Solved ProjectEuler/028,048

ProjectEuler/048/desc.yml

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+title: Find the last ten digits of 11 + 22 + ... + 10001000.
+url: http://projecteuler.net/problem=48
+
+desc: |
+  The series, 1^1 + 2^2 + 3^3 + ... + 10^10 = 10405071317.
+  Find the last ten digits of the series, 1^1 + 2^2 + 3^3 + ... + 1000^1000.
+  
+solution: |
+  Bruteforce
+
+solutions:
+  solve.php:
+    desc: Using BCMath to cope with large numbers
+    language: php
+  solve.rb:
+    desc: Basic solution
+    language: ruby

ProjectEuler/048/solve.php

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+<?php
+define('MAX',1000);
+$sum = 0;
+for($c =1; $c <= MAX;$c++) {
+	$sum = bcadd($sum,bcpow($c,$c));
+}
+echo substr($sum,-10,10);

ProjectEuler/048/solve.rb

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+MAX = 1000
+sum = 0 
+(1..MAX).each do |digit|
+	sum += digit**digit
+end
+puts sum.to_s[-10..-1]