Solved ProjectEuler/035

ProjectEuler/034/desc.yml

@@ -6,7 +6,7 @@ desc: |
   Find the sum of all numbers which are equal to the sum of the factorial of their digits.
   Note: as 1! = 1 and 2! = 2 are not sums they are not included.
 solution: Bruteforce
-solutons:
+solutions:
   solve.php:
     desc: Basic solution
     language: php

ProjectEuler/035/desc.yml

@@ -0,0 +1,11 @@
+title: How many circular primes are there below one million?
+url: http://projecteuler.net/problem=35
+
+desc: |
+  The number, 197, is called a circular prime because all rotations of the digits: 197, 971, and 719, are themselves prime.
+  There are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, and 97.
+  How many circular primes are there below one million?
+solution: Bruteforce
+solutions:
+
+

ProjectEuler/035/solve.php

@@ -0,0 +1,40 @@
+<?php
+define('MAX',1000000);
+
+function is_prime($prime) {    
+    $sqrt = sqrt($prime);
+    for ($i = 3; $i <= $sqrt; $i+=2){
+        if ($prime%$i == 0) return false; 
+    } 
+    return true; 
+}
+ 
+function fact($int){
+	static $facts = array(1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880);
+	return $facts[$int];
+}
+$primes = array(2);
+$primes_ordered = array(2);
+for($i=3;$i<MAX;$i+=2) {
+	if(is_prime($i)) {
+		//$primes[] = $i;
+		$temp = str_split($i);
+		sort($temp);
+		$primes_ordered[] = implode($temp);
+	}
+}
+echo "Found primes below 1mil\n";
+$result = 0;
+foreach($primes_ordered as $p) {
+	$counts = fact(strlen($p));
+	foreach($primes_ordered as $order) {
+		if($p == $order) {
+			$counts--;
+		}
+	}
+	if($counts == 0) {
+		//echo $p."\n";
+		$result++;
+	}
+}
+echo $result;

ProjectEuler/036/desc.yml

@@ -0,0 +1,18 @@
+title: Find the sum of all numbers less than one million, which are palindromic in base 10 and base 2.
+url: http://projecteuler.net/problem=36
+
+desc: |
+  The decimal number, 585 = 1001001001 (binary), is palindromic in both bases.
+  Find the sum of all numbers, less than one million, which are palindromic in base 10 and base 2.
+  (Please note that the palindromic number, in either base, may not include leading zeros.)
+
+solution: We can skip all even numbers since a even number would end in binary 10 - In reverse that would be 01 and we are told to skip leading 0's
+
+solutions:
+  solve.php:
+    desc: Basic solution
+    language: php
+  solve.rb:
+    desc: Basic solution in Ruby
+    language: ruby
+

ProjectEuler/036/solve.php

@@ -0,0 +1,12 @@
+<?php
+define('MAX',1000000);
+$result = 0;
+for($i=1;$i<MAX;$i+=2) {
+	if($i == strrev($i)) {
+		$bin = decbin($i);
+		if($bin == strrev($bin)) {
+			$result += $i;
+		}
+	}
+}
+echo $result;

ProjectEuler/036/solve.rb

@@ -0,0 +1,10 @@
+MAX = 1000000
+result = 0
+(1..MAX).step(2) { |i|
+	if(i.to_s == i.to_s.reverse) 
+		if(i.to_s(2) == i.to_s(2).reverse) 
+			result += i
+		end
+	end
+}
+print result