The task in Problem 32 of Project Euler is:
We shall say that an $n$-digit number is pandigital if it makes use of all the digits 1 to n exactly once; for example, the 5-digit number, 15234, is 1 through 5 pandigital. The product 7254 is unusual, as the identity, $39 \cdot 186 = 7254$, containing multiplicand, multiplier, and product is 1 through 9 pandigital. Find the sum of all products whose multiplicand/multiplier/product identity can be written as a 1 through 9 pandigital. HINT: Some products can be obtained in more than one way so be sure to only include it once in your sum.
How to solve it
We have to get a check, if a number is pandigital. It could look like this:
def isPandigitalString(string): """ Check if string contains a pandigital number. """ digits = len(string) if digits >= 10: return False for i in xrange(1,digits+1): if str(i) not in string: return False return True
We also need a check if a product of two numbers is 9-pandigital:
def gives9PandigitalProduct(a, b): numbers = str(a) + str(b) + str(a*b) if len(numbers) != 9: return False return isPandigitalString(numbers)
Now you need to figure out how to go through all possible combinations:
products =  for a in xrange(0, 100000): for b in xrange(a, 100000): if len(str(a*b) + str(a) + str(b)) > 9: break if gives9PandigitalProduct(a, b): products.append(a*b) print("%i x %i = %i" % (a, b, a*b)) print(sum(set(products)))
This is from Thaddeus Abiye from Ethiopia:
print sum(set(map(lambda x:int(x[0:4]),filter(lambda x:sorted([i for i in x])==map(str,range(1,10)),[str(a*b)+str(a)+str(b) for a in range(1,2000) for b in range(1,100)]))))
It needs one line and 173 characters, but I think it's hard to read.
Data about my solution
- It worked in less than a second.
- 28 LOC (including whitespaces and comments)
- 719 characters for this solution (including whitespace and comments)