def panDigitProducts( permList ):
"""Take each permutation of digits and slice into 3 pieces.
Check to see if each piece is a proper product.
"""
for ai in range(1,9):
for bi in range(ai):
for perm in permList:
i, j, k = perm[:ai], perm[ai:ai+bi], perm[ai+bi:]
if number(i)*number(j) == number(k):
yield number(k)
def substringDivisibility( i ):
"""The substring divisibility property. The number
is represented as a sequence of digits.
>>> from euler43 import substringDivisibility
>>> substringDivisibility( digits(1406357289) )
True
>>> substringDivisibility( digits(1406357298) )
False
"""
n2= number(i[1:4])
n3= number(i[2:5])
n5= number(i[3:6])
n7= number(i[4:7])
n11= number(i[5:8])
n13= number(i[6:9])
n17= number(i[7:10])
test = (
n2 % 2 == 0,
n3 % 3 == 0,
n5 % 5 == 0,
n7 % 7 == 0,
n11%11 == 0,
n13%13 == 0,
n17%17 == 0
)
return all( test )
def genPanDigitalProdConcat():
"""Get the list of pandigital products.
>>> from euler38 import genPanDigitalProdConcat
>>> pdpc= list( genPanDigitalProdConcat() )
>>> pdpc.sort()
>>> pdpc # doctest: +ELLIPSIS
[123456789, 192384576, 219438657, ..., 927318546, 932718654]
"""
for i in range(1000000):
seq= prodConcat(i)
if pandigital(seq):
yield number(seq)
def PDSD_gen():
"""Generate pan-digital numbers with the substring divisibility property.
The number is represented as a sequence of digits.
>>> from euler43 import PDSD_gen
>>> numbers= list( PDSD_gen() )
>>> 1406357289 in numbers
True
>>> numbers
[1406357289, 1430952867, 1460357289, 4106357289, 4130952867, 4160357289]
"""
pandigital10= Permutation(range(10))
for i in pandigital10.nextPerm():
if substringDivisibility(i):
yield number(i)
def reverseAndAdd( n ):
"""
>>> from euler55 import reverseAndAdd
>>> from euler04 import palindrome
>>> reverseAndAdd(47)
121
>>> palindrome( reverseAndAdd(47) )
True
>>> reverseAndAdd( reverseAndAdd( reverseAndAdd( 349 ) ) )
7337
>>> palindrome( reverseAndAdd( reverseAndAdd( reverseAndAdd( 349 ) ) ) )
True
"""
d = digits(n)
d.reverse()
return n + number(d)
def pandigitalPrimes(size):
"""Generate all pan-digital primes of a given size.
>>> from euler41 import pandigitalPrimes
>>> pd4 = list( pandigitalPrimes(4) )
>>> 2143 in pd4
True
>>> pd4
[1423, 2143, 2341, 4231]
"""
permN= Permutation( range(1,size+1) )
for p in permN.nextPerm():
ld= p[-1]
if ld % 2 == 0: continue # skip even numbers
if ld == 5: continue # skip 5's, also
n= number( p )
if isprime(n):
yield n
def answer():
for a,b,c in primeGapPerm():
if a == 1487 and b == 4817 and c == 8147:
continue
return number( digits(a) + digits(b) + digits(c) )
