AMC10 2024 A

AMC10 2024 A · Q3

AMC10 2024 A · Q3. It mainly tests Primes & prime factorization, Parity (odd/even).

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

能写成 $5$ 个不同质数之和的最小质数的各位数字之和是多少？

(A) 5 5

(B) 7 7

(D) 10 10

(E) 13 13

Answer

Correct choice: (B)

正确答案：（B）

Solution

Let the requested sum be $S.$ Recall that $2$ is the only even (and the smallest) prime, so $S$ is odd. It follows that the five distinct primes are all odd. The first few odd primes are $3,5,7,11,13,17,19,\ldots.$ We conclude that $S>3+5+7+11+13=39,$ as $39$ is a composite. The next possible value of $S$ is $3+5+7+11+17=43,$ which is a prime. Therefore, we have $S=43,$ and the sum of its digits is $4+3=\boxed{\textbf{(B) }7}.$

设所求和为 $S$。忆及 $2$ 是唯一偶质数（也是最小质数），故 $S$ 为奇数。因此，五个不同质数均为奇质数。前几个奇质数为 $3,5,7,11,13,17,19,\ldots$。由此，$S>3+5+7+11+13=39$，因 $39$ 为合数。下一个可能值为 $3+5+7+11+17=43$，它是质数。因此，$S=43$，其各位数字之和为 $4+3=\boxed{\textbf{(B) }7}$。

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