AMC10 2022 B

AMC10 2022 B · Q13

AMC10 2022 B · Q13. It mainly tests Quadratic equations, Primes & prime factorization.

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?

一对质数的正差等于$2$，这两个质数的立方正差等于$31106$。这两个质数之后的最小质数的各位数字之和是多少？

(A) 8 8

(B) 10 10

(D) 13 13

(E) 16 16

Answer

Correct choice: (E)

正确答案：（E）

Solution

Denote the two primes as $ a $ and $ b $. Then, $ a - b = 2 $ $ a^3 - b^3 = 31106 $ We see that $ a = 2 + b $ Now, we have $ (2+b)^3 - b^3 = 31106 $ We apply the binomial theorem (or just expand) to $ (2+b)^3 $, getting $ 8 + 12b + 6b^2 + b^3 - b^3 = 31106 $ $ \Rightarrow 8 + 12b + 6b^2 = 31106 $ Subtracting by 8 on both sides results in $ 6b^2 + 12b = 31098 $ $ b^2 + 2b - 5183 $ $ (b-71)(b+73) $ We see that $ b \in \{-73, 71\} $. We negate all negative values, and see that $ b = 71 $. Therefore $ a - b = 2 $, $ a = 2 + b $, $ a = 2 + 71 = 73 $. The next prime number greater than both of these is $79$, and therefore our answer is $ 7 + 9 = $ $\boxed{\textbf{(E) }16}$.

设两个质数为$$ a $$和$$ b $$。则 $$ a - b = 2 $$ $$ a^3 - b^3 = 31106 $$ 可见$$ a = 2 + b $$ 现$$ (2+b)^3 - b^3 = 31106 $$ 应用二项式定理（或直接展开）$$ (2+b)^3 $$，得 $$ 8 + 12b + 6b^2 + b^3 - b^3 = 31106 $$ $$ \Rightarrow 8 + 12b + 6b^2 = 31106 $$ 两边减$8$得 $$ 6b^2 + 12b = 31098 $$ $$ b^2 + 2b = 5183 $$ $$ (b-71)(b+73)=0 $$ 可见$$ b \in \{-73, 71\} $$。舍弃负值，得$$ b = 71 $$。因此$$ a = 73 $$。两者之后的下一个质数是$79$，各位数字和$7 + 9 = \boxed{\textbf{(E) }16}$。

Topics