def problem35():
"""
The number, 197, is called a circular prime because all rotations
of the digits: 197, 971, and 719, are themselves prime.
There are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17,
31, 37, 71, 73, 79, and 97.
How many circular primes are there below one million?
"""
count = 1
for i in range(3, 1000000):
m = str(i)
not_prime = False
for d in m:
if int(d) % 2 == 0:
not_prime = True
break
if not_prime:
continue
for i in range(0, len(m)):
m = m[1:] + m[0]
if not mlib.is_prime(int(m)):
not_prime = True
break
if not not_prime:
count += 1
return count
def problem3():
"""
The prime factors of 13195 are 5, 7, 13 and 29.
What is the largest prime factor of the number 600851475143 ?
"""
num = 600851475143
r = int(math.floor(math.sqrt(num)))
if r%2 == 0:
r += 1
for i in range(r, 0, -2):
if num%i == 0 and mlib.is_prime(i):
return i
def problem37():
"""
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.
"""
count = 0
n = 10
sum = 0
while count < 11:
n += 1
is_cprime = True
for d in str(n):
# 2 on the edge can be prime
if d in "0468":
is_cprime = False
p = n
while p > 0:
if not is_cprime or not mlib.is_prime(p):
is_cprime = False
break
p /= 10
p = n
while p > 0:
if not is_cprime or not mlib.is_prime(p):
is_cprime = False
break
p = p - int(str(p)[0]) * 10 ** (len(str(p)) - 1)
if is_cprime:
count += 1
sum += n
return sum
def problem7():
"""
By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13,
we can see that the 6^(th) prime is 13.
What is the 10001^(st) prime number?
"""
val = 1
count = 0
while True:
val += 1
if mlib.is_prime(val):
count += 1
if count == 10001:
return val
def problem41():
# must have < 9 digits, 9-pandigit is div by 9
digits = '12345678'
max_p = 0
for i in range(8, 1, -1):
gen = mlib.gen_permutation(digits[:i])
while True:
try:
p = gen.next()
if mlib.is_prime(int(p)) and int(p) > max_p:
max_p = int(p)
except StopIteration:
break
return max_p
def problem47():
max = 10**6
pmap = [0 for i in range(max)]
for i in range(2, int(math.sqrt(max))):
if not mlib.is_prime(i):
continue
k = i
while k < len(pmap):
pmap[k] += 1
k += i
for i in range(10, len(pmap), 4):
if pmap[i] == 4:
if pmap[i+1] == 4 and pmap[i+2] == 4 and pmap[i+3] == 4:
return i
if pmap[i-1] == 4 and pmap[i-2] == 4 and pmap[i-3] == 4:
return i-3
