def truncatable(prime):
dig = intToDigits(prime)
for i in range(1, len(dig)):
trunc_r = digitsToInt(dig[:i])
if trunc_r not in sieve:
return False
trunc_l = digitsToInt(dig[i:])
if trunc_l not in sieve:
return False
return True
def concatproduct(n):
digits = []
for i in range(1, 10):
digits += intToDigits(n * i)
if len(digits) >= 9:
break
if pandigital9(digitsToInt(digits)):
return digitsToInt(digits)
return 0
def familysize(p):
if not repeatingdigits(p):
return 1
dig = intToDigits(p)
bestfamily = [p]
for oldd in set(dig) - {dig[-1]}:
family = [p]
if len([d for d in dig if d == oldd]) < 2:
continue
for newd in set(range(10)) - {oldd}:
if newd == 0 and dig[0] == oldd:
continue
newp = digitsToInt([newd if d == oldd else d for d in dig])
if newp in sieve:
family.append(newp)
if len(family) > len(bestfamily):
bestfamily = family
for pr in bestfamily:
memoize[pr] = len(bestfamily)
if len(bestfamily) == 8:
print(bestfamily)
return len(bestfamily)
def rotations(n):
dig = intToDigits(n)
for i in range(1, len(dig)):
yield digitsToInt(dig[i:] + dig[:i])
def concatprime(p1,p2):
d1 = intToDigits(p1)
d2 = intToDigits(p2)
c1 = digitsToInt(d1+d2)
c2 = digitsToInt(d2+d1)
return isprime(c1) and isprime(c2)
def palindromicInBinary(n):
binary = bin(n)[2:]
for a, b in zip(binary, reversed(binary)):
if a != b:
return False
return True
if __name__ == "__main__":
tic()
S = 0
for i in range(1, 1000, 2):
digits = intToDigits(i)
p = digitsToInt(list(reversed(digits)) + digits) #even length: cbaabc
if palindromicInBinary(p):
S += p
p = digitsToInt(list(reversed(digits)) +
digits[1:]) #odd length: cbabc
if palindromicInBinary(p):
S += p
if i < 100: #center zeros: cb00bc
p = digitsToInt(list(reversed(digits)) + [0] + digits)
if palindromicInBinary(p):
S += p
p = digitsToInt(list(reversed(digits)) + [0, 0] + digits)
if palindromicInBinary(p):
S += p
if i < 10: #center zeros: cb00bc
p = digitsToInt(list(reversed(digits)) + [0, 0, 0] + digits)
from Euler.digits import digitsToInt
if __name__ == "__main__":
tic()
S = 0
d = [None] * 10
for d789 in range(17, 987, 17):
d[7] = d789 // 100
d[8] = (d789 // 10) % 10
d[9] = d789 % 10
if d[7] == d[8] or d[7] == d[9] or d[8] == d[9]:
d[7:] = [None] * 3
continue
for d6 in [i for i in range(10) if i not in d]:
d[6] = d6
if digitsToInt(d[6:9]) % 13 != 0:
d[6] = None
continue
for d5 in [i for i in [0, 5] if i not in d]:
d[5] = d5
if digitsToInt(d[5:8]) % 11 != 0:
d[5] = None
continue
for d4 in [i for i in range(10) if i not in d]:
d[4] = d4
if digitsToInt(d[4:7]) % 7 != 0:
d[4] = None
continue
for d3 in [i for i in range(0, 10, 2) if i not in d]:
d[3] = d3
#if digitsToInt(d[3:6])%5 != 0: #not necessary because d5 is 0 or 5
first digit must be a one.
assuming unique digits.
"""
from Euler.tictoc import tic, toc
from Euler.eprint import eprint
from Euler.digits import intToDigits, digitsToInt
def repeatingdigits(n):
from math import floor, log10
d = intToDigits(n)
return len(set(d)) <= floor(log10(n))
if __name__ == "__main__":
tic()
for x_ in range(10**4, 10**7):
dig = [1] + intToDigits(x_)
x = digitsToInt(dig)
if repeatingdigits(x):
continue
if sorted(intToDigits(2 * x)) == sorted(dig):
if sorted(intToDigits(3 * x)) == sorted(dig):
if sorted(intToDigits(4 * x)) == sorted(dig):
if sorted(intToDigits(5 * x)) == sorted(dig):
if sorted(intToDigits(6 * x)) == sorted(dig):
print(x)
toc()
exit()
#!/usr/bin/env python3.6
"""
PROBLEM: 024
AUTHOR: Dirk Meijer
STATUS: done
EXPLANATION:
sympy solution
"""
from Euler.tictoc import *
from Euler.digits import digitsToInt
from sympy.combinatorics import Permutation
if __name__ == "__main__":
tic()
p = Permutation.unrank_lex(10, rank=1_000_000 - 1).array_form
print(digitsToInt(p))
toc()
exit()