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?
"""
from modules.matlib import isPrime, nextPrime
mxl = 0 # maximum length of consecutive primes list
mxsm = 0 # sum of longest consecutive primes list
s = 2 # start
while s < 1000000: # while start is below 1 milion
cp = [] # consecutive primes
t = s # start of second loop
while (t + sum(cp) < 1000000) and isPrime(
t + sum(cp)
): # while sum of consecutive primes plus the last one is below 100
cp.append(t) # append prime to consecutive primes list
if len(cp) > mxl: # if list of consecutive primes is longest and sum of those consecutive primes is prime
mxl = len(cp) # lenght of current list
mxsm = sum(cp) # sum of consecutive primes
t = nextPrime(t) # continue with next prime
s = nextPrime(s) # continue with next prime
print(mxsm) # output sum of longest consecutive primes list
#!/usr/bin/python3
"""
Problem 10 - Summation of primes
The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
Find the sum of all the primes below two million.
"""
from modules.matlib import isPrime
sm = 2 # sum, starting with first prime
for x in range(3, 2000000, 2): # for all odd numbers below two million
if isPrime(x): # if is prime
sm += x # add it to sum
print(sm) # return result
def isTruncatablePrime(n):
n = str(n)
if len(n) < 2: return False
for i in range(1, len(n), 1):
if not isPrime(int(n[i:])) or not isPrime(int(n[:-i])): return False
return True
