Project Euler Problem 38

Pandigital multiples

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?

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Source code examples on Github

We are have to find pandigital number greater than 918273645. So first product in concatenation should start with 9.

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* N is not 9 (9  and (1,2,3,4,5) already produce 918273645)
* N = 92 .. 98   and (1,2,3,4) cannot produce 9-digit number
* N = 921 .. 987 and (1,2,3) too many digits
* N = 921 .. 987 and (1,2) too few digits
* N = 9213 .. 9876 and (1,2) -> 9-digit number
* no other combintions works

Erlang version

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#!/usr/bin/env escript
%% -*- erlang -*-
%%! -smp enable -sname p32
% vim:syn=erlang

-mode(compile).

main(_) -> io:format("Answer ~p ~n", [t38(9876)]).

t38(9212) -> 0;
t38(N)    -> case isPandigital(integer_to_list(N) ++ integer_to_list(N*2)) of
                {true, A} -> A;
                false     -> t38(N-1)
             end.

isPandigital(N) ->
    case {length(N), lists:usort(N)} of
        {9, "123456789"} -> {true, N}
        ;_               -> false
    end.

Python version

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#!/usr/bin/python

def t38():
    for n in range(9876,9213,-1):
        d = str(n) + str(n*2)
        if is_pandigital(d): return d
    return 0

def is_pandigital(n):
    if len(n) != 9:
        return False
    if "".join(sorted(set(n))) == "123456789":
        return True
    else:
        return False


print "Answer: %s" % t38()