http://betterexplained.com/articles/techniques-for-adding-the-numbers-1-to-100/ Learning shouldn't hurt. Let's share the insights that made difficult ideas click. Fri, 20 Nov 2009 14:09:06 -0800 http://wordpress.org/?v=2.8.4 hourly 1 http://betterexplained.com/articles/techniques-for-adding-the-numbers-1-to-100/#comment-254525 Cord Sun, 04 Oct 2009 03:34:52 +0000 http://betterexplained.com/articles/techniques-for-adding-the-numbers-1-to-100/#comment-254525 I stumbled across your page while looking for an answer to the same question posed in #30, but was intrigued by all the methods described here. I was trying to figure out a way to map a series of numbers (say 1 through 15) to a smaller result set of numbers (say 1 through 5) such that: 1 => 1 2,3 => 2 4,5,6 => 3 7,8,9,10 => 4 11,12,13,14,15 => 5 as I was pondering how to write a mathematical function to get a result (instead of doing brute iteration) I realized that this smacked of Gauss. Not being able to memorize formulas, I sat down with a piece of paper and tried summing 1 through 100. I cut the set in half, yielding 1 through 50 and 51 through 100. The two 'half' sets match such that you can make a pair using one number from each half-set that adds up to 101 (1 + 100, 2 + 99, 3 + 98, etc). There are exactly 50 such pairs (100/2), so the sum must be 101*50. Making a formula, this makes (X+1)*(X/2) which is the same as ((X+1)*X)/2, which brings us back into familiar territory. Thanks for an interesting article, and for reassurance that the math actually works. now ... how to invert a parabolic function ... *heads back to google* I stumbled across your page while looking for an answer to the same question posed in #30, but was intrigued by all the methods described here. I was trying to figure out a way to map a series of numbers (say 1 through 15) to a smaller result set of numbers (say 1 through 5) such that:
1 => 1
2,3 => 2
4,5,6 => 3
7,8,9,10 => 4
11,12,13,14,15 => 5
as I was pondering how to write a mathematical function to get a result (instead of doing brute iteration) I realized that this smacked of Gauss. Not being able to memorize formulas, I sat down with a piece of paper and tried summing 1 through 100. I cut the set in half, yielding 1 through 50 and 51 through 100. The two ‘half’ sets match such that you can make a pair using one number from each half-set that adds up to 101 (1 + 100, 2 + 99, 3 + 98, etc). There are exactly 50 such pairs (100/2), so the sum must be 101*50. Making a formula, this makes (X+1)*(X/2) which is the same as ((X+1)*X)/2, which brings us back into familiar territory.
Thanks for an interesting article, and for reassurance that the math actually works.
now … how to invert a parabolic function … *heads back to google*

]]> http://betterexplained.com/articles/techniques-for-adding-the-numbers-1-to-100/#comment-254402 Missy Sat, 03 Oct 2009 07:15:24 +0000 http://betterexplained.com/articles/techniques-for-adding-the-numbers-1-to-100/#comment-254402 My question is how to add/count every "1" found between 0 and any end point. End Point could be 10,000. Thanks. My question is how to add/count every “1″ found between 0 and any end point. End Point could be 10,000. Thanks.

]]> http://betterexplained.com/articles/techniques-for-adding-the-numbers-1-to-100/#comment-252121 Kalid Tue, 08 Sep 2009 21:53:44 +0000 http://betterexplained.com/articles/techniques-for-adding-the-numbers-1-to-100/#comment-252121 @nada: If you want to add 47 to 100, for example, you can do this: Add 1 to 100 (all the numbers: (100 * 101)/2) and the subtract the sum of 1 to 46 (all the numbers you don't want (46 * 47)/2). @nada: If you want to add 47 to 100, for example, you can do this: Add 1 to 100 (all the numbers: (100 * 101)/2) and the subtract the sum of 1 to 46 (all the numbers you don’t want (46 * 47)/2).

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