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AMC数学竞赛真题2016年B 15-16

2018-11-29 重点归纳

AMC10数学竞赛是美国高中数学竞赛中的一项，是针对高中一年级及初中三年级学生的数学测试，该竞赛开始于2000年，分A赛和B赛，于每年的2月初和2月中举行，学生可任选参加一项即可。不管是对高校申请还是今后在数学领域的发展都极其有利！那么接下来跟随小编来看一下AMC10数学竞赛真题以及官方解答吧：

Problem 15

All the numbers $1, 2, 3, 4, 5, 6, 7, 8, 9$ are written in a $3\times3$ array of squares, one number in each square, in such a way that if two numbers are consecutive then they occupy squares that share an edge. The numbers in the four corners add up to $18$ . What is the number in the center?

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Solution 1 - Trial and Error

Quick testing shows that $3~2~1$ $4~7~8$ $5~6~9$ is a valid solution. $3+1+5+9 = 18$ , and the numbers follow the given condition. The center number is found to be $\boxed{7}$ .

Solution 2

First let the numbers be $1 ~8~ 7$ $2 ~ 9 ~6$ $3 ~ 4~ 5$ with the numbers $1-8$ around the outsides and $9$ in the middle. We see that the sum of the four corner numbers is $16$ . If we switch $7$ and $9$ , then the corner numbers will add up to $18$ and the consecutive numbers will still be touching each other. The answer is $\boxed{7}$ .

Solution 3

Consecutive numbers share an edge. That means that it is possible to walk from $1$ to $9$ by single steps north, south, east, or west. Consequently, the squares in the diagram with different shades have different parity:

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But there are only four even numbers in the set, so the five darker squares must contain the odd numbers, which sum to $1+3+5+7+9=25.$ Therefore if the sum of the numbers in the corners is $18$ , the number in the center must be $7$ , which is answer $\textbf{(C)}$ .

Problem 16

The sum of an infinite geometric series is a positive number $S$ , and the second term in the series is $1$ . What is the smallest possible value of $S?$

$\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ \sqrt{5} \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$

Solution 1

The sum of an infinite geometric series is of the form: $\begin{split} S & = \frac{a_1}{1-r} \end{split}$ where $a_1$ is the first term and $r$ is the ratio whose absolute value is less than 1.

We know that the second term is the first term multiplied by the ratio. In other words:

$\begin{split} a_1 \cdot r & = 1 \\ a_1 & = \frac{1}{r} \end{split}$

Thus, the sum is the following:

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Since we want the minimum value of this expression, we want the maximum value for the denominator, $-r^2$ $+$ $r$ . The maximum x-value of a quadratic with negative $a$ is $\frac{-b}{2a}$ .

$\begin{split} r & = \frac{-(1)}{2(-1)} \\ r & = \frac{1}{2} \end{split}$

Plugging $r$ $=$ $\frac{1}{2}$ into the quadratic yields:

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Therefore, the minimum sum of our infinite geometric sequence is $\boxed{\textbf{(E)}\ 4}$ . (Solution by akaashp11)

Solution 2

After observation we realize that in order to minimize our sum $\frac{a}{1-r}$ with $a$ being the reciprocal of r. The common ratio $r$ has to be in the form of $1/x$ with $x$ being an integer as anything more than $1$ divided by $x$ would give a larger sum than a ratio in the form of $1/x$ .

The first term has to be $x$ , so then in order to minimize the sum, we have minimize $x$ .

The smallest possible value for $x$ such that it is an integer that's greater than $1$ is $2$ . So our first term is $2$ and our common ratio is $1/2$ . Thus the sum is $\frac{2}{1/2}$ or $\boxed{\textbf{(E)}\ 4}$ . Solution 2 by No_One

Solution 3

We can see that if $a$ is the first term, and $r$ is the common ratio between each of the terms, then we can get $amc数学竞赛报名官网$ Also, we know that the second term can be expressed as $a\cdot r$ notice if we multiply $S-Sr=a$ by $r$ , we would get $r(S-Sr)=ar \implies Sr-Sr^2=1 \implies Sr^2-Sr+1=0$ This quadratic has real solutions if the discriminant is greater than or equal to zero, or $S^2-4\cdot S \cdot 1 \ge 0$ This yields that $S\le 0$ or $S\ge 4$ . However, since we know that $S$ has to be positive, we can safely conclude that the minimum possible value of $S$ is $\boxed{\textbf{(E)}\ 4}$ .