def smallest_divisible(n):
"""
Finds the smallest positive number that is evenly divisible by all the
numbers from 1 to n.
"""
factors_list = []
for i in range(1, n):
ls = prime_factorization(i)
for item in factors_list:
if item in ls:
ls.remove(item)
factors_list.extend(ls)
product = 1
for x in factors_list:
product *= x
return product
1: 1
3: 1,3
6: 1,2,3,6
10: 1,2,5,10
15: 1,3,5,15
21: 1,3,7,21
28: 1,2,4,7,14,28
We can see that 28 is the first triangle number to have over five divisors.
What is the value of the first triangle number to have over five hundred
divisors?
'''
from prime import prime_sieve, prime_factorization
from operator import mul
primes = prime_sieve(int(500 ** .5) + 1)
n = 0
while True:
n += 1
triangle = n * (n + 1) / 2
factors = prime_factorization(triangle, primes)
if factors:
divisors = reduce(mul, (c + 1 for (factor, c) in factors))
if divisors > 500:
print triangle
break
def number_of_divisors(n):
return reduce(operator.mul,
[v + 1 for v in prime_factorization(n).values()], 1)
1: 1
3: 1,3
6: 1,2,3,6
10: 1,2,5,10
15: 1,3,5,15
21: 1,3,7,21
28: 1,2,4,7,14,28
We can see that 28 is the first triangle number to have over five divisors.
What is the value of the first triangle number to have over five hundred
divisors?
'''
from prime import prime_sieve, prime_factorization
from operator import mul
primes = prime_sieve(int(500**.5) + 1)
n = 0
while True:
n += 1
triangle = n * (n + 1) / 2
factors = prime_factorization(triangle, primes)
if factors:
divisors = reduce(mul, (c + 1 for (factor, c) in factors))
if divisors > 500:
print triangle
break
14 = 2 × 7
15 = 3 × 5
The first three consecutive numbers to have three distinct prime factors are:
644 = 2² × 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 prime import prime_sieve, prime_factorization
primes = prime_sieve(10**4)
n = 1000
c = []
while True:
n = n + 1
if len(prime_factorization(n)) == 4:
c.append(n)
else:
del c[:]
if len(c) == 4:
print c
break
def distinct_primes(x, n):
return len(prime_factorization(x).keys()) == n
def lcm(a,b): return (a*b) / gcd(a,b)
cin, cout = stdin.readline, stdout.write
START = time()
MOD = 10**9 + 7
LIMIT = 20000
def times(my_counter, time):
new_counter = Counter(my_counter)
for prime in new_counter:
new_counter[prime]*=time
return new_counter
f_f = [0 for i in xrange(LIMIT+1)]
f_f[0] = Counter()
for i in xrange(1, LIMIT+1):
f_f[i] = prime_factorization(i) + f_f[i-1]
denominator = [0]*(LIMIT+1)
denominator[0] = Counter()
for i in xrange(1, LIMIT+1):
denominator[i] = denominator[i-1] + times(f_f[i], 2)
cout('Done preprocessing\n')
def d_of_n(num):
numerator = times(f_f[num], num+1)
my_number = numerator - denominator[num]
d_of_n_ = 1
for prime, count in my_number.items():
num = (pow(prime, count+1, MOD) - 1 +MOD)%MOD
prime_pow_sum = (num* pow(prime-1, MOD-2, MOD))%MOD
d_of_n_ = (prime_pow_sum * d_of_n_)%MOD
def distinct_primes(x, n):
return len(prime_factorization(x).keys()) == n
14 = 2 × 7
15 = 3 × 5
The first three consecutive numbers to have three distinct prime factors are:
644 = 2² × 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 prime import prime_sieve, prime_factorization
primes = prime_sieve(10 ** 4)
n = 1000
c = []
while True:
n = n + 1
if len(prime_factorization(n)) == 4:
c.append(n)
else:
del c[:]
if len(c) == 4:
print c
break
def number_of_divisors(n):
return reduce(operator.mul, [v + 1 for v in prime_factorization(n).values()], 1)
