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AMC数学竞赛8真题2016年 17-18

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 17

An ATM password at Fred's Bank is composed of four digits from $0$ to $9$, with repeated digits allowable. If no password may begin with the sequence $9,1,1,$ then how many passwords are possible?

amc数学竞赛

Solution 1

For the first three digits, there are $10^3-1=999$ combinations since $911$ is not allowed. For the final digit, any of the $10$ numbers are allowed. $999 \cdot 10 = 9990 \rightarrow \boxed{\textbf{(D)}\ 9990}$

Solution 2

Counting the prohibited cases, we find that there are 10 of them. This is because we start with 9,1,1 and we can have any of the 10 digits for the last digit. So our answer is amc真题

Problem 18

In an All-Area track meet, $216$ sprinters enter a $100-$meter dash competition. The track has $6$ lanes, so only $6$ sprinters can compete at a time. At the end of each race, the five non-winners are eliminated, and the winner will compete again in a later race. How many races are needed to determine the champion sprinter?

amc竞赛

Solution

From any $n-$th race, only $\frac{1}{6}$ will continue on. Since we wish to find the total number of races, a column representing the races over time is ideal. Starting with the first race:\[\frac{216}{6}=36\]\[\frac{36}{6}=6\]\[\frac{6}{6}=1\]Adding all of the numbers in the second column yields amc数学竞赛

Solution 2

Every race eliminates $5$ players. The winner is decided when there is only $1$ runner left. Thus, $215$ players have to be eliminated. Therefore, we need $\frac{215}{5}$ games to decide the winner, or $\boxed{\textbf{(C)}\ 43}$

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