Project Euler Lab - Problem 50

#50 - Consecutive Prime Sum

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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?

This problem is taken from Project Euler, Problem 50.
Problem text © Project Euler, licensed under CC BY-NC-SA 4.0. Original: projecteuler.net/problem=50. Published Friday, 15th August 2003, 06:00 pm. Solved by 69,809 members at time of mirroring.

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