def main():
for prime in prime_sieve_lazy():
if prime < 10**4: # rewind to the first 5-digit prime
continue
s = str(prime)
# we're searching from low to high
for digit in "012":
count = s.count(digit)
# Some observaions:
# * count has to be a multiple of 3, otherwise some of the new
# primes will be divisible by 3.
# * the rightmost digit can't be replaced (it could only be "1")
if count and not count % 3 and not digit == "1" == s[-1]:
cnt = 1
for i in range(int(digit) + 1, 10):
if i - cnt > 2:
break
if is_prime(int(s.replace(digit, str(i), count))):
cnt += 1
if cnt == 8:
return prime
def main():
factors = []
for prime in prime_sieve_lazy():
if pow(10, 10**9, 9 * prime) == 1: # R(k) = (10**k - 1) // 9
factors.append(prime)
if len(factors) == 40:
return sum(factors)
def main():
factors = []
for prime in prime_sieve_lazy():
if pow(10, 10**9, 9*prime) == 1: # R(k) = (10**k - 1) // 9
factors.append(prime)
if len(factors) == 40:
return sum(factors)
def main():
lim = 1000000
res = 1
for prime in prime_sieve_lazy():
if res * prime > lim:
break
res *= prime
return res
def main():
res = cnt = 0
for p in prime_sieve_lazy():
if is_truncatable(p):
res += p
cnt += 1
if cnt == 11:
return res
def main():
lim = 1000000
res = 1
for prime in prime_sieve_lazy():
if res*prime > lim:
break
res *= prime
return res
def main():
nways = 5000
for num in count(10):
ways = [0]*(num+1)
ways[0] = 1
for p in prime_sieve_lazy():
if p > num:
break
for i in range(p, num+1):
ways[i] += ways[i-p]
if ways[num] > nways:
return num
def main():
nways = 5000
for num in count(10):
ways = [0] * (num + 1)
ways[0] = 1
for p in prime_sieve_lazy():
if p > num:
break
for i in range(p, num + 1):
ways[i] += ways[i - p]
if ways[num] > nways:
return num
def main():
lim = 1_000_000
primes = prime_sieve_lazy()
next(primes)
next(primes)
res = 0
# p1 + d * k == 0 (mod p2).
for p1, p2 in pairwise(primes):
if p1 > lim:
break
d = 10**(ceil(log10(p1)))
dinv = pow(d, p2 - 2, p2)
k = ((p2 - p1) * dinv) % p2
res += p1 + d * k
return res
def main():
lim = 1_000_000
primes = prime_sieve_lazy()
next(primes)
next(primes)
res = 0
# p1 + d * k == 0 (mod p2).
for p1, p2 in pairwise(primes):
if p1 > lim:
break
d = 10**(ceil(log10(p1)))
dinv = pow(d, p2 - 2, p2)
k = ((p2 - p1) * dinv) % p2
res += p1 + d * k
return res
def main():
for n, p in islice(enumerate(prime_sieve_lazy(), 1), 0, None, 2):
if 2*p*n > 10**10:
return n
def main():
for count, prime in enumerate(prime_sieve_lazy(), 1):
if count == 10001:
return prime
def main():
for n, p in islice(enumerate(prime_sieve_lazy(), 1), 0, None, 2):
if 2 * p * n > 10**10:
return n