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AMC数学竞赛真题2017年10A 1-2

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 1

What is the value of 美国数学竞赛amc10?

$\textbf{(A)}\ 70\qquad\textbf{(B)}\ 97\qquad\textbf{(C)}\ 127\qquad\textbf{(D)}\ 159\qquad\textbf{(E)}\ 729$


Solution 1

Notice this is the term $a_6$ in a recursive sequence, defined recursively as $a_1 = 3, a_n = 2a_{n-1} + 1.$ Thus:amc数学竞赛真题.

Solution 2

Starting to compute the inner expressions, we see the results are $1, 3, 7, 15, \ldots$. This is always $1$ less than a power of $2$. The only admissible answer choice by this rule is thus $\boxed{\textbf{(C)}\ 127}$.

Solution 3

Working our way from the innermost parenthesis outwards and directly computing, we have $\boxed{\textbf{(C) } 127}$.

Solution 4

If you distribute this you get a sum of the powers of $2$. The largest power of $2$ in the series is $64$, so the sum is amc数学竞赛.

Problem 2

Pablo buys popsicles for his friends. The store sells single popsicles for $$1$ each, $3$-popsicle boxes for $$2$ each, and $5$-popsicle boxes for $$3$. What is the greatest number of popsicles that Pablo can buy with $$8$?

$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 13\qquad\textbf{(E)}\ 15$

Solution

$$3$ boxes give us the most popsicles/dollar, so we want to buy as many of those as possible. After buying $2$, we have $$2$ left. We cannot buy a third $$3$ box, so we opt for the $$2$ box instead (since it has a higher popsicles/dollar ratio than the $$1$ pack). We're now out of money. We bought $5+5+3=13$ popsicles, so the answer is $\boxed{\textbf{(D) }13}$.

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