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	<title>Occam&#039;s Razor &#187; projecteuler</title>
	<atom:link href="http://developmentgeek.com/blog/tag/projecteuler/feed/" rel="self" type="application/rss+xml" />
	<link>http://developmentgeek.com/blog</link>
	<description>Fueled by Mt. Dew and a love for technology</description>
	<lastBuildDate>Tue, 13 Jul 2010 15:39:34 +0000</lastBuildDate>
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		<item>
		<title>Problem 019</title>
		<link>http://developmentgeek.com/blog/20100316/problem-019/</link>
		<comments>http://developmentgeek.com/blog/20100316/problem-019/#comments</comments>
		<pubDate>Tue, 16 Mar 2010 09:53:08 +0000</pubDate>
		<dc:creator>Kevin Fairchild</dc:creator>
				<category><![CDATA[Miscellaneous]]></category>
		<category><![CDATA[projecteuler]]></category>

		<guid isPermaLink="false">http://developmentgeek.com/blog/?p=158</guid>
		<description><![CDATA[You are given the following information, but you may prefer to do some research for yourself. * 1 Jan 1900 was a Monday. * Thirty days has September, April, June and November. All the rest have thirty-one, Saving February alone, Which has twenty-eight, rain or shine. And on leap years, twenty-nine. * A leap year [...]]]></description>
			<content:encoded><![CDATA[<blockquote><p>You are given the following information, but you may prefer to do some research for yourself.</p>
<p>    * 1 Jan 1900 was a Monday.<br />
    * Thirty days has September,<br />
      April, June and November.<br />
      All the rest have thirty-one,<br />
      Saving February alone,<br />
      Which has twenty-eight, rain or shine.<br />
      And on leap years, twenty-nine.<br />
    * A leap year occurs on any year evenly divisible by 4, but not on a century unless it is divisible by 400.</p>
<p>How many Sundays fell on the first of the month during the twentieth century (1 Jan 1901 to 31 Dec 2000)?</p></blockquote>
<p><strong>Solution:</strong><br />
<code>Function Prob019() As Long<br />
        Dim dt As Date = CDate("01/01/1901")<br />
        While dt < = CDate("12/31/2000")<br />
            If dt.DayOfWeek = System.DayOfWeek.Sunday Then Prob019 += 1<br />
            dt = dt.AddMonths(1)<br />
        End While<br />
    End Function</code></p>
<p><strong>Summary:</strong><br />
I'm not above cheating by using built-in functions <img src='http://developmentgeek.com/blog/wp-includes/images/smilies/icon_wink.gif' alt=';)' class='wp-smiley' /> </code></p>
]]></content:encoded>
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		<slash:comments>0</slash:comments>
		</item>
		<item>
		<title>Problem 014</title>
		<link>http://developmentgeek.com/blog/20100316/problem-014/</link>
		<comments>http://developmentgeek.com/blog/20100316/problem-014/#comments</comments>
		<pubDate>Tue, 16 Mar 2010 09:50:55 +0000</pubDate>
		<dc:creator>Kevin Fairchild</dc:creator>
				<category><![CDATA[Miscellaneous]]></category>
		<category><![CDATA[projecteuler]]></category>

		<guid isPermaLink="false">http://developmentgeek.com/blog/?p=155</guid>
		<description><![CDATA[The following iterative sequence is defined for the set of positive integers: n → n/2 (n is even) n → 3n + 1 (n is odd) Using the rule above and starting with 13, we generate the following sequence: 13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → [...]]]></description>
			<content:encoded><![CDATA[<blockquote><p>The following iterative sequence is defined for the set of positive integers:</p>
<p>n → n/2 (n is even)<br />
n → 3n + 1 (n is odd)</p>
<p>Using the rule above and starting with 13, we generate the following sequence:<br />
13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1</p>
<p>It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms. Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1.</p>
<p>Which starting number, under one million, produces the longest chain?</p>
<p>NOTE: Once the chain starts the terms are allowed to go above one million.</p></blockquote>
<p><strong>Solution:</strong><br />
<code>Function Prob014() As Long<br />
        Prob014 = 0<br />
        Dim NumWithLargestChain As Long = 0<br />
        Dim ChainCount As Long = 0<br />
        For StartNumber As Long = 999999 To 1 Step -1<br />
            Dim tmpNum As Long = StartNumber<br />
            Dim tmpCount As Long = 0<br />
            Do<br />
                tmpCount += 1<br />
                If tmpCount > ChainCount Then<br />
