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美国数学竞赛AMC8真题2014年 23-24

2018-09-21 重点归纳

美国数学竞赛AMC8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年级的考生加入到美国数学竞赛AMC8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美国数学竞赛AMC8的官方真题以及官方解答吧：

Problem 22

Three members of the Euclid Middle School girls' softball team had the following conversation.

Ashley: I just realized that our uniform numbers are all $2$ -digit primes.

Brittany : And the sum of your two uniform numbers is the date of my birthday earlier this month.

Caitlin: That's funny. The sum of your two uniform numbers is the date of my birthday later this month.

Ashley: And the sum of your two uniform numbers is today's date.

What number does Caitlin wear?

$AMC数学竞赛$

Solution

The maximum amount of days any given month can have is 31, and the smallest two digit primes are 11, 13, and 17. There are a few different sums that can be deduced from the following numbers, which are 24, 30, and 28, all of which represent the three days. Therefore, since Brittany says that the other two people's uniform numbers are earlier, so that means Caitlin and Ashley's numbers must add up to 24. Similarly, Caitlin says that the other two people's uniform numbers is later, so the sum must add up to 30. This leaves 28 as today's date. From this, Caitlin was referring to the uniform wearers 13 and 17, telling us that her number is 11, giving our solution as $美国数学竞赛AMC$

Problem 23

One day the Beverage Barn sold $252$ cans of soda to $100$ customers, and every customer bought at least one can of soda. What is the maximum possible median number of cans of soda bought per customer on that day?

$amc数学竞赛报名$

Solution

In order to maximize the median, we need to make the first half of the numbers as small as possible. Since there are $100$ people, the median will be the average of the $50\text{th}$ and $51\text{st}$ largest amount of cans per person. To minimize the first 49, they would each have one can. Subtracting these $49$ cans from the $252$ cans gives us $203$ cans left to divide among $51$ people. Taking $\frac{203}{51}$ gives us $3$ and a remainder of $50$ . Seeing this, the largest number of cans the $50th$ person could have is $3$ , which leaves $4$ to the rest of the people. The average of $3$ and $4$ is $3.5$ . Thus our answer is $美国数学竞赛AMC8考试真题$ .