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AMC数学竞赛8真题2016年 19-20

2018-08-28 重点归纳

AMC数学竞赛8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 19

The sum of $25$ consecutive even integers is $10,000$. What is the largest of these $25$ consecutive integers?

amc数学竞赛

Solution

Let $n$ be the 13th consecutive even integer that's being added up. By doing this, we can see that the sum of all 25 even numbers will simplify to $25n$ since amc真题. Now, $25n=10000 \rightarrow n=400$ Remembering that this is the 13th integer, we wish to find the 25th, which is $400+2(25-13)=\boxed{\textbf{(E)}\ 424}$.

Problem 20

The least common multiple of $a$ and $b$ is $12$, and the least common multiple of $b$ and $c$ is $15$. What is the least possible value of the least common multiple of $a$ and $c$?

amc竞赛

Solution

We wish to find possible values of $a$,$b$, and $c$. By finding the greatest common factor of $12$ and $15$, algebraically, it's some multiple of $b$and from looking at the numbers, we are sure that it is 3, thus $b$ is 3. Moving on to $a$ and $c$, in order to minimize them, we wish to find the least such that the least common multiple of $a$ and $3$ is $12$$\rightarrow 4$. Similarly with $3$ and $c$, we obtain $5$. The least common multiple of $4$and $5$ is $20 \rightarrow \boxed{\textbf{(A)} 20}$

以上就是小编对AMC竞赛官方真题以及解析的介绍,希望对你有所帮助,更多AMC真题下载请持续关注AMC数学竞赛网