AMC 8 2006 Problem 25
AMC8 2006 年第 25 题
奇偶性★★★★☆
Barry wrote 6 different numbers, one on each side of 3 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?
Bary 在 3 张牌的两面各写了一个数，共写了 6 个不同的数字，然后把牌放在桌子上，如图所示。每张卡片上的两个数字之和都相等。没有翻开的三个面上的数字都是质数。则这 3 个没翻开的面上的质数的平均值是多少？
- A.13
- B.14
- C.15
- D.16
- E.17
正确答案 · Correct Answer
B
解析 · Solution
解法1
常规解法 · 奇偶定位 2
可见数 44、38 为偶数，59 为奇数。设每张卡两数之和都是 ，则三个隐藏质数为 S-44、S-59、S-38。要让三者都是质数，必有且只有一个是偶质数 2，且它只能与奇数 59 配对（否则会出现大于 2 的偶质数）。故 S=59+2=61，隐藏数为 17、2、23，平均数 。
巧解 · 设元 + 平均数公式
设 44、59、38 背面的质数为 、、。由 x+44=y+59=z+38 得 z=x+6、y=x-15，于是平均数 。三者皆为质数只能取 y=x-15=2，即 x=17（则 z=23），平均数 17-3=14。
Solution 1
Elementary · use parity to place the 2
The visible numbers 44 and 38 are even while 59 is odd. With common sum , the hidden primes are S-44, S-59, S-38. To keep all three prime, exactly one is the even prime 2, and it must pair with the odd 59 (otherwise an even hidden number greater than 2 appears). So S=61, the hidden primes are 17, 2, 23, and their average is .
Slick · set a variable + mean formula
Let the hidden numbers on the 44, 59, 38 cards be , , . From x+44=y+59=z+38 we get z=x+6 and y=x-15, so the average is . All three are prime only when y=x-15=2, i.e. x=17 (so z=23); the average is 17-3=14.
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