##Take the number 192 and multiply it by each of 1, 2, and 3: ## ## 192 1 = 192 ## 192 2 = 384 ## 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 max = 987654321 t = [] from tools import is_pandigital_1to9 for i in range(1,100000): n = 1 p = "" while len(p)<9: p = p + str(n*i) n = n+1 if is_pandigital_1to9(p): t.append(p) print t[len(t)-1]
# The Fibonacci sequence is defined by the recurrence relation: # # Fn = Fn1 + Fn2, where F1 = 1 and F2 = 1. # It turns out that F541, which contains 113 digits, is the first # Fibonacci number for which the last nine digits are 1-9 pandigital # (contain all the digits 1 to 9, but not necessarily in order). # And F2749, which contains 575 digits, is the first Fibonacci # number for which the first nine digits are 1-9 pandigital. # # Given that Fk is the first Fibonacci number for which the first # nine digits AND the last nine digits are 1-9 pandigital, find k. from tools import is_pandigital_1to9 f1 = 1 f2 = 1 k = 2 loop = True while loop: temp = f1 f1 += f2 f2 = temp k += 1 if is_pandigital_1to9(str(f1)[0: 9]): if is_pandigital_1to9(str(f1)[-9:]): print k loop = False