Solved ProjectEuler/010: php, ruby, c

ProjectEuler/010/desc.yml

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+title: Calculate the sum of all the primes below two million.
+url: http://projecteuler.net/problem=10
+
+desc: |
+  The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
+  Find the sum of all the primes below two million.
+  
+solution: |
+  Use a primenumber tester
+
+solutions:
+  solve.php:
+    desc: Basic solution
+    language: php
+  solve.rb:
+    desc: Basic solution
+    language: ruby
+  solve.c:
+    desc: ANSI C solution compiled with gcc-4.3.4
+    language: c

ProjectEuler/010/solve.c

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+#include <stdio.h>
+#include <math.h>
+
+int is_prime( int prime) {
+
+	int max_test = sqrt( prime );
+
+	int i;
+    for (i = 3; i <= max_test; i+=2){
+        if (prime % i == 0){
+			return 0; 
+		}
+    } 
+    return 1; 
+}
+
+int main( )
+{
+	long long sum = 2+3;
+	long number;
+	for(number = 5; number < 2000000; number+=2) {
+		if (is_prime(number))
+		{
+			sum += number;
+		}
+	}
+	printf("%llu",sum);
+}

ProjectEuler/010/solve.php

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+<?php
+
+function is_prime($prime) {    
+    $sqrt = sqrt($prime);
+    for ($i = 3; $i <= $sqrt; $i+=2){
+        if ($prime%$i == 0) return false; 
+    } 
+    return true; 
+} 
+ 
+$sum = 2+3;
+for($number = 5; $number < 2000000; $number+=2) {
+	if (is_prime($number))
+	{
+		$sum += $number;
+	}
+}
+echo $sum;

ProjectEuler/010/solve.rb

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+def is_prime(prime)     
+    sqrt = Math.sqrt(prime)
+	i=3
+	while(i <= sqrt) 
+        if (prime % i == 0) 
+			return false
+		end
+		i+= 2
+    end
+    return true
+end
+ 
+number = 3
+sum = 5
+until(number >= 2000000) 
+	number += 2
+	if(is_prime(number))
+		sum += number
+	end
+end
+puts sum;