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AMC数学竞赛真题2016年12B 7-8

2018-09-06 重点归纳

AMC12是针对高中学生的数学测验，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。其主要目的在于激发学生对数学的兴趣，参予AMC12的学生应该不难发现测验的问题都很具挑战性，但测验的题型都不会超过学生的学习范围。这项测验希望每个考生能从竞赛中享受数学。那么接下来跟随小编来看一下AMC12的官方真题以及官方解答吧：

Problem 7

Josh writes the numbers $1,2,3,\dots,99,100$ . He marks out $1$ , skips the next number $(2)$ , marks out $3$ , and continues skipping and marking out the next number to the end of the list. Then he goes back to the start of his list, marks out the first remaining number $(2)$ , skips the next number $(4)$ , marks out $6$ , skips $8$ , marks out $10$ , and so on to the end. Josh continues in this manner until only one number remains. What is that number?

$amc真题$

Solution

Following the pattern, you are crossing out...

Time 1: Every non-multiple of $2$

Time 2: Every non-multiple of $4$

Time 3: Every non-multiple of $8$

Following this pattern, you are left with every multiple of $64$ which is only $\boxed{\textbf{(D)}64}$ .

Problem 8

A thin piece of wood of uniform density in the shape of an equilateral triangle with side length $3$ inches weighs $12$ ounces. A second piece of the same type of wood, with the same thickness, also in the shape of an equilateral triangle, has side length of $5$ inches. Which of the following is closest to the weight, in ounces, of the second piece?

$amc数学竞赛$

Solution

By: dragonfly

We can solve this problem by using similar triangles, since two equilateral triangles are always similar. We can then use

$\left(\frac{3}{5}\right)^2=\frac{12}{x}$ .

We can then solve the equation to get $x=\frac{100}{3}$ which is closest to $美国数学竞赛$

Solution 2

Another approach to this problem, very similar to the previous one but perhaps explained more thoroughly, is to use proportions. First, since the thickness and density are the same, we can set up a proportion based on the principle that $d=\frac{m}{V}$ , thus $dV=m$ .

However, since density and thickness are the same and $A$ is proportional to $s^2$ (recognizing that the area of an equilateral triangle is $\frac{(s)^2\sqrt{3}}{4}$ ), we can say that $m$ is proportional to $s^2$ .

Then, by increasing s by a factor of $\frac{5}{3}$ , $s^2$ is increased by a factor of $\frac{25}{9}$ , thus $m=12*\frac{25}{9}$ or $\boxed{\textbf{(D)}\ 33.3}$ .