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AMC数学竞赛真题2016年12B 22-23

2018-09-12 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 22

For a certain positive integer $n$ less than $1000$ , the decimal equivalent of $\frac{1}{n}$ is $0.\overline{abcdef}$ , a repeating decimal of period of $6$ , and the decimal equivalent of $\frac{1}{n+6}$ is $0.\overline{wxyz}$ , a repeating decimal of period $4$ . In which interval does $n$ lie?

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Solution

Solution by e_power_pi_times_i

If $\frac{1}{n} = 0.\overline{abcdef}$ , $n$ must be a factor of $999999$ . Also, by the same procedure, $n+6$ must be a factor of $9999$ . Checking through all the factors of $999999$ and $9999$ that are less than $1000$ , we see that $n = 297$ is a solution, so the answer is $amc数学竞赛$ .

Note: $n = 27$ is also a solution, which invalidates this method. However, we need to examine all factors of $999999$ that are not factors of $99999$ , $999$ , or $99$ , or $9$ . Additionally, we need $n+6$ to be a factor of $9999$ but not $999$ , $99$ , or $9$ . Indeed, $297$ satisfies these requirements.

Problem 23

What is the volume of the region in three-dimensional space defined by the inequalities $|x|+|y|+|z|\le1$ and $|x|+|y|+|z-1|\le1$

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Solution 1 (Non Calculus)

The first inequality refers to the interior of a regular octahedron with top and bottom vertices $(0,0,1),\ (0,0,-1)$ . Its volume is $8\cdot\tfrac16=\tfrac43$ . The second inequality describes an identical shape, shifted $+1$ upwards along the $Z$ axis. The intersection will be a similar octahedron, linearly scaled down by half. Thus the volume of the intersection is one-eighth of the volume of the first octahedron, giving an answer of $\textbf{(A) }\tfrac16$ .

Solution 2 (Calculus)

Let $z\rightarrow z-1/2$ , then we can transform the two inequalities to $美国数学竞赛AMC真题$ and $|x|+|y|+|z+1/2|\le1$ . Then it's clear that $-1/2\le z \le 1/2$ , consider $0 \le z \le 1/2$ , $|x|+|y|\le 1/2-z$ , then since the area of $|x|+|y|\le k$ is $2k^2$ , the volume is $\int_{0}^{1/2}2k^2 \,dk=\frac{1}{12}$ . By symmetry, the case when $\frac{-1}{2}\le z\le0$ is the same. Thus the answer is $\frac{1}{6}$ .