def right_truncs():
ans = [2, 3, 5, 7]
l = 0
changed = True
while changed:
l += 1
changed = False
work_seq = [s for s in ans if math.floor(math.log10(s)) == l - 1]
for s in work_seq:
for x in (1, 3, 7, 9):
n = 10 * s + x
if is_prime(n):
ans.append(n)
changed = True
return {s for s in ans if s > 10}
def main():
p = 0
for i, d in enumerate(diags()):
p += len([n for n in d if intlib.is_prime(n)])
r = p/(5+4*i)
if r < 0.1: return 2*i + 3
def is_left_truncatable(n):
s = str(n)
while s:
if not is_prime(int(s)): return False
s = s[1:]
return True
def is_right_truncatable(n):
while n > 0:
if not is_prime(n): return False
n = n // 10
return True
if __name__ == '__main__':
'''
4<= n <=9 を探索すれば良い.
1+2+3+4+5 = 15 より, 5ケタのpandigital numberは3の倍数
となるから, 調べなくて良い.
1+2+3+4+5+6 = 21
1 + ... + 8 = 36
1 + ... + 9 = 45
より6, 8, 9ケタも同様に対象外.
7ケタのうち, 末尾が3,7の数を調べれば良い.
'''
max_prime = 2143
for n in itertools.permutations('1234'):
n = int(''.join(n))
if intlib.is_prime(n):
max_prime = max((n, max_prime))
for n in itertools.permutations('124567'):
n = int(''.join(n) + '3')
if intlib.is_prime(n):
max_prime = max((n, max_prime))
for n in itertools.permutations('123456'):
n = int(''.join(n) + '7')
if intlib.is_prime(n):
max_prime = max((n, max_prime))
print(max_prime)