2006 AMC 8 第 25 题

先试着解答 2006 AMC 8 第 25 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2006 AMC 8 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

25.

Barry 写了 66 个不同的数,分别写在 33 张卡片的两面,并把卡片如图摆在桌上。每张卡片两面数字之和都相等。三个隐藏面上的数都是质数。隐藏的质数的平均数是多少?
Barry wrote 66 different numbers, one on each side of 33 cards, and laid the cards on a table, as shown. The sums of the two numbers on each of the three cards are equal. The three numbers on the hidden sides are prime numbers. What is the average of the hidden prime numbers?
1313
1414
1515
1616
1717
答案:B
知识点:质数奇偶性平均数
小提示:
每张卡片的公共和对三个可见数必须有相同奇偶性。
The common card sum must have the same parity for all three visible numbers.
大提示:
使用 22 是唯一偶质数这一事实。
Use the fact that 22 is the only even prime.
解答:
公共和必须是奇数。如果公共和是偶数,那么 4444 和 3838 背后的隐藏数都必须是偶质数,但偶质数只有一个。
所以奇数 5959 背后的隐藏质数必须是 22,公共和为 59+2=6159+2=61。
另外两个隐藏质数是 61−44=1761-44=17 和 61−38=2361-38=23。
隐藏质数的平均数为 2+17+233=14\dfrac{2+17+23}{3}=14。
所以正确答案是 B。
The common sum must be odd. If the common sum were even, then the hidden numbers behind 4444 and 3838 would both have to be even primes, but there is only one even prime.
So the prime hidden behind the odd visible number 5959 must be 22, making the common sum 59+2=6159+2=61.
The other two hidden primes are 61−44=1761-44=17 and 61−38=2361-38=23.
The average of the hidden primes is 2+17+233=14\dfrac{2+17+23}{3}=14.
Thus, B is the correct answer.
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