Summation Problem: Find Lowest Non-Negative Value

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Homework Statement



A set contains numbers from 1-100. What is the least non-negative value that one can form by putting a + or - in front of each number, and summing the values?

Homework Equations



there are a few general summation formulas which I know...

The Attempt at a Solution



The problem that I am facing is the purpose:

by adding a + or - in front of each number

Is the question asking me to perform an alternating series? +, -, +, -, +, -
or solely one for 1+2+3+4+5... and one for 1-2+3-4+5...

Please give me some guidance... thanks
 
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You can add a + or a - anywhere, so you can partition the set and then proceed to + or - accordingly. If the set is the integers in [0,100], then you can pair integers a,b such that a+b=100. Hopefully I am understanding the question right.
 
Hmmm let me update the question if anyone is confused.

A set contains numbers from 1-100. What is the least non-negative value that one can form by putting a + or - in front of each number, and summing the values?

I think this is a bit more clear.Hmm I am sorry, VeeEight. I read about partition, but I still don't understand its application.
Now i just rewrite the question, so i will wait for your confirmation.
 
My idea was that if the your set is {0, 1, 2, ..., 100}, then you can say 0+100=100, 1+99=100, 2+98=... and so on. Thus, you can take 100/2 terms in this form and take a second class of 100/2 terms and add them up and subtract the two classes.
 
We want to consider every such expression. For instance:

1 + 2 + 3 + 4 + ... + 100
where every sign is +.

1 - 2 - 3 + 4 - 5 + 6 - 7 + 8 + 9 + 10 - 11 + ... + 100
where every prime has a - sign.

For each number, we choose a + or a -, and then we evaluate the resulting sum/difference. There's no way to evaluate these in general other than going through each and adding/subtracting all 100 numbers.

However, we're looking for the smallest non-negative sum, so if we can show a way to sum these numbers and get zero, we'd be finished.

Just start experimenting with the signs on the first several numbers, and you should find a nice pattern.

Big hint:
Note that 1 - 2 - 3 + 4 = 5 - 6 - 7 + 8 = 0.
 
wooo this is really really impressive. i wish i learn this at my young age. my friend told me he learned this since he was in Math Olympics team.

lol thanks guys.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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