def answer():
# Lower bound is 1487
found = 0
primes = primes_to(10000, 1487)
for a in range(len(primes)):
for b in range(a+1, len(primes)):
prime_a = primes[a]
prime_b = primes[b]
c = prime_b + (prime_b - prime_a)
if c < 10000 and is_prime(c):
if is_perm(prime_a, prime_b) and is_perm(prime_a, c):
found = str(prime_a) + str(prime_b) + str(c)
break
if found:
break
return found
def answer():
# ratios = []
# for n in range(2, 10000000):
# # Euler's Product Formula
# primes = list(set(prime_factors(n)))
# product = 1
# for p in primes:
# product *= (1 - (1/p))
# if is_perm(n, product*n):
# ratios.append(1/product)
# return ratios.index(min(ratios)) + 2
"""
The above solution would work, but it takes far longer than the minute required
so the best is to analyze the code. It seems to minimize the ratio 1/product
we should maximize the product and take advantage of the fact that it is the
product of two large distinct primes close to the sqrt(10**7)
"""
ratios = {}
primes = primes_to(3800,2200)
for a in range(len(primes)):
for b in range(a+1, len(primes)):
n = primes[a] * primes[b]
if n < 10000000:
phi = (primes[a] - 1) * (primes[b] - 1)
ratio = n/phi
if is_perm(n, phi):
ratios[ratio] = n
return ratios[min(ratios.keys())]
def answer():
x = 100
while True:
if is_perm(x, 2*x) and is_perm(x, 3*x) and is_perm(x, 4*x) and is_perm(x, 5*x) and is_perm(x, 6*x):
return x
x += 1