AMC8 2006

AMC8 2006 · Q25

AMC8 2006 · Q25. It mainly tests Linear inequalities, Logic puzzles.

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?

Barry在3张卡片的每面写了一个不同的数字，并如图所示将卡片放在桌上。三张卡片上两面的数字之和相等。隐藏面的三个数字是素数。隐藏素数的平均数是多少？

(A) 13 13

(B) 14 14

(D) 16 16

(E) 17 17

Answer

Correct choice: (B)

正确答案：（B）

Solution

(B) There are one odd and two even numbers showing. Because all primes other than 2 are odd and the sum of an even number and an odd number is odd, the common sum must be odd. That means 2 must be opposite 59 and the common sum is $2+59=61$. The other two hidden numbers are $61-44=17$ and $61-38=23$. The average of 2, 17 and 23 is $\frac{2+17+23}{3}=\frac{42}{3}=14$.

(B) 现在显示的是一个奇数和两个偶数。因为除了 2 以外的所有质数都是奇数，而偶数与奇数之和为奇数，所以公共和必定是奇数。这意味着 2 必须与 59 相对，公共和为 $2+59=61$。另外两个隐藏的数字是 $61-44=17$ 和 $61-38=23$。2、17 和 23 的平均数是 $\frac{2+17+23}{3}=\frac{42}{3}=14$。

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