Comments on: Monday Math Madness 6 Solutions http://samjshah.com/2008/05/22/monday-math-madness-6-solutions/ Wed, 28 Jul 2010 15:46:26 +0000 hourly 1 http://wordpress.com/ By: Answer to my generalized MMM6 question « Continuous Everywhere but Differentiable Nowhere http://samjshah.com/2008/05/22/monday-math-madness-6-solutions/#comment-141 Answer to my generalized MMM6 question « Continuous Everywhere but Differentiable Nowhere Thu, 22 May 2008 23:39:09 +0000 http://samjshah.wordpress.com/?p=175#comment-141 [...] About ← Monday Math Madness 6 Solutions [...] [...] About ← Monday Math Madness 6 Solutions [...]

]]> By: samjshah http://samjshah.com/2008/05/22/monday-math-madness-6-solutions/#comment-139 samjshah Thu, 22 May 2008 20:53:29 +0000 http://samjshah.wordpress.com/?p=175#comment-139 Oh my gosh, that's awesome. I too love different solutions to the same problem. Like this one! Three different solutions, essentially the same, but different ways of approaching it. http://samjshah.com/2008/05/15/algebraic-manipulation-is-overrated/ Oh my gosh, that’s awesome. I too love different solutions to the same problem.

Like this one! Three different solutions, essentially the same, but different ways of approaching it.

http://samjshah.com/2008/05/15/algebraic-manipulation-is-overrated/

]]> By: Ξ http://samjshah.com/2008/05/22/monday-math-madness-6-solutions/#comment-138 Ξ Thu, 22 May 2008 19:37:40 +0000 http://samjshah.wordpress.com/?p=175#comment-138 This is neat -- I solved the first problem essentially the way you described (the second way) but did the second problem completely differently. What I noticed with the hamburgers was that the numbers formed a Fibonacci Sequence: there was 1 way to eat 1 lb of meat, 2 ways to eat 2 lbs, 3 ways to eat 3 lbs, 5 ways to eat 4 lbs, 8 ways to eat 5 lbs, etc. In general, the number of ways to eat n pounds of meat is the sum of the number of ways to eat (n-1) lbs and the number of ways to eat (n-2) lbs [because you could start by eating a 1lb burger and then eat the remaining (n-1) lbs in however many ways, or you could start by eating a 2-lb burger and then eat the remaining (n-2) lbs in however many ways.]. What I think is really neat is that Pascal's Triangle ties together our two different solutions to this problem. The jth entry in the ith row (starting with 0 rather than 1) is i choose j -- you can see the beginning in <a href="http://commons.wikimedia.org/wiki/Image:TrianguloPascalC.svg" rel="nofollow">this picture on wikipedia</a>. But if you add diagonals in Pascal's Triangle, you get Fibonacci numbers, like in <a href="http://goldennumber.net/pascal.htm" rel="nofollow">this picture</a>. So your formula is just adding those entries of Pascal's triangle, and my method gave the sum in terms of Fibonacci's number. I found that referred to as the Fibonacci Binomial Representation on <a href="http://binomial.csuhayward.edu/IdentitiesNamed.htm" rel="nofollow">this site</a> (but I'm not sure I can typeset it in the comment -- hence the hyperlink). Thanks for sharing your method -- I love it when there are really different ways to look at the same thing. This is neat — I solved the first problem essentially the way you described (the second way) but did the second problem completely differently. What I noticed with the hamburgers was that the numbers formed a Fibonacci Sequence: there was 1 way to eat 1 lb of meat, 2 ways to eat 2 lbs, 3 ways to eat 3 lbs, 5 ways to eat 4 lbs, 8 ways to eat 5 lbs, etc. In general, the number of ways to eat n pounds of meat is the sum of the number of ways to eat (n-1) lbs and the number of ways to eat (n-2) lbs [because you could start by eating a 1lb burger and then eat the remaining (n-1) lbs in however many ways, or you could start by eating a 2-lb burger and then eat the remaining (n-2) lbs in however many ways.].

What I think is really neat is that Pascal’s Triangle ties together our two different solutions to this problem.

The jth entry in the ith row (starting with 0 rather than 1) is i choose j — you can see the beginning in this picture on wikipedia. But if you add diagonals in Pascal’s Triangle, you get Fibonacci numbers, like in this picture. So your formula is just adding those entries of Pascal’s triangle, and my method gave the sum in terms of Fibonacci’s number. I found that referred to as the Fibonacci Binomial Representation on this site (but I’m not sure I can typeset it in the comment — hence the hyperlink).

Thanks for sharing your method — I love it when there are really different ways to look at the same thing.

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