def check_consecutive_numbers_having_distinct_prime_factors(start, n):
for i in xrange(start, start+n):
distinct_pf = set()
pf = prime_factors(i)
for p in pf:
distinct_pf.add(p)
if len(distinct_pf) != n:
return False
return True
def count_divisors(n):
pfs = prime_factors(n)
size = max(pfs)
factor_count = [0] * (size+1)
for i in pfs:
factor_count[i] += 1
count = 1
for i in factor_count:
if i > 0:
count *= (i + 1)
return count
def eul88(num): found_lengths = [] found_values = [] solution = [] for i in range(4,num*2): used = False factors = util.prime_factors(i) factor_sets = util.recuse_over_factors(factors,[[]]) lengths = map(lambda x:util.solve_length_for_numbers(x),factor_sets) for length in lengths: if not length in found_lengths and length <= 12000: found_lengths.append(length) solution.append([length,i]) if len(found_lengths) == 12000: break if not used: found_values.append(i) used = True print i,used return solution,found_values
def main():
# The period of a repeating decimal is the multiplicative order of 10
# modulo the denominator, i.e. the lowest number k such that 10**k = 1
# modulo the denominator.
# Note that if the prime decomposition of n consists only of 2 and 5, it
# will be a terminating decimal and can be immediately excluded.
# Additionally, the period of the repeating decimal is always less than the
# denominator.
max_period = 0
answer = 0
for n in range(2, 1000):
if all(x == 2 or x == 5 for x in prime_factors(n)):
continue
for k in range(1, n):
if (10**k) % n == 1:
if k > max_period:
answer = n
max_period = k
break
print(answer)
def solution():
N = 600851475143
pfs = prime_factors(N)
result = pfs[-1]
return result
def totient(n):
pf = set(util.prime_factors(n))
return n - 1 + sum([(-1)**i * ((n - 1) / reduce(lambda x, y: x * y, comb))
for i in range(1,
len(pf) + 1)
for comb in itertools.combinations(pf, i)])
def phi(n):
r = 1.0
for p in prime_factors(n):
r *= 1.0 - 1.0 / float(p)
r *= n
return r
def main():
print(max(prime_factors(600851475143)))
def bruteForce():
startTime = time.time()
N = 10000000
minResult = N
index = 0
roundStartTime = time.time()
for i in range(2, N):
if i % 100000 == 0:
print('{0} done, this round takes {1} seconds, elapsed {2} seconds'.format(i, time.time()-roundStartTime, time.time()-startTime))
roundStartTime = time.time()
phi_i = phi(i)
current = float(i) / float(phi(i))
if isPermutation(i, phi_i) and current < minResult:
minResult = current
index = i
print('found {}, phi_i: {}, min: {}, prime factors: {}'.format(i, phi_i, minResult, prime_factors(i)))
return index
def phi(n):
r = 1.0
for p in prime_factors(n):
r *= (1.0 - 1.0/float(p))
r *= n
return int(round(r))
def main():
print max(prime_factors(600851475143))
def square_free(x):
return len(prime_factors(x)) == len(unique_prime_factors(x))
def problem_3(n): return max(prime_factors(n))
def largest_prime_factor(n):
return max(util.prime_factors(n))
from itertools import count
from util import prime_factors
for n in count(2*3*5*7):
if all(map(lambda i: len(prime_factors(n+i)) == 4, range(4))):
print(n)
break
