Block Tower

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/