                    ChainCount = tmpCount<br />
                    NumWithLargestChain = StartNumber<br />
                End If<br />
                If tmpNum = 1 Then Exit Do<br />
                If tmpNum Mod 2 = 0 Then<br />
                    tmpNum = tmpNum / 2<br />
                Else<br />
                    tmpNum = (tmpNum * 3) + 1<br />
                End If<br />
            Loop<br />
        Next<br />
        Return NumWithLargestChain<br />
End Function</code></p>
<p><strong>Summary:</strong><br />
Thankfully, this was a lot easier than the previous problem I did.  I skipped problem 12 and 13, but I&#8217;ll probably come back to them at some point. Problem 12 dealt with triangle numbers, which I&#8217;m not really familiar with. Problem 13 dealt with adding up a hundred 50-digit numbers, which seems like it would be easy, but I either would need a crazy-big numeric datatype or I&#8217;d need to come up with a new way to add the stuff up through some other fashion.</p>
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		<item>
		<title>Problem 011</title>
		<link>http://developmentgeek.com/blog/20100313/problem-011/</link>
		<comments>http://developmentgeek.com/blog/20100313/problem-011/#comments</comments>
		<pubDate>Sat, 13 Mar 2010 21:15:32 +0000</pubDate>
		<dc:creator>Kevin Fairchild</dc:creator>
				<category><![CDATA[Miscellaneous]]></category>
		<category><![CDATA[projecteuler]]></category>

		<guid isPermaLink="false">http://developmentgeek.com/blog/?p=150</guid>
		<description><![CDATA[In the 20×20 grid below, four numbers along a diagonal line have been marked in red. 08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08 49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 [...]]]></description>
			<content:encoded><![CDATA[<blockquote><p>In the 20×20 grid below, four numbers along a diagonal line have been marked in red.</p>
<p>08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08<br />
49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00<br />
81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65<br />
52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91<br />
22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80<br />
24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50<br />
32 98 81 28 64 23 67 10 <font color="Red">26</font> 38 40 67 59 54 70 66 18 38 64 70<br />
67 26 20 68 02 62 12 20 95 <font color="Red">63</font> 94 39 63 08 40 91 66 49 94 21<br />
24 55 58 05 66 73 99 26 97 17 <font color="Red">78</font> 78 96 83 14 88 34 89 63 72<br />
21 36 23 09 75 00 76 44 20 45 35 <font color="Red">14</font> 00 61 33 97 34 31 33 95<br />
78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92<br />
16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57<br />
86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58<br />
19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40<br />
04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66<br />
88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69<br />
04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36<br />
20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16<br />
20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54<br />
01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48</p>
<p>The product of these numbers is 26 × 63 × 78 × 14 = 1788696.</p>
<p>What is the greatest product of four adjacent numbers in any direction (up, down, left, right, or diagonally) in the 20×20 grid?</p></blockquote>
<p><strong>Solution:</strong><br />
<code>Function Prob011() As Long<br />
        Prob011 = 0<br />
        Dim strGrid = ""<br />
        strGrid &#038;= " 08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08"<br />
        strGrid &#038;= " 49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00"<br />
        strGrid &#038;= " 81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65"<br />
        strGrid &#038;= " 52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91"<br />
        strGrid &#038;= " 22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80"<br />
        strGrid &#038;= " 24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50"<br />
        strGrid &#038;= " 32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70"<br />
        strGrid &#038;= " 67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21"<br />
        strGrid &#038;= " 24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72"<br />
        strGrid &#038;= " 21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95"<br />
