def problem51():
primes = primesUpTo(10 ** 7)
for p in primes[1229:]:
m = str(p)
for i in ['0', '1', '2']:
if m.count(i) == 3 and primes8(m, i):
return p
def problem35():
from PE_primes import primesUpTo, isPrime
# count is 1 is we are not count 2 later
count = 1
# remove all primes that contain an even number, since those will have a cyclic shift that is even
primesList = {p for p in primesUpTo(10**6) if isMadeOfOdd(p)}
for p in primesList:
if all(q in primesList for q in iCyclicShifts(p)):
count += 1
return count
def problem49():
primes = primesUpTo(10000)
for a in dropwhile(lambda x: x <= 1487, primes):
# the bound comes from wanting c = 2b - a ≤ 10000
for b in dropwhile(lambda x: x <= a, primes):
if b >= (10000 + a) / 2:
break
c = b + (b - a)
# Do the same digits first since I think that is the most time
# consuming part
if sameDigits(a, b) and sameDigits(b, c) and isPrime(c):
return int(str(a) + str(b) + str(c))
def problem70():
GOAL = 10 ** 7
primes = primesUpTo(2 * int(GOAL ** (1 / 2)))
record = 1000000000
for index, p in enumerate(primes[1:]):
for q in primes[index:]:
if q > GOAL // p:
break
n = p * q
ph = (p - 1) * (q - 1)
if isPermutation(n, ph) and n / ph < record:
record = n / ph
recordN = n
return recordN
def problem50(): GOAL = 10**6 listOfPrimes = primesUpTo(GOAL) recordLength = 21 maxPrime = 953 for index in count(): for length, running in enumerate(accumulate(listOfPrimes[index:])): if length + 1 > recordLength and isPrime(running): recordLength = length + 1 maxPrime = running if running > GOAL: break # If there is no hope, break out if sum(listOfPrimes[index: index+recordLength]) > GOAL: break return maxPrime
def problem46():
primesList = primesUpTo(10 ** 4)
for l in range(1, 10 ** 4):
n = 2 * l + 1
flag = True
if not (n in primesList):
a = 1
while 2 * a ** 2 < n:
q = n - 2 * a ** 2
if q in primesList:
flag = False
break
a += 1
else:
flag = False
if flag:
return n
def problem27():
record = 0
ab = 0
checker = Orderedlist(primesUpTo(1998))
# We know there is a record of 80
# this removes many possibilities for low values of b
# since the most number of primes for f(n) is b-1
for b in checker[80:1000]:
for p1 in checker[0:1000 + b + 1]:
a = p1 - b - 1
n = 2
p2 = a + p1 + (2 * n - 1)
while isPrime(abs(p2)):
n += 1
p2 = a + p2 + (2 * n - 1)
if n > record:
record = n
ab = a * b
return ab
def problem7(): ''' We have a crude bound on the number of primes given by x/ln(x) Thus we need 10001 < x/ln(x) and 200000 works''' primes = primesUpTo( 200000) return primes[10001-1]
def problem10(): return sum(primesUpTo(2*10**6))