Question
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, , 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.
Haskell
import Data.List (nub, sort)
candidates :: [(String, Integer)]
candidates = [(concatMap show [a, b, a*b], a*b) | a <- [1..2000], b <- [1..50]]
pandigital :: String -> Bool
pandigital = (== "123456789") . sort
main :: IO ()
main = print $ sum $ nub [p | (digits, p) <- candidates, pandigital digits]
Python
#!/usr/bin/env python
def is_pandigital(*args, **kwargs):
num = sorted(''.join(str(arg) for arg in args))
try:
if kwargs['length'] and len(num) != kwargs['length']:
return False
except KeyError:
pass
for i in range(len(num)):
if str(i+1) != str(num[i]):
return False
return True
def main():
pandigitals = set()
total = 0
for multiplicand in range(1, 5000):
for multiplier in range(1, 100):
product = multiplicand * multiplier
if is_pandigital(multiplicand, multiplier, product, length=9):
pandigitals.add(product)
print(sum(pandigitals))
if __name__ == "__main__":
main()