        strGrid &#038;= " 78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92"<br />
        strGrid &#038;= " 16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57"<br />
        strGrid &#038;= " 86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58"<br />
        strGrid &#038;= " 19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40"<br />
        strGrid &#038;= " 04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66"<br />
        strGrid &#038;= " 88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69"<br />
        strGrid &#038;= " 04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36"<br />
        strGrid &#038;= " 20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16"<br />
        strGrid &#038;= " 20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54"<br />
        strGrid &#038;= " 01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48"</p>
<p>        Dim CharPerLine As Integer = 60<br />
        Dim Offset As Integer = 3</p>
<p>        'Across<br />
        For f As Integer = 0 To 19<br />
            Dim Val1 As Integer = 0<br />
            Dim Val2 As Integer = 0<br />
            Dim Val3 As Integer = 0<br />
            Dim Val4 As Integer = 0<br />
            Dim StartPos As Integer = (f * CharPerLine)<br />
            Dim GroupCount As Integer = 1<br />
            While GroupCount < = 17<br />
                Val1 = CInt(strGrid.Substring(StartPos, Offset).Trim)<br />
                Val2 = CInt(strGrid.Substring(StartPos + (Offset), Offset).Trim)<br />
                Val3 = CInt(strGrid.Substring(StartPos + (Offset * 2), Offset).Trim)<br />
                Val4 = CInt(strGrid.Substring(StartPos + (Offset * 3), Offset).Trim)<br />
                If Val1 * Val2 * Val3 * Val4 > Prob011 Then Prob011 = Val1 * Val2 * Val3 * Val4<br />
                StartPos += Offset<br />
                GroupCount += 1<br />
            End While<br />
        Next</p>
<p>        'Down<br />
        For f As Integer = 0 To 16<br />
            Dim Val1 As Integer = 0<br />
            Dim Val2 As Integer = 0<br />
            Dim Val3 As Integer = 0<br />
            Dim Val4 As Integer = 0<br />
            Dim StartPos As Integer = (f * CharPerLine)<br />
            Dim GroupCount As Integer = 1<br />
            While GroupCount < = 20<br />
                Val1 = CInt(strGrid.Substring(StartPos, Offset).Trim)<br />
                Val2 = CInt(strGrid.Substring(StartPos + CharPerLine, Offset).Trim)<br />
                Val3 = CInt(strGrid.Substring(StartPos + (CharPerLine * 2), Offset).Trim)<br />
                Val4 = CInt(strGrid.Substring(StartPos + (CharPerLine * 3), Offset).Trim)<br />
                If Val1 * Val2 * Val3 * Val4 > Prob011 Then Prob011 = Val1 * Val2 * Val3 * Val4<br />
                StartPos += Offset<br />
                GroupCount += 1<br />
            End While<br />
        Next</p>
<p>        'Diag-Left<br />
        For f As Integer = 0 To 16<br />
            Dim Val1 As Integer = 0<br />
            Dim Val2 As Integer = 0<br />
            Dim Val3 As Integer = 0<br />
            Dim Val4 As Integer = 0<br />
            Dim StartPos As Integer = (f * CharPerLine)<br />
            Dim GroupCount As Integer = 1<br />
            While GroupCount < = 17<br />
                Val1 = CInt(strGrid.Substring(StartPos, Offset).Trim)<br />
                Val2 = CInt(strGrid.Substring(StartPos + CharPerLine + Offset, Offset).Trim)<br />
                Val3 = CInt(strGrid.Substring(StartPos + (CharPerLine * 2) + (Offset * 2), Offset).Trim)<br />
                Val4 = CInt(strGrid.Substring(StartPos + (CharPerLine * 3) + (Offset * 3), Offset).Trim)<br />
                If Val1 * Val2 * Val3 * Val4 > Prob011 Then Prob011 = Val1 * Val2 * Val3 * Val4<br />
                StartPos += Offset<br />
                GroupCount += 1<br />
            End While<br />
        Next</p>
<p>        'Diag-Right<br />
        For f As Integer = 0 To 16<br />
            Dim Val1 As Integer = 0<br />
            Dim Val2 As Integer = 0<br />
            Dim Val3 As Integer = 0<br />
            Dim Val4 As Integer = 0<br />
            Dim StartPos As Integer = (Offset * 3) + (f * CharPerLine)<br />
            Dim GroupCount As Integer = 1<br />
            While GroupCount < = 17<br />
                Val1 = CInt(strGrid.Substring(StartPos, Offset).Trim)<br />
                Val2 = CInt(strGrid.Substring(StartPos + CharPerLine - Offset, Offset).Trim)<br />
