def solve(num):
primes = list(sieveOfEratosthenes.getPrimes(num))
primefactors = map(factors, range(2, num + 1), [primes] * (num - 1))
minfactors = {}
for p in primes:
minfactors[p] = 0
for factor in primefactors:
for key in factor.keys():
if factor[key] > minfactors[key]:
minfactors[key] = factor[key]
product = 1
for prime in minfactors.keys():
product *= math.pow(prime, minfactors[prime])
return product
'''
A faster approach that explicitly makes use of the notion that prime
factorizations are unique and that for a prime p < N, the product
p^a (where a in the least integer such that p^(a+1) exceeds N) will be a
factor in at least one factorization of a positive integer n <= N.
The algorithm steps through the primes below the given limit and
calculates the exponent a (described above). Then you simply multiply
all the primes ^ their respective a's together to get the number.
'''
import sys
import math
import sieveOfEratosthenes
limit = int(sys.argv[1])
primes = list(sieveOfEratosthenes.getPrimes(limit))
product = 1
for p in primes:
count = 1
while(math.pow(p,count) <= limit):
product *= p
count += 1
print product
