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	<title>Comments on: Misleading picture</title>
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	<description>mathematical dialogues aimed to confuse</description>
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		<title>By: Alan</title>
		<link>http://easyquestion.net/thinkagain/2009/06/22/misleading-picture/#comment-28895</link>
		<dc:creator>Alan</dc:creator>
		<pubDate>Tue, 23 Jun 2009 03:30:23 +0000</pubDate>
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		<description>First, it doesn&#039;t matter if it is 2000 or 2009.  They have an equal number of solutions; one can get the 2009 solutions by adding 9 1 gram weights to the 2000 solutions and vice versa.

To find the number of ways to balance a 2000g cat, I&#039;ll first assume that we have only 1000g and 100g weights.  One can use 2 1000g weights, 1 1000g weight and 10 100g weights, or 20 100g weights.  Let me abbreviate those as (2,0), (1,10), and (0,20)

Now, lets add the 10g weights.  We have the above 3 ways, plus an additional 30 ways derived by replacing 100g weights with 10 10g weights each.  There are 33 ways to weigh a 2000g cat with 10g, 100g, and 1000g weights; 
3 have no 10g weights (2,0,0), (1,10,0), (0,20,0)
2 have 10 10g weights (1,9,10), (0,19,10)
2 have 20 10g weights (1,8,20), (0,18,20)
2 have 30 10g weights (1,7,30), (0,17,30)
...
2 have 100 10g weights (1,0,100), (0,10,100)
1 has 110 10g weights (0,9,110)
1 has 120 10g weights (0,9,110)
...
1 has 190 10g weights (0,1,190)
1 has 200 10g weights (0,0,200)

When we add the 1g weights, we add 10 unique ways to balance the cat for each 10g weight in the above listing; or 
10 * sum (1+2+...20) +
10 * sum (1+2+...10)
= 550 + 2100

So there are 2683 total ways to balance a 2009 g cat with 1g, 10g, 100g, and 1000g weights.

There must be a slick formula for this in the form of ax**3 + bx**2 + cx + d.</description>
		<content:encoded><![CDATA[<p>First, it doesn&#8217;t matter if it is 2000 or 2009.  They have an equal number of solutions; one can get the 2009 solutions by adding 9 1 gram weights to the 2000 solutions and vice versa.</p>
<p>To find the number of ways to balance a 2000g cat, I&#8217;ll first assume that we have only 1000g and 100g weights.  One can use 2 1000g weights, 1 1000g weight and 10 100g weights, or 20 100g weights.  Let me abbreviate those as (2,0), (1,10), and (0,20)</p>
<p>Now, lets add the 10g weights.  We have the above 3 ways, plus an additional 30 ways derived by replacing 100g weights with 10 10g weights each.  There are 33 ways to weigh a 2000g cat with 10g, 100g, and 1000g weights;<br />
3 have no 10g weights (2,0,0), (1,10,0), (0,20,0)<br />
2 have 10 10g weights (1,9,10), (0,19,10)<br />
2 have 20 10g weights (1,8,20), (0,18,20)<br />
2 have 30 10g weights (1,7,30), (0,17,30)<br />
&#8230;<br />
2 have 100 10g weights (1,0,100), (0,10,100)<br />
1 has 110 10g weights (0,9,110)<br />
1 has 120 10g weights (0,9,110)<br />
&#8230;<br />
1 has 190 10g weights (0,1,190)<br />
1 has 200 10g weights (0,0,200)</p>
<p>When we add the 1g weights, we add 10 unique ways to balance the cat for each 10g weight in the above listing; or<br />
10 * sum (1+2+&#8230;20) +<br />
10 * sum (1+2+&#8230;10)<br />
= 550 + 2100</p>
<p>So there are 2683 total ways to balance a 2009 g cat with 1g, 10g, 100g, and 1000g weights.</p>
<p>There must be a slick formula for this in the form of ax**3 + bx**2 + cx + d.</p>
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