                Val3 = CInt(strGrid.Substring(StartPos + (CharPerLine * 2) - (Offset * 2), Offset).Trim)<br />
                Val4 = CInt(strGrid.Substring(StartPos + (CharPerLine * 3) - (Offset * 3), Offset).Trim)<br />
                If Val1 * Val2 * Val3 * Val4 > Prob011 Then Prob011 = Val1 * Val2 * Val3 * Val4<br />
                StartPos += Offset<br />
                GroupCount += 1<br />
            End While<br />
        Next<br />
End Function</code></p>
<p><strong>Summary:</strong><br />
I had a LOT of fun figuring out a solution to this one!  I was pretty sure there was a way to use a Matrix for this, but &#8212; mainly because I can&#8217;t remember the last time I&#8217;ve actually used one &#8212; I skipped the matrix and went with a slightly less elegant way.</p>
]]></content:encoded>
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		</item>
		<item>
		<title>Problem 010</title>
		<link>http://developmentgeek.com/blog/20100311/problem-010/</link>
		<comments>http://developmentgeek.com/blog/20100311/problem-010/#comments</comments>
		<pubDate>Thu, 11 Mar 2010 19:13:10 +0000</pubDate>
		<dc:creator>Kevin Fairchild</dc:creator>
				<category><![CDATA[Miscellaneous]]></category>
		<category><![CDATA[projecteuler]]></category>

		<guid isPermaLink="false">http://developmentgeek.com/blog/?p=144</guid>
		<description><![CDATA[The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17. Find the sum of all the primes below two million. Solution: Function Prob010() As Long For k As Integer = 2 To 2000000 If IsPrime(k) Then Prob010 += k Next End Function Summary: They are making this WAY [...]]]></description>
			<content:encoded><![CDATA[<blockquote><p>The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.</p>
<p>Find the sum of all the primes below two million.</p></blockquote>
<p><strong>Solution:</strong><br />
<code>Function Prob010() As Long<br />
        For k As Integer = 2 To 2000000<br />
            If <a href="http://developmentgeek.com/blog/20100226/getting-to-know-eu-ler/">IsPrime</a>(k) Then Prob010 += k<br />
        Next<br />
End Function</code></p>
<p><strong>Summary:</strong><br />
They are making this WAY too easy.  I&#8217;m not sure if it&#8217;s just the nature of this sort of problem or whether my IsPrime routine needs some tweaking, but this problem takes the longest compared to all of the others I&#8217;ve done up to this point.  Even so, it&#8217;s only 10-seconds, which is well within the target range allowed by Project Euler.</p>
]]></content:encoded>
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		</item>
		<item>
		<title>Problem 009</title>
		<link>http://developmentgeek.com/blog/20100311/problem-009/</link>
		<comments>http://developmentgeek.com/blog/20100311/problem-009/#comments</comments>
		<pubDate>Thu, 11 Mar 2010 09:11:27 +0000</pubDate>
		<dc:creator>Kevin Fairchild</dc:creator>
				<category><![CDATA[Miscellaneous]]></category>
		<category><![CDATA[projecteuler]]></category>

		<guid isPermaLink="false">http://developmentgeek.com/blog/?p=141</guid>
		<description><![CDATA[A Pythagorean triplet is a set of three natural numbers, a < b < c, for which, a^(2) + b^(2) = c^(2) For example, 3^(2) + 4^(2) = 9 + 16 = 25 = 5^(2). There exists exactly one Pythagorean triplet for which a + b + c = 1000. Find the product abc. Solution: [...]]]></description>
			<content:encoded><![CDATA[<blockquote><p>A Pythagorean triplet is a set of three natural numbers, a  < b  < c, for which,<br />
a^(2) + b^(2) = c^(2)</p>
<p>For example, 3^(2) + 4^(2) = 9 + 16 = 25 = 5^(2).</p>
<p>There exists exactly one Pythagorean triplet for which a + b + c = 1000.<br />
Find the product abc.</p></blockquote>
<p><strong>Solution:</strong><br />
<code>Function Prob009() As Long<br />
        For a As Integer = 0 To 1000<br />
            For b As Integer = 0 To 1000<br />
                For c As Integer = 0 To 1000<br />
                    If (a + b + c) = 1000 AndAlso a < b AndAlso b < c AndAlso System.Math.Pow(a, 2) + System.Math.Pow(b, 2) = System.Math.Pow(c, 2) Then Return a * b * c<br />
                Next<br />
            Next<br />
        Next<br />
End Function</code></p>
<p><strong>Summary:</strong><br />
Pretty easy...</code></p></blockquote>
]]></content:encoded>
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		</item>
		<item>
		<title>Problem 008</title>
		<link>http://developmentgeek.com/blog/20100311/problem-008/</link>
