Project Euler Problem 50
Again, surprised how effective the brute force solution was
Revision 1:
Old solution stopped working for some reason. Re-did it starting from the biggest possible space this time
Problem:
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?
1.8.11
