Various incomplete solutions

ProjectEuler/024/desc.yml

@@ -6,7 +6,8 @@ desc: |
   012   021   102   120   201   210
   What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?
 
+solution: Use factoring
 solutions:
-    solve.php:
-      desc: Basic Solution
-      language: php
+  solve.php:
+    desc: Basic Solution
+    language: php

ProjectEuler/025/desc.yml

@@ -3,10 +3,8 @@ url: http://projecteuler.net/problem=25
 
 desc: |
   The Fibonacci sequence is defined by the recurrence relation:
-
   Fn = Fn1 + Fn2, where F1 = 1 and F2 = 1.
   Hence the first 12 terms will be:
-
   F1 = 1
   F2 = 1
   F3 = 2
@@ -20,11 +18,13 @@ desc: |
   F11 = 89
   F12 = 144
   The 12th term, F12, is the first term to contain three digits.
-
   What is the first term in the Fibonacci sequence to contain 1000 digits? 
 solution: Bruteforce
 
 solutions:
   solve.php:
-    desc: Basic Solution
-    language: php
+    desc: Basic Solution - needs BCMath
+    language: php
+  solve.rb:
+    desc: Basic solution
+    language: ruby

ProjectEuler/029/desc.yml

@@ -0,0 +1,20 @@
+title: How many distinct terms are in the sequence generated by a^b for 2 <= a <= 100 and 2 <= b <= 100?
+url: http://projecteuler.net/problem=29
+
+desc: |
+  Consider all integer combinations of a^b for 2 <= a  5 and 2 <= b <= 5:
+  2^2=4, 2^3=8, 2^4=16, 2^5=32
+  3^2=9, 3^3=27, 3^4=81, 3^5=243
+  4^2=16, 4^3=64, 4^4=256, 4^5=1024
+  5^2=25, 5^3=125, 5^4=625, 5^5=3125
+  If they are then placed in numerical order, with any repeats removed, we get the following sequence of 15 distinct terms:
+  4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125
+  How many distinct terms are in the sequence generated by a^b for 2 <= a <= 100 and 2 <= b <= 100?
+solution: Bruteforce
+solutions:
+  solve.php:
+    desc: Basic Solution - needs BCMath
+    language: php
+  solve.rb:
+    desc: Basic solution
+    language: ruby

ProjectEuler/029/solve.php

@@ -0,0 +1,12 @@
+<?php
+define('A_START',2);
+define('A_END',100);
+define('B_START',2);
+define('B_END',100);
+
+for($a=A_START;$a<=A_END;$a++) {
+	for($b=B_START;$b<=B_END;$b++) {
+		$array[] = bcpow($a,$b);
+	}
+}
+echo count(array_unique($array));

ProjectEuler/029/solve.rb

@@ -0,0 +1,21 @@
+
+MAX = 6*9**5
+value = 2;
+total = 0;
+for value in (2..MAX) do 
+	a = 0
+	
+end
+while($value<354294) {
+	$a = 0;
+	for($t=0;$t<strlen($value);$t++) {
+		$a += pow(substr((string)$value,$t,1),5);
+	}
+	
+	
+	if($value == $a) { $total += $value;}
+	
+$value++;
+}
+echo $total;
+

ProjectEuler/030/desc.yml

@@ -0,0 +1,16 @@
+title: Find the sum of all the numbers that can be written as the sum of fifth powers of their digits.
+url: http://projecteuler.net/problem=30
+
+desc: |
+  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.
+solution: We can safely assume that we don't have to search high values higher than 354294. Because 6*9^5 = 354294 is far from the value of 999999
+solutions:
+  solve.php:
+    desc: Basic Solution
+    language: php

ProjectEuler/030/solve.php

@@ -0,0 +1,16 @@
+<?php
+define('POWER',5);
+define('START',2);
+define('END', 6*pow(9,POWER) );
+
+$result = 0;
+for($value = START; $value < END; $value++ ) {
+	$cmp = 0;
+	for($c= 0, $len = strlen($string), $string = (string)$value; $c< $len; $c++)
+	{
+		$cmp += pow($string[$c],POWER);
+	}
+	
+	$result += ($value == $cmp) ? $value : 0;
+}
+echo $result;

ProjectEuler/031/desc.yml

@@ -0,0 +1,10 @@
+title: Investigating combinations of English currency denominations.
+url: http://projecteuler.net/problem=31
+
+desc: |
+  In England the currency is made up of pound, £ and pence, p, and there are eight coins in general circulation:
+  1p, 2p, 5p, 10p, 20p, 50p, ñ (100p) and £ (200p).
+  It is possible to make £ in the following way:
+  1£ + 150p + 220p + 15p + 12p + 31p
+  How many different ways can £ be made using any number of coins?
+solution: Define the target value as 200 to simplify

ProjectEuler/031/solve.php

@@ -0,0 +1,4 @@
+<?php
+$start = 200;
+
+for($c = 0; $c > $left; $c++);

ProjectEuler/032/desc.yml

@@ -0,0 +1,10 @@
+title: Find the sum of all numbers that can be written as pandigital products.
+url: http://projecteuler.net/problem=32
+
+desc: |
+  We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once; for example, the 5-digit number, 15234, is 1 through 5 pandigital.
+  The product 7254 is unusual, as the identity, 39 x 186 = 7254, containing multiplicand, multiplier, and product is 1 through 9 pandigital.
+  Find the sum of all products whose multiplicand/multiplier/product identity can be written as a 1 through 9 pandigital.
+  HINT: Some products can be obtained in more than one way so be sure to only include it once in your sum.
+solution: Bruteforce
+

ProjectEuler/032/solve.php

@@ -0,0 +1,6 @@
+<?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; }
+}

ProjectEuler/034/desc.yml

@@ -0,0 +1,13 @@
+title: Find the sum of all numbers which are equal to the sum of the factorial of their digits.
+url: http://projecteuler.net/problem=34
+
+desc: |
+  145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145.
+  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:
+  solve.php:
+    desc: Basic solution
+    language: php
+

ProjectEuler/034/solve.php

@@ -0,0 +1,16 @@
+<?php
+$num = 3;
+$result =0;
+for($num = 3; $num < 99999; $num++) {
+	$sum = 0;
+	foreach(str_split($num) as $digit) {
+		$sum += fact($digit);
+	}
+	if($sum == $num) { $result += $sum;}
+}
+echo $result;
+
+function fact($int){
+	static $facts = array(1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880);
+	return $facts[$int];
+}