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	<title>Free Lancers Unite &#187; Math</title>
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	<description>&#34;look upon my works ye mighty and despair&#34;</description>
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		<title>Book Review: Mathematics – A Very Short Intoduction.</title>
		<link>http://freelancersunite.net/books/book-review-mathematics-short-introductionmathematics-short-intoduction/</link>
		<comments>http://freelancersunite.net/books/book-review-mathematics-short-introductionmathematics-short-intoduction/#comments</comments>
		<pubDate>Mon, 12 Jul 2010 03:18:27 +0000</pubDate>
		<dc:creator>Serinox</dc:creator>
				<category><![CDATA[Books]]></category>
		<category><![CDATA[book review]]></category>
		<category><![CDATA[Math]]></category>

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		<description><![CDATA[(This post is the first in a new section on this site. This will probably become a regular item on the site as I read through the library of books I have laying about. As I have a lot of math related books around right now most of the book reviews will be math related [...]]]></description>
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		<title>FSharp Miller-Rabin primality test</title>
		<link>http://freelancersunite.net/math/fsharp-miller-rabin-primality-test/</link>
		<comments>http://freelancersunite.net/math/fsharp-miller-rabin-primality-test/#comments</comments>
		<pubDate>Wed, 21 Apr 2010 00:38:37 +0000</pubDate>
		<dc:creator>Serinox</dc:creator>
				<category><![CDATA[Math]]></category>
		<category><![CDATA[.net4]]></category>
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		<description><![CDATA[In preparation for some upcoming project euler problems I&#8217;ve been looking for faster methods of determining if a number is prime. One such method I&#8217;ve come across was the Miller-Rabin primality test which is a probabilistic test for determining if a given number is probably prime. The higher the value of s passed to the [...]]]></description>
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		<title>Project Euler Problem #142!</title>
		<link>http://freelancersunite.net/project_euler/project-euler-problem-142/</link>
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		<pubDate>Sat, 26 Dec 2009 17:01:19 +0000</pubDate>
		<dc:creator>Serinox</dc:creator>
				<category><![CDATA[Project Euler]]></category>
		<category><![CDATA[c#]]></category>
		<category><![CDATA[csharp]]></category>
		<category><![CDATA[Math]]></category>
		<category><![CDATA[project euler]]></category>

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		<description><![CDATA[Problem #142 say: Find the smallest x + y + z with integers x > y > z > 0 such that x + y, x − y, x + z, x − z, y + z, y − z are all perfect squares. Now one can brute force this by looping over x, y [...]]]></description>
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		<title>Project Euler Problem #145!</title>
		<link>http://freelancersunite.net/project_euler/project-euler-problem-145/</link>
		<comments>http://freelancersunite.net/project_euler/project-euler-problem-145/#comments</comments>
		<pubDate>Wed, 02 Dec 2009 04:44:36 +0000</pubDate>
		<dc:creator>Serinox</dc:creator>
				<category><![CDATA[Project Euler]]></category>
		<category><![CDATA[c#]]></category>
		<category><![CDATA[Math]]></category>
		<category><![CDATA[programming]]></category>
		<category><![CDATA[project euler]]></category>

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		<description><![CDATA[Problem #145 says: Some positive integers n have the property that the sum [ n + reverse(n) ] consists entirely of odd (decimal) digits. For instance, 36 + 63 = 99 and 409 + 904 = 1313. We will call such numbers reversible; so 36, 63, 409, and 904 are reversible. Leading zeroes are not [...]]]></description>
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		<title>Project Euler Problem #188!</title>
		<link>http://freelancersunite.net/project_euler/project-euler-problem-188/</link>
		<comments>http://freelancersunite.net/project_euler/project-euler-problem-188/#comments</comments>
		<pubDate>Thu, 19 Nov 2009 16:34:24 +0000</pubDate>
		<dc:creator>Serinox</dc:creator>
				<category><![CDATA[Project Euler]]></category>
		<category><![CDATA[Math]]></category>
		<category><![CDATA[project euler]]></category>
		<category><![CDATA[python]]></category>

		<guid isPermaLink="false">http://freelancersunite.net/?p=585</guid>
		<description><![CDATA[Problem #188 says: The hyperexponentiation or tetration of a number a by a positive integer b, denoted by a↑↑b or ^(b)a, is recursively defined by: a↑↑1 = a, a↑↑(k+1) = a^((a↑↑k)). Thus we have e.g. 3↑↑2 = 3^(3) = 27, hence 3↑↑3 = 3^(27) = 7625597484987 and 3↑↑4 is roughly 10^(3.6383346400240996*10^12). Find the last 8 [...]]]></description>
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