Project Euler Problem 30

Digit fifth powers.

Surprisingly there are only three numbers that can be written as the sum of fourth powers of their digits:

            1634 = 1^4 + 6^4 + 3^4 + 4^4
            8208 = 8^4 + 2^4 + 0^4 + 8^4
            9474 = 9^4 + 4^4 + 7^4 + 4^4

As 1 = 1^4 is not a sum it is not included.
The sum of these numbers is 1634 + 8208 + 9474 = 19316.
Find the sum of all the numbers that can be written as the sum of fifth powers of their digits.

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Source code examples on Github

A number x with n digits and the sum s of the fifth powers of its digits satisfy x>=10^(n-1) and s<=n^5 respectively.
Thus it must be 10^(n-1)<=n^5, which implies 1<=n<=6, and x<=6*9^5.

Python version

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#!/usr/bin/python

def power_of_digits(n, p):
    s = 0
    while n > 0:
        d = n % 10
        n = n / 10
        s = s + pow(d, p)
    return s


print sum(n for n in xrange(2, 6*9**5) if power_of_digits(n, 5) == n)

Perl version

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#!/usr/bin/perl -l

use strict;
use warnings;
use List::Util qw/sum/;

print sum grep { $_ == sum map $_**5, /./g } 2..6*9**5;

Erlang version

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#!/usr/bin/env escript
%% -*- erlang -*-
%%! -smp enable -sname p30
% vim:syn=erlang

-mode(compile).

main(_) -> 
    Answer = lists:sum(lists:filter(fun(N)-> sump(N,10,0) == N end, lists:seq(2,6*p5(9)))),
    io:format("Answer ~p ~n", [Answer]).

p5(D) -> D*D*D*D*D.

sump(0,_,A) -> A;
sump(N,B,A) when N < B -> A+p5(N);
sump(N,B,A) -> sump(N div B, B, A + p5(N rem B)).