Added a few generic problems\nSolved Euler027

ProjectEuler/026/desc.yml

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+title: Find the value of d < 1000 for which 1/d contains the longest recurring cycle in its decimal fraction part.
+url: http://projecteuler.net/problem=26
+
+desc: |
+  A unit fraction contains 1 in the numerator. The decimal representation of the unit fractions with denominators 2 to 10 are given:
+  1/2	= 	0.5
+  1/3	= 	0.(3)
+  1/4	= 	0.25
+  1/5	= 	0.2
+  1/6	= 	0.1(6)
+  1/7	= 	0.(142857)
+  1/8	= 	0.125
+  1/9	= 	0.(1)
+  1/10	= 	0.1
+  Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle. It can be seen that 1/7 has a 6-digit recurring cycle.
+  Find the value of d < 1000 for which 1/d contains the longest recurring cycle in its decimal fraction part.
+
+solution: Bruteforce
+
+solutions:
+  solve.php:
+    desc: Basic Solution - needs BCMath
+    language: php

ProjectEuler/026/solve.php

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+<?php
+define('MAX',1000);
+
+$max_len = 0;
+for($div = MAX; $div > 2; $div--) {
+	$result = bcdiv('1',(string)$div,2000);
+	
+	for($len = 2; $len < 1000; $len++) {
+		if(substr($result,10,$len) === substr($result,10+$len,$len)) {
+			if($len > $max_len) {
+				echo "Div: ".$div. " Len: ".$len."\n";
+				$max_len = $len;
+				$max_div = $div;
+			}
+			break;
+		}
+	}
+
+}
+echo $max_div;

ProjectEuler/026/solve.rb

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+DIGITS = 1000
+
+max = 10 ** (DIGITS-1)
+fcurrent = 1
+fprev = 1
+term = 2
+while fcurrent < max do 
+	term = term + 1
+	fnext = fcurrent + fprev
+	fprev = fcurrent;
+	fcurrent = fnext;
+end
+puts term

ProjectEuler/027/desc.yml

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+title: Quadratic primes
+url: http://projecteuler.net/problem=27
+
+desc: |
+  Euler discovered the remarkable quadratic formula:
+  n<> + n + 41
+  It turns out that the formula will produce 40 primes for the consecutive values n = 0 to 39. However, when n = 40, 402 + 40 + 41 = 40(40 + 1) + 41 is divisible by 41, and certainly when n = 41, 41<34> + 41 + 41 is clearly divisible by 41.
+  The incredible formula  n<> - 79n + 1601 was discovered, which produces 80 primes for the consecutive values n = 0 to 79. The product of the coefficients, -79 and 1601, is -126479.
+  Considering quadratics of the form:
+  n<> + an + b, where |a| < 1000 and |b| < 1000
+  where |n| is the modulus/absolute value of n
+  e.g. |11| = 11 and |-4| = 4
+  Find the product of the coefficients, a and b, for the quadratic expression that produces the maximum number of primes for consecutive values of n, starting with n = 0
+
+solution: Bruteforce
+
+solutions:
+  solve.php:
+    desc: Basic Solution - needs BCMath
+    language: php

ProjectEuler/027/solve.php

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+<?php
+function is_prime($prime) {  
+	if($prime == 1) 
+		return true;
+	if($prime == 2)
+		return true;
+    $sqrt = sqrt($prime);
+    for ($i = 3; $i <= $sqrt; $i+=2){
+        if ($prime%$i == 0) return false; 
+    } 
+    return true; 
+} 
+$high = 0;
+$max_primes = 0;
+for($a=-999; $a<1000; $a++) {
+	for($b=-999; $b<1000; $b=$b+2) {
+		
+		$n=0;
+		while(true) {
+		echo $n*$n + $a * $n + $b . "\n";
+			if(!is_prime($n*$n + $a * $n + $b)) {
+				if($n > $max_primes) {
+					$max_primes = $n;
+					echo "max: $n a: $a b: $b\n";
+				}
+				break;
+			}
+			$n++;
+		}
+	
+	}
+}

ProjectEuler/027/solve.rb

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+DIGITS = 1000
+
+max = 10 ** (DIGITS-1)
+fcurrent = 1
+fprev = 1
+term = 2
+while fcurrent < max do 
+	term = term + 1
+	fnext = fcurrent + fprev
+	fprev = fcurrent;
+	fcurrent = fnext;
+end
+puts term