def pe0012():
print 'Project Euler Problem 12 https://projecteuler.net/problem=12'
p = eulerlib.sieve_of_atkin(65500)
n = 3
Dn = 2
cnt = 0
while cnt <= 500:
n = n + 1
n1 = n
if n1 % 2 == 0:
n1 = n1 / 2
Dn1 = 1
for i in range(0, p[-1]):
if p[i] * p[i] > n1:
Dn1 *= 2
break
exponent = 1
while n1 % p[i] == 0:
exponent += 1
n1 /= p[i]
if exponent > 1:
Dn1 *= exponent
if n1 == 1:
break
cnt = Dn * Dn1
Dn = Dn1
print 'Answer is ' + str(n*(n-1)/2)
def pe0003():
print 'Project Euler Problem 3 https://projecteuler.net/problem=3'
num = 600851475143
pf = eulerlib.sieve_of_atkin(int(math.sqrt(num)))
for p in reversed(pf):
if num % p == 0:
print 'Answer is ' + str(p)
return
def pe0026():
print 'Project Euler Problem 26 https://projecteuler.net/problem=26'
primes = eulerlib.sieve_of_atkin(1000)[::-1]
ans = 0
for p in primes:
x = 1
while (10 ** (x + 2) // p) % 1000 != (10 ** (2 * x + 2) // p) % 1000:
x += 1
if x + 1 == p:
ans = p
break
print 'Answer is ' + str(ans)
def pe0050():
print 'Project Euler Problem 50 https://projecteuler.net/problem=50'
primes = eulerlib.sieve_of_atkin(4000)
for i in range(len(primes) - 1, 1, -1):
start = 0
while True:
prime_sum = sum(primes[start:start + i])
if prime_sum > 1000000:
break
if eulerlib.is_prime(prime_sum):
print 'Answer is %d' % prime_sum
return
start += 1
def pe0049():
print 'Project Euler Problem 49 https://projecteuler.net/problem=49'
primes = [p for p in eulerlib.sieve_of_atkin(9999) if p > 1000]
for i in range(0, len(primes) - 1):
count = 1
prime_perm = [primes[i]]
for j in range(i + 1, len(primes)):
if sorted(str(primes[i])) == sorted(str(primes[j])):
count += 1
prime_perm.append(primes[j])
if count == 3 and abs(prime_perm[1] - prime_perm[0]) == abs(prime_perm[2] - prime_perm[1]):
print 'Answer is ' + ''.join(str(c) for c in prime_perm)
return
def pe0077():
print 'Project Euler Problem 77 https://projecteuler.net/problem=77'
primes = eulerlib.sieve_of_atkin(100)
print primes
ways = [0] * 100
ways[0] = 1
for p in primes:
for z in range(p, 100):
ways[z] = ways[z] + ways[z - p]
for w in ways:
if w >= 5000:
print 'Answer is %d' % ways.index(w)
break
def pe0046():
print 'Project Euler Problem 46 https://projecteuler.net/problem=46'
primes = eulerlib.sieve_of_atkin(10000)
i = 33
keep_trying = True
while keep_trying:
j = 0
keep_trying = False
while i >= primes[j]:
if eulerlib.is_twice_square(i - primes[j]):
keep_trying = True
break
j += 1
i += 2
print 'Answer is %d' % (i - 2)
def pe0037():
print 'Project Euler Problem 37 https://projecteuler.net/problem=37'
primes = eulerlib.sieve_of_atkin(1000000)
trunc_primes = []
possible_primes = []
for i in primes:
if i < 100 or len(set(['0', '2', '4', '5', '6', '8']).intersection(set(str(i)))) == 0:
possible_primes.append(i)
for p in possible_primes:
for z in range(1, len(str(p))):
if int(str(p)[:z]) not in primes or int(str(p)[z:]) not in primes:
break
if z == len(str(p)) - 1:
trunc_primes.append(p)
print trunc_primes
print 'Answer is %d' % sum(trunc_primes)
def pe0069():
print 'Project Euler Problem 69 https://projecteuler.net/problem=69'
# To have higher n / phi(n), phi(n) should be smaller
# phi(n) = n * sum(1 - 1/p), where p is all prime factors of n
# hence if we have more prime factors for a number then that will have less phi(n)
limit = 1000001
primes = eulerlib.sieve_of_atkin(50)
prod = 1
pnum = []
for p in primes:
prod *= p
if prod < limit:
pnum.append(p)
else:
break
print 'Answer is ' + str(reduce(mulop, pnum, 1))
def pe0035():
print 'Project Euler Problem 35 https://projecteuler.net/problem=35'
circular_primes = [2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, 97]
primes = eulerlib.sieve_of_atkin(1000000)
possible_primes = []
for i in primes:
if i > 100 and len(set(['0', '2', '4', '5', '6', '8']).intersection(set(str(i)))) == 0:
possible_primes.append(i)
for p in possible_primes:
r = list(eulerlib.get_num_digits_rotated(p))
for j in range(0, len(r)):
if r[j] not in primes:
break
if j == len(r) - 1:
for z in r:
circular_primes.append(z)
print 'Answer is %d' % len(set(circular_primes))
def pe0070():
print 'Project Euler Problem 70 https://projecteuler.net/problem=70'
