'''
import time
t = time.time()
largest = 0
for n in range(2,10):
upper_bound = 10000 #exclusive
lower_bound = 1
if n == 2:
lower_bound = 5000
elif n == 3:
upper_bound = 334
lower_bound = 100
else:
upper_bound = 100
for i in range(lower_bound, upper_bound):
num = ''
for x in range(1,n+1):
num += str(i*x)
num = int(num)
if num > largest and cf.is_pandigital(num, 9):
largest = num
print largest
print time.time() - t
'''
Created on 2013-01-21
@author: paymahn
'''
'''
A 1-9 pandigital product is a 2-digit number
times a 3-digit number, or a 1-digit number
times a 4-digit number, with a product less
than 1e4. (3 2 and 4 1 also works, but don't
give any extra products.)
'''
import CommonFunctions as cf
found_number = set()
for i in range(1,100):
for j in range(100,10000):
num_str = str(i) + str(j) + str(i*j)
if len(num_str) == 9 and cf.is_pandigital(int(num_str), 9):
found_number.add(i*j)
print sum(found_number)
'''
Created on 2013-01-22
@author: paymahn
'''
'''
Turns out that 9 and 8 digit pandigital numbers are always divisible by 3
'''
import CommonFunctions as cf
import math
max = 7654322
primes = cf.generate_primes_less_than(max)
for i in reversed(primes):
if cf.is_pandigital(i, int(math.log10(i)) + 1):
print i
break;