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.
Haskell
import Data.List (nub, sort)
candidates :: [(String, Integer)]
= [(concatMap show [a, b, a*b], a*b) | a <- [1..2000], b <- [1..50]]
candidates
pandigital :: String -> Bool
= (== "123456789") . sort
pandigital
main :: IO ()
= print $ sum $ nub [p | (digits, p) <- candidates, pandigital digits] main
$ 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):
= sorted(''.join(str(arg) for arg in args))
num
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():
= set()
pandigitals = 0
total for multiplicand in range(1, 5000):
for multiplier in range(1, 100):
= multiplicand * multiplier
product if is_pandigital(multiplicand, multiplier, product, length=9):
pandigitals.add(product)print(sum(pandigitals))
if __name__ == "__main__":
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
real 0m2.417s
user 0m2.377s
sys 0m0.040s