diff --git a/ProjectEuler/038/desc.yml b/ProjectEuler/038/desc.yml
new file mode 100644
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+++ b/ProjectEuler/038/desc.yml
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+title: What is the largest 1 to 9 pandigital that can be formed by multiplying a fixed number by 1, 2, 3, ... ?
+url: http://projecteuler.net/problem=38
+
+desc: |
+    Take the number 192 and multiply it by each of 1, 2, and 3:
+    192 x 1 = 192
+    192 x 2 = 384
+    192 x 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?
+      
+solution: |
+  Bruteforce
+
+solutions:
+  solve.php:
+    desc: Basic solution
+    language: php
diff --git a/ProjectEuler/038/solve.php b/ProjectEuler/038/solve.php
new file mode 100644
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+++ b/ProjectEuler/038/solve.php
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+<?php
+function pandigital($number) {
+		$array = count_chars($number,1);
+		ksort($array);
+		if($array == array(49=>1,50=>1,51=>1,52=>1,53=>1,54=>1,55=>1,56=>1,57=>1)) { return true;} else { return false; }
+}
+
+while(true) {
+$value++;
+
+	$test = array(1,2);
+	$key = 2;
+	while(true) {
+		$new = array();
+		foreach($test as $p) {
+			$new[] = $value * $p;
+		}
+		$var = implode('',$new);
+		if(strlen($var) != 9) { break; }
+		if(strlen($var) == 9 AND pandigital($var)) { $good[] = $var; break; }
+		$test[] = $key++;
+	}
+	if($value > 9999) { break; }
+}
+
+rsort($good);
+echo $good[0];
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diff --git a/ProjectEuler/104/desc.yml b/ProjectEuler/104/desc.yml
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+++ b/ProjectEuler/104/desc.yml
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+title: Finding Fibonacci numbers for which the first and last nine digits are pandigital.
+url: http://projecteuler.net/problem=104
+
+desc: |
+  The Fibonacci sequence is defined by the recurrence relation:
+  Fn = Fn1 + Fn2, where F1 = 1 and F2 = 1.
+  It turns out that F541, which contains 113 digits, is the first Fibonacci number for which the last nine digits are 1-9 pandigital (contain all the digits 1 to 9, but not necessarily in order). And F2749, which contains 575 digits, is the first Fibonacci number for which the first nine digits are 1-9 pandigital.
+  Given that Fk is the first Fibonacci number for which the first nine digits AND the last nine digits are 1-9 pandigital, find k.
+  
+solution: |
+  Bruteforce
+
+solutions:
+  solve.php:
+    desc: Bruteforce - Sooooo slow
+    language: php
diff --git a/ProjectEuler/104/solve.php b/ProjectEuler/104/solve.php
new file mode 100644
index 0000000..f5116d0
--- /dev/null
+++ b/ProjectEuler/104/solve.php
@@ -0,0 +1,20 @@
+<?php
+
+function pandigital($num) {
+	for ($i = 1; $i <= 9; $i++) {
+		if(strpos($n,(string)$i) === false) {
+			return false;
+		}
+	}
+	return true;
+}
+$var = 1;
+$k = 1;
+while(true) {
+	$k++;
+	$new = bcadd($var,$prev);
+	$prev = $var;
+	$var = $new;
+	
+	if(pandigital(substr($var,-9)) AND pandigital(substr($var,0,9))) { echo $k; die; }
+}
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