def is_truncatable(num: int) -> bool:
if not primes.is_prime(num):
return False
mult = 10
while mult < num:
if not primes.is_prime(num % mult) or not primes.is_prime(num // mult):
return False
mult *= 10
return True
def solution():
t_primes = []
left_right = [3, 7]
right_left = [2, 5, 3, 7]
while len(t_primes) < 11:
left_right = [
add_digit(n, d, 1) for n in left_right for d in range(1, 10)
]
right_left = [
add_digit(n, d, 0) for n in right_left for d in range(1, 10)
]
left_right = [p for p in left_right if is_prime(p)]
right_left = [p for p in right_left if is_prime(p)]
t_primes += [p for p in left_right if p in right_left]
return sum(t_primes)
def problem_50():
limit = 10**6
# find the smallest list of primes which will sum up to at least the limit we set above
for x in count(1):
primes = list(get_primes_under(10**x))
if sum(primes) > limit:
break
# starting with the longest possible sequence (every element in the list), and then working
# to smaller sequence lengths
for length in reversed(range(1, len(primes) + 1)):
# Use a "sliding window" approach. Starting at the low end of the list, grab the first
# `length` elements and check it. Then slide the window right 1 element, and grab those,
# and check it, etc
for start in range(len(primes)-length):
candidate = sum(primes[start:start+length])
# if the candidate sum is bigger than the limit, no point in checking any more windows
# of size `length`. Break the inner loop, and reduce the length
if candidate >= limit:
break
# if the candidate sum is prime, we're done!
if is_prime(candidate):
print(candidate)
return
def nth_prime(n):
"""Returns nth prime"""
current = 1
primes = 0
while primes < n:
current += 1
if is_prime(current):
primes += 1
return current
def check(num): if '200' not in str(num): return False for i in xrange(len(str(num))): for j in range(0, 10): new_num = int(str(num)[:i] + str(j) + str(num)[i + 1:]) if is_prime(new_num): return False return True
def is_circular_prime(n):
""" Returns True is n is a circular prime: that is, if n itself is prime, as are all rotations
of its digits. """
# if n itself is not prime, it's certainly not a circular prime
if not is_prime(n):
return False
# convert n to a list of digits so we can rotate it
n = list(str(n))
# For every rotation of the digits of n, convert that back into an int, and check primality
# If any of them are not, return False
rotated_number = lambda list, i: int(''.join(rotate(list, i)))
if any(not is_prime(rotated_number(n, i)) for i in range(1, len(n))):
return False
# We made it here, so n and all rotations of n are prime
return True
def solution():
# primes_list = seive_of_erat(1000000000)
# primes_set = set(primes_list)
n = 1
all_corners = 5
corner_primes = 3
while corner_primes / all_corners >= 0.1:
n += 1
end = (2*n + 1) * (2*n + 1)
all_corners += 4
corner_primes += len([1 for c in get_corners(n) if is_prime(c)])
return 2*n + 1
def solution():
digits = [1, 2, 3, 4, 5, 6, 7]
best = 0
current = digits
while len(digits) < 10:
conc = int("".join([str(d) for d in current]))
if is_prime(conc):
best = conc
current = next_perm(current)
if not current:
digits.append(digits[-1] + 1)
current = digits
return best
def main():
prime_count = 0
total_count = 1
num = 1
for side_length in itertools.count(3, 2):
for _ in range(4):
total_count += 1
num += side_length - 1
if primes.is_prime(num):
prime_count += 1
if prime_count / total_count < .1:
return side_length
def can_be_written_as_prime_plus_twice_square(n):
""" Returns True if n can be written as the sum of a prime and twice a square number, or
returns False otherwise. """
if is_prime(n):
return True
# For every prime under n, find what value of x satisfies: n = prime + 2x^2
# If any of those values x are an integer, then n can be written as prime + 2x square number
for prime in get_primes_under(n):
x = ((n - prime) / 2) ** 0.5
if x % 1 == 0:
return True
# If we get here, n cannot be written as prime + 2x square number
return False
def count_primes_of_quadratic(a, b):
""" Returns the number of consecutive primes produced by the quadratic equation n² + an + b,
with n starting at 0. """
n = 0
count = 0
while True:
if is_prime(n*n + a*n + b):
count += 1
n += 1
else:
break
return count
def ana_solve(self):
solution = 0
primeList = [5, 7, 11, 13, 17, 19, 23]
combos = combinations(primeList, 3)
primeCombos = []
for c in combos:
s = sum(c)
if primes.is_prime(s):
primeCombos.append((c, s))
counts = Counter([pc[1] for pc in primeCombos])
bigC = []
for c in counts:
if counts[c] >= 4:
bigC.append(c)
candidates = []
for pc in primeCombos:
if pc[1] in bigC:
candidates.append(pc)
candidates.sort(key=lambda x: x[1])
print(candidates)
print('analytical solution = ', solution)
def problem_58():