		<comments>http://developmentgeek.com/blog/20100311/problem-008/#comments</comments>
		<pubDate>Thu, 11 Mar 2010 09:01:11 +0000</pubDate>
		<dc:creator>Kevin Fairchild</dc:creator>
				<category><![CDATA[Miscellaneous]]></category>
		<category><![CDATA[projecteuler]]></category>

		<guid isPermaLink="false">http://developmentgeek.com/blog/?p=138</guid>
		<description><![CDATA[Find the greatest product of five consecutive digits in the 1000-digit number. 73167176531330624919225119674426574742355349194934 96983520312774506326239578318016984801869478851843 85861560789112949495459501737958331952853208805511 12540698747158523863050715693290963295227443043557 66896648950445244523161731856403098711121722383113 62229893423380308135336276614282806444486645238749 30358907296290491560440772390713810515859307960866 70172427121883998797908792274921901699720888093776 65727333001053367881220235421809751254540594752243 52584907711670556013604839586446706324415722155397 53697817977846174064955149290862569321978468622482 83972241375657056057490261407972968652414535100474 82166370484403199890008895243450658541227588666881 16427171479924442928230863465674813919123162824586 17866458359124566529476545682848912883142607690042 24219022671055626321111109370544217506941658960408 07198403850962455444362981230987879927244284909188 84580156166097919133875499200524063689912560717606 05886116467109405077541002256983155200055935729725 71636269561882670428252483600823257530420752963450 Solution: Function Prob008() As Long Dim strInput As New System.Text.StringBuilder strInput.Append("73167176531330624919225119674426574742355349194934") strInput.Append("96983520312774506326239578318016984801869478851843") strInput.Append("85861560789112949495459501737958331952853208805511") strInput.Append("12540698747158523863050715693290963295227443043557") strInput.Append("66896648950445244523161731856403098711121722383113") strInput.Append("62229893423380308135336276614282806444486645238749") strInput.Append("30358907296290491560440772390713810515859307960866") strInput.Append("70172427121883998797908792274921901699720888093776") strInput.Append("65727333001053367881220235421809751254540594752243") strInput.Append("52584907711670556013604839586446706324415722155397") strInput.Append("53697817977846174064955149290862569321978468622482") strInput.Append("83972241375657056057490261407972968652414535100474") strInput.Append("82166370484403199890008895243450658541227588666881") [...]]]></description>
			<content:encoded><![CDATA[<blockquote><p>Find the greatest product of five consecutive digits in the 1000-digit number.</p>
<p>73167176531330624919225119674426574742355349194934<br />
96983520312774506326239578318016984801869478851843<br />
85861560789112949495459501737958331952853208805511<br />
12540698747158523863050715693290963295227443043557<br />
66896648950445244523161731856403098711121722383113<br />
62229893423380308135336276614282806444486645238749<br />
30358907296290491560440772390713810515859307960866<br />
70172427121883998797908792274921901699720888093776<br />
65727333001053367881220235421809751254540594752243<br />
52584907711670556013604839586446706324415722155397<br />
53697817977846174064955149290862569321978468622482<br />
83972241375657056057490261407972968652414535100474<br />
82166370484403199890008895243450658541227588666881<br />
16427171479924442928230863465674813919123162824586<br />
17866458359124566529476545682848912883142607690042<br />
24219022671055626321111109370544217506941658960408<br />
07198403850962455444362981230987879927244284909188<br />
84580156166097919133875499200524063689912560717606<br />
05886116467109405077541002256983155200055935729725<br />
71636269561882670428252483600823257530420752963450</p></blockquote>
<p><strong>Solution:</strong><br />
<code>Function Prob008() As Long<br />
        Dim strInput As New System.Text.StringBuilder<br />
        strInput.Append("73167176531330624919225119674426574742355349194934")<br />
        strInput.Append("96983520312774506326239578318016984801869478851843")<br />
        strInput.Append("85861560789112949495459501737958331952853208805511")<br />
        strInput.Append("12540698747158523863050715693290963295227443043557")<br />
        strInput.Append("66896648950445244523161731856403098711121722383113")<br />
        strInput.Append("62229893423380308135336276614282806444486645238749")<br />
        strInput.Append("30358907296290491560440772390713810515859307960866")<br />
