2022 AMC 10B 第 13 题

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

小提示：

对立方差进行因式分解

Factor the difference of cubes

大提示：

设这一对孪生质数为

p

和

p+2

Let the twin primes be

p

and

p+2

解答：

由于两个质数相差

2

，可以将它们写成

m-1,m+1

，其中

m

是它们的平均数。

于是

(m+1)^3-(m-1)^3 = 31106\text{，}

即

m^3+3m^2+3m+1

-(m^3-3m^2+3m-1)

= 6m^2+2 = 31106\text{。}

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

小提示：

大提示：

解答：

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