def problem131(): seen = 0 for m in count(2): factors = set(factorize(m)) n = m**3 for k in range(0,n,product(factors)**2): if any( k % p != 0 for p in factors): continue if k % product(factors) != 0: print(k) if (n**3 - k**3) % n**2 == 0 and isPrime((n**3 - k**3) // n**2): p = (n**3 - k**3) // n**2 if p > 10**6: return seen print(n, p, k) seen += 1
def problem231(): n = 20000000 c = 15000000 total = 1 for factor in genFactors(n): prod = product(factor) if n - c < prod <= n: total += sum(factor) if 1<= prod <= c: total -= sum(factor) return total
def problem108a():
recordN = 10**9
recordSolutions = 0
for n in genProducts(10**6,[2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59]):
n = product(n)
if n > 10:
m = basicSolutions(n)
#if m >= 1000 and n < recordN:
if m > recordSolutions:
print(n,m)
recordN = n
recordSolutions = m
def problem127():
GOAL = 120000
rad = {} # rad[6] = {2,3}, radn[8] = {2}
for primes in genFactors(GOAL):
rad[product(primes)] = (set(primes), product(set(primes)))
def relprime(s, t):
return s & t == set()
found = 0
total = 0
for b in range(1, GOAL):
for a in range(1, min(b, GOAL - b)):
c = a + b
x, y, z = rad[a], rad[b], rad[c]
if x[0] & y[0] != set():
continue
if x[1] * y[1] * z[1] < c:
found += 1
total += c
return total
def problem108():
def update(minimum, n=1, divs=1, index=0, power=50):
# Worst than what we found so far or used up all the primes
if n >= minimum or index >= len(primes):
return minimum
# We found n with enough divisors
if divs >= 2*(4*10**6):
return n
# the current prime we are looking at
p = primes[index]
# All the possible powers
for e in range(1, power+1):
minimum = min(minimum, update(minimum, n * p**e, divs * (2*e+1), index+1, e))
return minimum
primes = primesUpTo(50)
minimum = product(primes)
return update(minimum)
def problem108():
for n in genProducts(10**6,[2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59]):
if numberOfSolutions(product(n)) > 10**3:
return product(n)
def divisors(n):
if type(n) == type(0):
return {product(a) for a in powerset(factorize(n))}
if type(n) in [type(()), type(set())]:
return {product(a) for a in powerset(n)}
A little observation shows us that
(5k)! ≡g 5!^k * k!
So our function f(n) is defined recursively:
f(0) = 1
f(n = 5k + j) = g((5k + 1)(5k + 2)...(5k + j) * 12k * f(k))
'''
def f(n):
''' Returns the last 5 trailing non-zero digits of n!'''
if n == 0:
return 1
r = product(5 * (n // 5) + i + 1 for i in range(n % 5))
r *= pow(12, n // 5, 10 ** 5) * f(n // 5)
while r % 10 == 0:
r //= 10
return r % 10 ** 5
def problem160():
return f(10 ** 12)
from cProfile import run
if __name__ == "__main__":
# run("problem160()")
print(problem160() == 16576)
def PSR(primes):
return max([product(c) for c in powerset(primes) if product(c) < product(primes) ** (1 / 2)])
def divisors(n):
if type(n) == type(0):
return {product(a) for a in powerset(factorize(n))}
if type(n) in [type(()), type(set())]:
return {product(a) for a in powerset(n)}