        strInput.Append("70172427121883998797908792274921901699720888093776")<br />
        strInput.Append("65727333001053367881220235421809751254540594752243")<br />
        strInput.Append("52584907711670556013604839586446706324415722155397")<br />
        strInput.Append("53697817977846174064955149290862569321978468622482")<br />
        strInput.Append("83972241375657056057490261407972968652414535100474")<br />
        strInput.Append("82166370484403199890008895243450658541227588666881")<br />
        strInput.Append("16427171479924442928230863465674813919123162824586")<br />
        strInput.Append("17866458359124566529476545682848912883142607690042")<br />
        strInput.Append("24219022671055626321111109370544217506941658960408")<br />
        strInput.Append("07198403850962455444362981230987879927244284909188")<br />
        strInput.Append("84580156166097919133875499200524063689912560717606")<br />
        strInput.Append("05886116467109405077541002256983155200055935729725")<br />
        strInput.Append("71636269561882670428252483600823257530420752963450")<br />
        Dim Group1 As Integer = 0<br />
        Dim Group2 As Integer = 0<br />
        Dim Group3 As Integer = 0<br />
        Dim Group4 As Integer = 0<br />
        Dim Group5 As Integer = 0<br />
        Dim StringPos As Integer = 0<br />
        Do While StringPos < strInput.Length - 4<br />
            Group1 = strInput.ToString.Substring(StringPos, 1)<br />
            Group2 = strInput.ToString.Substring(StringPos + 1, 1)<br />
            Group3 = strInput.ToString.Substring(StringPos + 2, 1)<br />
            Group4 = strInput.ToString.Substring(StringPos + 3, 1)<br />
            Group5 = strInput.ToString.Substring(StringPos + 4, 1)<br />
            If Group1 * Group2 * Group3 * Group4 * Group5 > Prob008 Then Prob008 = Group1 * Group2 * Group3 * Group4 * Group5<br />
            StringPos += 1<br />
        Loop<br />
End Function</code></p>
<p><strong>Summary:</strong><br />
It isn&#8217;t pretty, but it works <img src='http://developmentgeek.com/blog/wp-includes/images/smilies/icon_smile.gif' alt=':)' class='wp-smiley' /> </p>
]]></content:encoded>
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		<slash:comments>0</slash:comments>
		</item>
		<item>
		<title>Problem 007</title>
		<link>http://developmentgeek.com/blog/20100310/problem-007/</link>
		<comments>http://developmentgeek.com/blog/20100310/problem-007/#comments</comments>
		<pubDate>Wed, 10 Mar 2010 15:19:08 +0000</pubDate>
		<dc:creator>Kevin Fairchild</dc:creator>
				<category><![CDATA[Miscellaneous]]></category>
		<category><![CDATA[projecteuler]]></category>

		<guid isPermaLink="false">http://developmentgeek.com/blog/?p=131</guid>
		<description><![CDATA[By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6^(th) prime is 13. What is the 10001^(st) prime number? Solution: Function Prob007() As Long Dim k As Integer = 1 Dim PrimeCount As Integer = 0 Dim TargetPrime As Integer = 10001 While PrimeCount < [...]]]></description>
			<content:encoded><![CDATA[<blockquote><p>By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6^(th) prime is 13.</p>
<p>What is the 10001^(st) prime number?</p></blockquote>
<p><strong>Solution:</strong><br />
<code>Function Prob007() As Long<br />
        Dim k As Integer = 1<br />
        Dim PrimeCount As Integer = 0<br />
        Dim TargetPrime As Integer = 10001<br />
        While PrimeCount < = TargetPrime<br />
            k += 1<br />
            If <a href="http://developmentgeek.com/blog/20100226/getting-to-know-eu-ler/">IsPrime(k) Then PrimeCount += 1<br />
            If PrimeCount = TargetPrime Then Return k<br />
        End While<br />
End Function</code></p>
<p><strong>Summary:</strong><br />
Simple and gets the job done.</p>
]]></content:encoded>
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		</item>
		<item>
		<title>Problem 006</title>
		<link>http://developmentgeek.com/blog/20100309/problem-006/</link>
		<comments>http://developmentgeek.com/blog/20100309/problem-006/#comments</comments>
		<pubDate>Tue, 09 Mar 2010 12:03:55 +0000</pubDate>
		<dc:creator>Kevin Fairchild</dc:creator>
				<category><![CDATA[Miscellaneous]]></category>
		<category><![CDATA[projecteuler]]></category>

		<guid isPermaLink="false">http://developmentgeek.com/blog/?p=127</guid>
