Wednesday, January 25, 2012

The Last Post

The last post I wrote was 1777 words. It mentions something that happened in 1776. If I subtracted one word, it would be a weird sort of fractal.

I had the (stupid, conceptualist) idea (which I probably won't pursue for reasons which I explain below) to write a book describing the years 0 to 2012 in which each year's description would be allotted exactly the number of words that the year itself represented (measured from the year 0).

That is, the book would start with a blank page for the year 0. You could only describe 1 A.D. (or C.E.) with one word, 5 C.E. with five words, 1000 A.D. with one thousand words and this current year with exactly two thousand and twelve words.

Okay, maybe I will try it. If I get bored enough. I mean bored like a monk in a monastery. Because do the math.

So how many words would be in this book?

Does that sound like a hard math puzzle to you?

It's very easy. I remember this trick from mathematics.

When the great mathematician Gauss was a young boy, his teacher asked the class to do the sum of the numbers between 1 and 100.

Gauss wrote his answer on his slate immediately and turned it over while the other kids were busy adding one to two, three to three, six to four, etc.

How did he do it? He saw this number series could be paired off; 100 + 1= 101 , 2 + 99 = 101, etc.

So you have 50 pairs of numbers adding up to 101 = 5050.

In general, to find the sum of all the numbers from 1 to N:

1 + 2 + 3 + 4 + . . . . + N = (1 + N)*(N/2)

So this theoretical book with a conceptualist base would have a wordcount of: (1 + 2012)*(2012/2) words or 2013 x 1006 = 2,025,078 words!

Holy Shit!

I think I'll pass on that book!

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