Date: 05/05/99 at 16:32:16 From: Kathy Koos Subject: Searching for famous mathematician based on unsolvable problem Hello Dr. Math, I ask this question for my daughter. She was given an assignment that neither my husband nor I could solve (we both have had collegiate level calculus, etc.). With the help of a friend (Ph.D., J.D.), we were able to reach a solution. The problem involves a deriving a formula for solving sequential numbers: If a tower has a center with 6 blocks and four adjacent wings with blocks in descending order (5, 4, 3, 2, 1), then how many blocks are there? How many blocks are there if the tower is 12 blocks? 15 blocks, "n" blocks? Derive the formula. I seem to recall that there was a mathematician who was asked to compute all the numbers from 1 to 100. He was able to do so in a very brief time because he derived the formula to do so. His name, however, escapes me. I would like to find out who he was so I can instruct my daughter, thereby making math more "personal." Funny stories always seem to help her understand. Also, I want to build her confidence because she was unable to solve what we perceive to be a very difficult problem! Any help you could give me would be appreciated. Thank you! Kathleen
Date: 05/06/99 at 12:16:32 From: Doctor Peterson Subject: Re: Searching for famous mathematician based on unsolvable problem Hi, Kathy. You didn't ask for a solution to this, or tell me what age your daughter is and what she knows already, but I'll say a little about it anyway. One way I see to approach this assumes you know the formula for triangular numbers. Each "wing" is a triangular number (the sum of 1, 2, ... to n = n(n+1)/2); so the total in the tower is n (the central part) plus 4 triangular wings of height n-1: n + 4 [(n-1)n/2] = n + 2n^2 - 2n = 2n^2 - n Another approach uses one of the methods I like for proving the triangular number formula. Build a tower with n = 1 (just one block), then build a tower with n = 2 by adding a block on top of each one, then another on each of the four sides; and so on. Each layer you add has four more blocks than the previous layer, so the tower for n will be the sum of n of these layers: 1 + 5 + 9 + 13 + 17 + 21 (for n = 6) where the nth term is 1 + 4(n-1). Twice this sum will be 1 + 5 + 9 + 13 + 17 + 21 21 + 17 + 13 + 9 + 5 + 1 --------------------------- 22 + 22 + 22 + 22 + 22 + 22 = 6*22 or for any n, n times the sum of the first and last terms: n * [1 + 1 + 4(n-1)] = n * (4n-2) = 4n^2 - 2n and the number of blocks in the tower is half this, giving the same formula as before. >I seem to recall that there was a mathematician who was asked to compute all the numbers from 1 to 100... Yes, it's a good story. The boy who did the sum was the young Carl Friedrich Gauss, whom you can read about here: http://www-groups.dcs.stand.ac.uk/~history/Mathematicians/Gauss.html Here's an excerpt: At the age of seven, Carl Friedrich started elementary school, and his potential was noticed almost immediately. His teacher,Buttner and his assistant, Martin Bartels, were amazed when Gauss summed the integers from 1 to 100 instantly by spotting that the sum was 50 pairs of numbers each pair summing to 101. You might like to look for more details (or different versions of the story!) in your library, in a good biography, or perhaps even in an encyclopedia. He didn't originate this formula, of course, but he did create a lot of modern mathematics before he was finished! - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
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