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Topic: Finger Guages: Sum of consecutive integers (Read 2828 times) |
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sigmashot
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Finger Guages: Sum of consecutive integers
« on: Jun 24th, 2008, 11:11am » |
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This is a question regarding the method of solving the finger guage problem on the "medium" riddles page. I took a look to see if there was another thread about it, but could not find one. Reply with a link to the thread if this is a dupe.
Basically, the problem is to find the sum of consecutive integers that equal 86, given the integers 1 to 24. Solution follows
20+21+22+23
I had some success mathematically representing the problem, where
x=the first number in the series
n=the number of integers in the series
z=the target sum (in the first case, 86)
The sum of a string of integers is the product of:
the sum of the first and last digits in the series
and:
half of the number of integers in the series.
(x + (x+n-1)) * n/2 = z
Which can be written:
.5n^2 + ((2x-1)/2)n = z
So, I'm trying to find a way to understand the problem better. It has been almost ten years since I took any higher level math, and the tools I have at my disposal (meaning: the operations I can still remember how to perform) are limited to algebra. Are there any problems that run parallel to this one? What I seem to have is an equation with three variables, so I can use brute force to find all positive integer solutions, etc., but I am looking for another way to look at this.
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towr
wu::riddles Moderator
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Re: Finger Guages: Sum of consecutive integers
« Reply #1 on: Jun 24th, 2008, 12:04pm » |
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There's an older thread here.
Note that some of the text there is hidden the old-fashioned way, by giving it the color of the background. So just select any suspicious gaps in text to see if anything's written there.
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| « Last Edit: Jun 24th, 2008, 12:04pm by towr » |
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Wikipedia, Google, Mathworld, Integer sequence DB
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sigmashot
Newbie
Posts: 8
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Re: Finger Guages: Sum of consecutive integers
« Reply #2 on: Jun 24th, 2008, 1:19pm » |
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Thank You.
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