# These define the north-east and the south-west spokes, for consecutive n starting at n=0
ne_spoke = lambda n: (2*n + 1)**2
sw_spoke = lambda n: 4*(n**2) + 1
# These define the north-west and the south-east spokes, for consecutive n starting at n=1
nw_spoke = lambda n: 4*(n**2) - 6*n + 3
se_spoke = lambda n: 4*(n**2) - 10*n + 7
all_spoke_values = set()
prime_spoke_values = set()
for n in count(0):
for value in (ne_spoke(n), sw_spoke(n), nw_spoke(n+1), se_spoke(n+1)):
all_spoke_values.add(value)
if is_prime(value):
prime_spoke_values.add(value)
if n > 1 and (len(prime_spoke_values) / len(all_spoke_values)) < 0.10:
spiral_side_length = 2*n + 1
print(spiral_side_length)
break
def concats_to_prime(p1, p2):
""" Check if both concatenations of primes p1 and p2 (p1+p2, and p2+p1), form a prime. """
p1, p2 = str(p1), str(p2)
t1, t2 = int(p1 + p2), int(p2 + p1)
return all(is_prime(t) for t in (t1, t2))
# It was proposed by Christian Goldbach that every odd composite number can be written as the sum of a prime and twice a square. # 9 = 7 + 2×1**2 # 15 = 7 + 2×2**2 # 21 = 3 + 2×3**2 # 25 = 7 + 2×3**2 # 27 = 19 + 2×2**2 # 33 = 31 + 2×1**2 # It turns out that the conjecture was false. # What is the smallest odd composite that cannot be written as the sum of a prime and twice a square? from utils.primes import primesfrom2to, is_prime num = 1 while True: found = False num += 2 if is_prime(num):continue pl = primesfrom2to(num) for p in pl: for i in range(1, int(((num - p) / 2)**0.5) + 1): if num == p + 2 * i**2: found = True break if found: break if not found: print(num) break
def is_circular_prime(num: int) -> bool:
for rotation in rotate(num):
if not is_prime(rotation):
return False
return True
def test_evens_are_not_prime(self):
self.assertFalse(is_prime(4))
self.assertFalse(is_prime(8))
def test_2_is_prime(self):
self.assertEquals(True, is_prime(2))
# We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once. For example, 2143 is a 4-digit pandigital and is also prime.
# What is the largest n-digit pandigital prime that exists?
from utils.primes import is_prime
from itertools import permutations
ans = 0
p = ['1','2','3']
for i in range(4,10):
p += [str(i)]
for s in permutations(p):
v = int(''.join(s))
if is_prime(v):
# print(v)
ans = max(ans, v)
print(ans)
def test_13_is_prime(self):
self.assertTrue(is_prime(13))
# There are no arithmetic sequences made up of three 1-, 2-, or 3-digit primes, exhibiting this property, but there is one other 4-digit increasing sequence.
# What 12-digit number do you form by concatenating the three terms in this sequence?
from itertools import permutations
from utils.primes import primesfrom2to, is_prime
found = False
for p in primesfrom2to(10000):
if p < 1000:continue
perms = tuple(permutations(list(str(p))))
if ('1','4','8','7') in perms:continue
for s1 in perms:
n1 = int(''.join(s1))
if n1 < 1000:continue
if not is_prime(n1):continue
for s2 in perms:
if s2 == s1:continue
n2 = int(''.join(s2))
if n2 < 1000:continue
if not is_prime(n2):continue
for s3 in perms:
if s3 == s1 or s3 == s2:continue
n3 = int(''.join(s3))
if n3 < 1000:continue
if not is_prime(n3):continue
if abs(n1 - n2) == abs(n2 - n3):
found = True
if found:break
if found:break
if found:break
# 15 = 3 × 5 # The first three consecutive numbers to have three distinct prime factors are: # 644 = 2**2 × 7 × 23 # 645 = 3 × 5 × 43 # 646 = 2 × 17 × 19. # Find the first four consecutive integers to have four distinct prime factors. What is the first of these numbers? from utils.primes import is_prime from utils.common import prime_factors num = 2 * 3 * 5 * 7 while True: if is_prime(num+3): num += 4 continue if is_prime(num+2): num += 3 continue if is_prime(num+1): num += 2 continue n1 = len(set(prime_factors(num))) == 4 n2 = len(set(prime_factors(num+1))) == 4 n3 = len(set(prime_factors(num+2))) == 4 n4 = len(set(prime_factors(num+3))) == 4 if n1 and n2 and n3 and n4: print(num) break
def main():
for x in itertools.count(9, 2):
if not primes.is_prime(x) and not is_goldbach_num(x):
return x
def test_negatives_are_notprime(self):
self.assertFalse(is_prime(-3))
self.assertFalse(is_prime(-7))
# , where |a|<1000 and |b|≤1000 # where |n| # is the modulus/absolute value of n # e.g. |11|=11 and |−4|=4 # Find the product of the coefficients, a # and b, for the quadratic expression that produces the maximum number of primes for consecutive values of n, starting with n=0. from utils.primes import is_prime rangea = 1000 rangeb = 1001 max_n = 0 for a in range(-rangea, rangea): for b in range(-rangeb, rangeb): n = 0 while True: num = n**2 + a * n + b if not is_prime(num): if n >= max_n: max_n = n max_a = a max_b = b # print(max_a, max_b, max_n) break n += 1 # print(max_a, max_b, max_n) print(max_a * max_b)