# To have lower n / phi(n), phi(n) should be bigger.
# Only when n is prime it's bigger with phi(n) = n - 1, but n - 1 will not be permutation of n
# phi(n) = n * sum(1 - 1/p), where p is all prime factors of n
# Hence if we have less prime factors for a number then that will have higher phi(n), hence go for 2 prime factors.
# phi(p1 x p2) = phi(p1) x phi(p2) = (p1 - 1) x (p2 - 1)
# It's rational to start from sqrt(10^7)
limit = 10 ** 7
primes = eulerlib.sieve_of_atkin(2 * int(math.sqrt(limit)))
ans = 1
min_num = float(10 ** 7)
for p1, p2 in combinations(primes, 2):
if p1 * p2 < limit:
n = p1 * p2
phi = (p1 - 1) * (p2 - 1)
tmp = float(n) / phi
if sorted(str(n)) == sorted(str(phi)) and tmp < min_num:
min_num = tmp
ans = n
print 'Answer is ' + str(ans)
def pe0060():
print 'Project Euler Problem 60 https://projecteuler.net/problem=60'
# Trial and error to find the uppoerbound as 10000 for sensible execution time
primes = eulerlib.sieve_of_atkin(10000)
# removed 2 & 5 from primes since numbers ending with 2 & 5 are not primes
primes.remove(2)
primes.remove(5)
def isPrimePair(x, y):
if eulerlib.is_prime(int(str(x) + str(y))) and eulerlib.is_prime(int(str(y) + str(x))):
return True
else:
return False
def find5thpair(primes):
for i in range(len(primes)):
for j in range(i + 1, len(primes)):
pairs = []
if isPrimePair(primes[i], primes[j]):
pairs.append(primes[i])
pairs.append(primes[j])
for a in range(j + 1, len(primes)):
if isPrimePair(pairs[0], primes[a]) and isPrimePair(pairs[1], primes[a]):
pairs.append(primes[a])
for b in range(a + 1, len(primes)):
if isPrimePair(pairs[0], primes[b]) and isPrimePair(pairs[1], primes[b]) and \
isPrimePair(pairs[2], primes[b]):
pairs.append(primes[b])
for c in range(b + 1, len(primes)):
if isPrimePair(pairs[0], primes[c]) and \
isPrimePair(pairs[1], primes[c]) and isPrimePair(pairs[2], primes[c]) and \
isPrimePair(pairs[3], primes[c]):
pairs.append(primes[c])
return pairs
result = find5thpair(primes)
print 'Answer is %d' % sum(result)
def pe0027():
print 'Project Euler Problem 27 https://projecteuler.net/problem=27'
primes = eulerlib.sieve_of_atkin(1000)[::-1]
for i in range(0, len(primes)):
primes.append(-primes[i])
cons_big = 0
coeff = [0, 0]
for b in primes:
for a in range(-999, 1000):
n, pc = 0, 0
while True:
t = (n * n) + (a * n) + b
n += 1
if t < 0:
break
elif eulerlib.is_prime(t):
pc += 1
if cons_big < pc:
cons_big = pc
coeff[0] = a
coeff[1] = b
else:
break
print 'Answer is %d' % (coeff[0] * coeff[1])
def pe0010():
print 'Project Euler Problem 10 https://projecteuler.net/problem=10'
p = eulerlib.sieve_of_atkin(2000000)
print 'Answer is ' + str(sum(p))