# Project Euler Problem 32 Solution

## 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, $39 \times 186 = 7254$, 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.

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]
$ghc -O2 -o pandigital pandigital.hs$ time ./pandigital
real   0m0.045s
user   0m0.045s
sys    0m0.000s

## 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):
main()
$time python3 pandigital-products.py real 0m1.922s user 0m1.922s sys 0m0.000s ## Ruby #!/usr/bin/env ruby puts (1..4999).flat_map { |a| (1..99).map do |b| [a.to_s + b.to_s + (a*b).to_s, a*b] end }.select { |p| p[0].length == 9 && p[0].each_char.sort.join == "123456789" }.map { |p| p[1] }.uniq.reduce(:+) $ time ruby pandigital.rb
sys    0m0.040s