Question

能写成 $5$ 个不同质数之和的最小质数，其各位数字之和是多少？

What is the sum of the digits of the smallest prime that can be written as a sum of $5$ distinct primes?

$5$

$7$

$9$

$10$

$11$

答案：B

知识点：质数奇偶性最优化

难度评级：1050

小提示：

若使用 $2$ ，和会是大于 $2$ 的偶数，因此不可能是质数；应使用五个奇质数。

Using $2$ makes the sum even and larger than $2,$ so it cannot be prime; use five odd primes

大提示：

五个最小奇质数之和为 $39$ ，不是质数；再增加到下一个可能的质数和。

The five smallest odd primes sum to $39,$ which is not prime; increase to the next prime sum

解答：

如果 $2$ 是五个质数之一，则总和是大于 $2$ 的偶数，因此是合数。所以五个质数都必须是奇数。五个最小奇质数之和为 $3 + 5 + 7 + 11 + 13 = 39$ $= 3 \cdot 13$ ，不是质数。下一个可能的和是把 $13$ 换成下一个质数 $17$ ；若改动更靠前的项，为保持五个质数互不相同，增加量至少同样大。这样得到 $3 + 5 + 7 + 11 + 17 = 43$ ，它是质数，其各位数字之和为 $4 + 3 = 7$ ，正确答案是 B。

Suppose $2$ is one of the five primes. Then the total is even and bigger than $2,$ so it is composite. Thus all five primes must be odd. The five smallest odd primes give $3 + 5 + 7 + 11 + 13 = 39$ $= 3 \cdot 13,$ which is not prime. The next possible sum is obtained by replacing $13$ with the next prime, $17;$ changing any earlier term forces at least as large an increase to keep the primes distinct. This gives $3 + 5 + 7 + 11 + 17 = 43,$ which is prime. Its digit sum is $4 + 3 = 7.$ Thus, B is the correct answer.

2024 AMC 10A 第 3 题

答案：B

难度评级：1050

小提示：

大提示：

解答：

其他年份的第 3 题