没有两条线段在五边形的内部相交。什么是否−64？（请注意问题的要求！您已被警告！）
3. (AIME-2010-I-7) 定义一个有序的三元组(一种,乙,C)集合的最小相交如果|一种∩乙|=|乙∩C|=|C∩一种|=1和一种∩乙∩C=∅. 例如，(1,2,2,3,1,3,4)是最小相交的三元组。让否是集合的最小相交有序三元组的数量，其中每个集合都是1,2,3,4,5,6,7. 求余数时否除以 1000 。
笔记：|小号|表示集合中元素的数量小号.
4. (HMMT Feb-2016-Guts-6) 考虑一个2×n点网格和路径组成2n−1连接所有这些的直线段2n点，从左下角开始，到右上角结束。如果每个点只通过一次并且没有两条线段相交，则这样的路径称为有效路径。有多少条有效路径n=2016 ?
5. (AIME-2016-I-8) 对于排列p=(一种1,一种2,…,一种9)的数字1,2,…,9， 让秒(p)表示三个 3 位数的总和一种1一种2一种3,一种4一种5一种6， 和一种7一种8一种9. 让米是的最小值秒(p)条件是秒(p)是 0 。让n表示排列的数量p和秒(p)=米. 寻找|米−n|.
6. (HMMT Feb-2010-Combinatorics-4) Manya 有一堆85=1+4+16+64由 4 层组成的块（k从上数第 th 层有4k−1块; 见下图）。每个块都位于 4 个较小的块上，每个块的尺寸都是较大块的一半。劳拉一次从这个堆栈中移除一个块，只移除当前没有块的块。找出 Laura 可以从 Manya 的堆叠中恰好移除 5 个方块的方法数（移除它们的顺序很重要）。

## 有限元方法代写

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## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。