Question

一对质数的正差为 $2$ ，它们的立方的正差为 $31106$ 。大于这两个质数的最小质数的各位数字之和是多少？

The positive difference between a pair of primes is equal to $2,$ and the positive difference between the cubes of the two primes is $31106.$ What is the sum of the digits of the least prime that is greater than those two primes?

$\ 8$

$\ 10$

$\ 11$

$\ 13$

$\ 16$

答案：E

知识点：质数立方和与立方差数字

难度评级：1140

解答：

设两个质数相差 $2$ ，分别为 $m-1,m+1$ ，中间数为 $m$ 。

于是也就是 $(m+1)^3-(m-1)^3 = 31106 ,$ $m^3+3m^2+3m+1$ $-(m^3-3m^2+3m-1)$ $= 6m^2+2 = 31106.$

因此 $m^2= 5184$ ，所以 $m=72$ 。

因此两个质数为 $71,73$ 。大于它们的最小质数是 $79$ ，其数字和为 $16$ 。

所以答案是 E。

Since the primes are $2$ away from each other, we can make them equal to $m-1,m+1,$ where $m$ is their average.

Then, $(m+1)^3-(m-1)^3 = 31106 ,$ making $m^3+3m^2+3m+1$ $-(m^3-3m^2+3m-1)$ $= 6m^2+2 = 31106.$

Therefore, $m^2= 5184,$ so $m=72.$

The primes are therefore $71,73.$ The least prime greater than both of those is $79,$ and its digit sum is $16.$

Thus, the answer is E .

2022 AMC 10B 第 13 题

答案：E

难度评级：1140

解答：

其他年份的第 13 题