		<description><![CDATA[The sum of the squares of the first ten natural numbers is, 1^(2) + 2^(2) + &#8230; + 10^(2) = 385 The square of the sum of the first ten natural numbers is, (1 + 2 + &#8230; + 10)^(2) = 55^(2) = 3025 Hence the difference between the sum of the squares of the [...]]]></description>
			<content:encoded><![CDATA[<blockquote><p>The sum of the squares of the first ten natural numbers is,<br />
1^(2) + 2^(2) + &#8230; + 10^(2) = 385</p>
<p>The square of the sum of the first ten natural numbers is,<br />
(1 + 2 + &#8230; + 10)^(2) = 55^(2) = 3025</p>
<p>Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 − 385 = 2640.</p>
<p>Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.</p></blockquote>
<p><strong>Solution:</strong><br />
<code>Function Prob006() As Long<br />
        Dim SumOfSquares As Long = 0<br />
        For k As Integer = 1 To 100<br />
            SumOfSquares += System.Math.Pow(k, 2)<br />
        Next<br />
        Dim Sum As Long = 0<br />
        For k As Integer = 1 To 100<br />
            Sum += k<br />
        Next<br />
        Dim SquareOfSum As Long = System.Math.Pow(Sum, 2)<br />
        Return SquareOfSum - SumOfSquares<br />
End Function</code></p>
<p><strong>Summary:</strong></p>
]]></content:encoded>
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		</item>
		<item>
		<title>Problem 005</title>
		<link>http://developmentgeek.com/blog/20100308/problem-005/</link>
		<comments>http://developmentgeek.com/blog/20100308/problem-005/#comments</comments>
		<pubDate>Mon, 08 Mar 2010 18:25:03 +0000</pubDate>
		<dc:creator>Kevin Fairchild</dc:creator>
				<category><![CDATA[Miscellaneous]]></category>
		<category><![CDATA[projecteuler]]></category>

		<guid isPermaLink="false">http://developmentgeek.com/blog/?p=124</guid>
		<description><![CDATA[2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder. What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20? Solution: Function Prob005() As Integer Dim FoundResult As Boolean = False Dim k As Integer [...]]]></description>
			<content:encoded><![CDATA[<blockquote><p>2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.</p>
<p>What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?</p></blockquote>
<p><strong>Solution:</strong><br />
<code>Function Prob005() As Integer<br />
        Dim FoundResult As Boolean = False<br />
        Dim k As Integer = 0<br />
        While FoundResult = False<br />
            k += 1<br />
            FoundResult = True<br />
            For f As Integer = 1 To 20<br />
                If k Mod f <> 0 Then<br />
                    FoundResult = False<br />
                End If<br />
            Next<br />
            If FoundResult = True Then Return k<br />
        End While<br />
End Function</code></p>
<p><strong>Summary:</strong><br />
This was pretty easy, but it seems like the problems are building upon each other, so I&#8217;m sure the next few are going to be a major pain&#8230;</p>
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		</item>
		<item>
		<title>Problem 004</title>
		<link>http://developmentgeek.com/blog/20100308/problem-004/</link>
		<comments>http://developmentgeek.com/blog/20100308/problem-004/#comments</comments>
		<pubDate>Mon, 08 Mar 2010 05:07:36 +0000</pubDate>
		<dc:creator>Kevin Fairchild</dc:creator>
				<category><![CDATA[Miscellaneous]]></category>
		<category><![CDATA[projecteuler]]></category>

		<guid isPermaLink="false">http://developmentgeek.com/blog/?p=108</guid>
		<description><![CDATA[A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 × 99. Find the largest palindrome made from the product of two 3-digit numbers. Solution: Function Prob004() As Integer For k As Integer = 100 To 999 For f As Integer = [...]]]></description>
			<content:encoded><![CDATA[<blockquote><p>A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 × 99.</p>
<p>Find the largest palindrome made from the product of two 3-digit numbers.</p></blockquote>
<p><strong>Solution:</strong><br />
<code>Function Prob004() As Integer<br />
        For k As Integer = 100 To 999<br />
            For f As Integer = 100 To 999<br />
                Dim intResult As Integer = k * f<br />
                Dim strResult As String = intResult.ToString<br />
                If strResult = StrReverse(strResult) AndAlso intResult > Prob004 Then<br />
                    Prob004 = intResult<br />
                End If<br />
            Next<br />
        Next<br />
End Function</code></p>
<p><strong>Summary:</strong><br />
This one was pretty quick and fun.</p>
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</rss>
