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AMC数学竞赛真题2017年10A 13-14

2018-08-27 重点归纳

AMC10数学竞赛是美国高中数学竞赛中的一项,是针对高中一年级及初中三年级学生的数学测试,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。不管是对高校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AMC10数学竞赛真题以及官方解答吧:

Problem 13

Define a sequence recursively by $F_{0}=0,~F_{1}=1,$ and $F_{n}=$ the remainder when $F_{n-1}+F_{n-2}$ is divided by $3,$ for all $n\geq 2.$ Thus the sequence starts $0,1,1,2,0,2,\ldots$ What is $F_{2017}+F_{2018}+F_{2019}+F_{2020}+F_{2021}+F_{2022}+F_{2023}+F_{2024}?$

amc数学竞赛

Solution

A pattern starts to emerge as the function is continued. The repeating pattern is $0,1,1,2,0,2,2,1\ldots$ The problem asks for the sum of eight consecutive terms in the sequence. Because there are eight numbers in the repeating sequence, we just need to find the sum of the numbers in the sequence, which is $\boxed{\textbf{(D)}\ 9}$

Problem 14

Every week Roger pays for a movie ticket and a soda out of his allowance. Last week, Roger's allowance was $A$ dollars. The cost of his movie ticket was $20\%$ of the difference between $A$ and the cost of his soda, while the cost of his soda was $5\%$ of the difference between $A$ and the cost of his movie ticket. To the nearest whole percent, what fraction of $A$ did Roger pay for his movie ticket and soda?

amc数学竞赛真题

Solution

Let $m$ = cost of movie ticket
Let $s$ = cost of soda

We can create two equations:

$m = \frac{1}{5}(A - s)$

$s  = \frac{1}{20}(A - m)$

Substituting we get: 

amc10 

which yields:
$m = \frac{19}{99}A$

Now we can find s and we get:

$s = \frac{4}{99}A$

Since we want to find what fraction of $A$ did Roger pay for his movie ticket and soda, we add $m$ and $s$ to get:

$\frac{19}{99}A + \frac{4}{99}A \implies \boxed{\textbf{(D)}\ 23\%}$

以上就是小编对AMC10数学竞赛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


2017年AMC10A数学竞赛整套真题及解析,请点击下方查看:

AMC竞赛10A 01-02   AMC竞赛10A 03-04   AMC竞赛10A 05-06   AMC竞赛10A 07-08   AMC竞赛10A 09-10

AMC真题10A 11-12   AMC真题10A 13-14   AMC真题10A 15-16   AMC真题10A 17-18   AMC真题10A 19-20

AMC数学竞赛10A 21-22   AMC数学竞赛10A 23-24   AMC数学竞赛10A 25