def ans():
largest = 0
for i in range(10000):
string = ''
for product in [i * j for j in range(1, 10)]:
string += str(product)
if 9 < len(string):
break
if is_pandigital(string, to=9) and largest < int(string):
largest = int(string)
return largest
def ans():
products = set()
for i in range(1, 2000):
for j in range(i, 2000):
product = i * j
string = str(i) + str(j) + str(product)
length = len(string)
if 9 < length:
break
if is_pandigital(int(string), to=9):
products.add(product)
return sum(products)
#print r
#print
#print s[0:9]
#print s[-9:]
#print
#print k
#break
while True:
r = fibo.next()[-9:]
k += 1
if k == 541:
print r
print 'trololo'
print 'lala', pandigital.is_pandigital(r)
if len(str(r)) < 9:
continue
s = str(r)
if k % 1000 == 0:
print k
if pandigital.is_pandigital(s):
f = str(fibonacci.fibonacci(k))
print k, s, f[0:9]
if pandigital.is_pandigital(f[0:9]):
print r
print
print f[0:9]
print f[-9:]
print
print k
from pandigital import is_pandigital
assert is_pandigital(1) == False
assert is_pandigital(1, 2) == False
assert is_pandigital('10', 2) == True
assert is_pandigital('10', 2, True) == False
assert is_pandigital('110', 2, unique=True) == False
assert is_pandigital(1234567890) == True
assert is_pandigital(1234567890, zeroless=True) == False
assert is_pandigital(123, b=4, zeroless=True) == True
assert is_pandigital('021', b=3) == True
assert is_pandigital('021', b=3, unique=True) == True
def solution():
"""i only bother searching for i = 2 as this gave a correct solution"""
x = {int(str(n*1)+str(n*2)) for n in range(0,9999) if (len(str(n*1)+str(n*2))==9 and pandigital.is_pandigital({n*1, n*2}))}
return max(x)
# 192 × 3 = 576
#
# By concatenating each product we get the 1 to 9 pandigital, 192384576.
# We will call 192384576 the concatenated product of 192 and (1,2,3)
#
# The same can be achieved by starting with 9 and multiplying by 1, 2, 3, 4, and 5,
# giving the pandigital, 918273645, which is the concatenated product of 9 and (1,2,3,4,5).
#
# What is the largest 1 to 9 pandigital 9-digit number that can be formed as the
# concatenated product of an integer with (1,2, ... , n) where n > 1?
from pandigital import is_pandigital
largest_pandigital = 0
# The largest number with pandigital potential that can be multiplied to a total of nine digits
# one of the given numbers starts with 9, so we must limit ourselves to start with numbers that start with 9
for i in range(9876, 8, -1):
lp = str(i)
if '0' in str(
i
): # if 0 is in the number it does not fit the given definition of pandigital
continue
for n in range(2, 6): # our largest n is 5, as shown in the example
lp += str(i * n)
if len(lp) >= 9:
break
if is_pandigital(lp) and int(lp) > largest_pandigital:
largest_pandigital = int(lp)
print(largest_pandigital) # 932718654
import pandigital
results = set()
for i in range(1, 10000000001):
if i % 1000 == 0:
print(i, results)
for j in range(1, 1000000001):
r = i * j
if len(str(i) + str(j) + str(r)) > 9:
break
if j % 100 == 0:
print('#', i, j, r)
if pandigital.is_pandigital(str(i) + str(j) + str(r)):
print(i, j, r)
results.add(r)
