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 7 - 10001st prime
By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.
What is the 10 001st prime number?
"""
from modules.matlib import nextPrime
n = 2 # first prime
for i in range(10000):
n = nextPrime(n) # find next prime 10000 times
print(n) # output result
Problem 37 - Truncatable primes
The number 3797 has an interesting property. Being prime itself, it is possible to continuously remove digits from left to right, and remain prime at each stage: 3797, 797, 97, and 7. Similarly we can work from right to left: 3797, 379, 37, and 3.
Find the sum of the only eleven primes that are both truncatable from left to right and right to left.
NOTE: 2, 3, 5, and 7 are not considered to be truncatable primes.
"""
from modules.matlib import isPrime
from modules.matlib import nextPrime
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
sm = 0 # sum of truncatable primes
c = 0 # count of truncatable primes
p = 11 # truncatable prime candidate
while c < 11:
if isTruncatablePrime(p):
c += 1
sm += p
p = nextPrime(p)
print(sm)
