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2015 AMC 8考题13-14

2018-08-06 重点归纳

AMC 8数学竞赛专为8年级及以下的初中学生设计，但近年来的数据显示，越来越多小学4-6年级的考生加入到AMC 8级别的考试行列中，而当这些学生能在成绩中取得“A”类标签，则是对孩子数学天赋的优势证明，不管是对美高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AMC 8的官方真题以及官方解答吧：

Problem 13

How many subsets of two elements can be removed from the set {1,2,3,4,5,6,7,8,9,10,11} so that the mean (average) of the remaining numbers is 6?

$\textbf{(A)}\text{ 1}\qquad\textbf{(B)}\text{ 2}\qquad\textbf{(C)}\text{ 3}\qquad\textbf{(D)}\text{ 5}\qquad\textbf{(E)}\text{ 6}$

Solution

Since there will be 9 elements after removal, and their mean is 6, we know their sum is 54. We also know that the sum of the set pre-removal is 66. Thus, the sum of the 2 elements removed is 66-54=12. There are only （D）5 subsets of 2 elements that sum to 12: {1.11}，{2,10}，{3,9}，{4,8}，{5,7}.

Problem 14

Which of the following integers cannot be written as the sum of four consecutive odd integers?

$\textbf{(A)}\text{ 16}\quad\textbf{(B)}\text{ 40}\quad\textbf{(C)}\text{ 72}\quad\textbf{(D)}\text{ 100}\quad\textbf{(E)}\text{ 200}$

Solution 1

Let our 4 numbers be n,n+2,n+4,n+6, where n is odd. Then our sum is 4n+12. The only answer choice that cannot be written as 4n+12, where $n$ is odd, is （D）100.

Solution 2

If the four consecutive odd integers are 2n-3,2n-1,2n+1 and 2n+3 then the sum is 8n. All the integers are divisible by 8 except （D）100.

Solution 3

If the four consecutive odd integers are a,a+2,a+4 and a+6, the sum is 4a+12, and 4a+12 divided by 4 gives a+3. This means that a+3 must be even. The only integer that does not give an even integer when divided by 4 is 100, so the answer is （D）100.

以上就是小编对AMC 8数学竞赛官方真题以及解析的介绍，希望对你有所帮助，更多学习资料请持续关注AMC数学竞赛网！