Solved ProjectEuler/028,048

ProjectEuler/005/solve.lua

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+i=0
+while(true) do
+	i = i + 20 
+	div = 19
+	while((i % div) == 0) do
+		div = div - 1
+		if(div == 0) then
+			print(i)
+			os.exit()
+		end
+	end
+end

ProjectEuler/028/desc.yml

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+title: What is the sum of both diagonals in a 1001 by 1001 spiral?
+url: http://projecteuler.net/problem=28
+
+desc: |
+  Starting with the number 1 and moving to the right in a clockwise direction a 5 by 5 spiral is formed as follows:
+  21 22 23 24 25
+  20  7  8  9 10
+  19  6  1  2 11
+  18  5  4  3 12
+  17 16 15 14 13
+  It can be verified that the sum of the numbers on the diagonals is 101.
+  What is the sum of the numbers on the diagonals in a 1001 by 1001 spiral formed in the same way?
+  
+solution: |
+  Step around in the spiral (add 2 - 3 times, then add 4 - 3times, then 6 - times etc)
+
+solutions:
+  solve.php:
+    desc: Basic solution
+    language: php
+  solve.rb:
+    desc: Basic solution
+    language: ruby
+  solve.c:
+    desc: ANSI C solution (Tested with TCC)
+    language: c
+  solve.js:
+    desc: NodeJS solution
+    language: javascript
+  solve.lua:
+    desc: Basic solution
+    language: lua

ProjectEuler/028/solve.c

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+#include <stdio.h>
+#define SIZE 1001
+#define SIDES 4
+
+int main( )
+{
+
+	int sum = 1;
+	int result=1;
+	int addition;
+	for(addition = 2; addition <= SIZE; addition+=2) {
+		int cside;
+		for(cside = 0; cside < SIDES; cside++) {
+			sum += addition;
+			result += sum;
+		}
+	}
+	printf( "%i", result );
+}

ProjectEuler/028/solve.js

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+SIZE = 1001;
+SIDES = 4
+
+sum = 1;
+result=1;
+for(addition = 2; addition <= SIZE; addition+=2) {
+	for(cside = 0; cside < SIDES; cside++) {
+		sum += addition;
+		result += sum;
+	}
+}
+
+console.log(result);

ProjectEuler/028/solve.lua

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+SIZE = 1001;
+SIDES = 4;
+	sum = 1;
+	result=1;
+	for addition = 2,SIZE,2 do
+		for cside=0,(SIDES-1) do
+			sum = sum + addition;
+			result = result + sum;
+		end
+	end
+
+print(result)

ProjectEuler/028/solve.php

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+<?php
+define('SIZE',1001);
+define('SIDES',4);
+$sum = 1;
+$result=1;
+for($addition = 2; $addition <= SIZE; $addition+=2) {
+	for($cside = 0; $cside < SIDES; $cside++) {
+		$sum += $addition;
+		$result += $sum;
+	}
+}
+echo $result;

ProjectEuler/028/solve.rb

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+
+SIZE = 1001
+SIDES = 4
+sum = 1;
+result=1;
+(2...SIZE).step(2) { |add|
+	SIDES.downto(1) { 
+		sum += add;
+		result += sum;
+	}
+}
+puts result

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]