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Problem 50

1. Problem

Consecutive Prime Sum

The prime 41, can be written as the sum of six consecutive primes:

\[41 = 2 + 3 + 5 + 7 + 11 + 13\]

This is the longest sum of consecutive primes that adds to a prime below one-hundred.

The longest sum of consecutive primes below one-thousand that adds to a prime, contains 21 terms, and is equal to 953.

Which prime, below one-million, can be written as the sum of the most consecutive primes?

连续素数的和

素数 41 可以写成六个连续素数的和:

\[41 = 2 + 3 + 5 + 7 + 11 + 13\]

在小于一百的素数中,41 能够被写成最多的连续素数的和。

在小于一千的素数中,953 能够被写成最多的连续素数的和,共包含连续 21 个素数。

在小于一百万的素数中,哪个素数能够被写成最多的连续素数的和?

2. Solution