Tricky math terms

[llijnnasil]

Hey, I just found this interesting exercise and I'd like to know how to solve it.

The exercise:
Between the numbers 4, 5, 6, ...... , n, you can put minus or plus signs ( - , +) to create
a term.
(n means any natural number) and (the numbers must be in order)

Which possible numbers for n can you put so that the term has the value 0 ?

I've already found a term which has the value 0, but how to find ALL the possible values for n????

example:
4-5-6+7 = 0
so in this example, the value for n is 7.

THANK YOU SO MUCH FOR HELP !

 

[HallsofIvy]

So given "n" you want to find p and a sequence of "+" and "-" so that \(\displaystyle (n- p)\pm (n-p-1)\pm (n-p-2)\pm \cdot\cdot\cdot \pm (n-1)\pm n= 0\)?

 

[llijnnasil]

So given "n" you want to find p and a sequence of "+" and "-" so that \(\displaystyle (n- p)\pm (n-p-1)\pm (n-p-2)\pm \cdot\cdot\cdot \pm (n-1)\pm n= 0\)?

I'm not sure what you mean, n stands for the last number in your term.
You always have to start the term with the number 4.
The numbers must be subsequent, I mean they have to be in order.
4,5,6,7,8,9,10 ........

 

[HallsofIvy]

Oh, so you want to start with "4" and choose signs so that \(\displaystyle 4\pm 5\pm \cdot\cdot\cdot\pm (n-1)\pm n= 0\) for every n. That is the same as asking whether, for every n, the numbers, 4, 5, ..., n can be divided into two sets which have the same sum.

 

[Ishuda]

In the particular case {4, 5, 6, 7}, there are two and in fact the number of cases must be even for any set of numbers since minus zero is still zero. Thus in your example you have
4-5-6+7 = 0
so you would also have
-(4-5-6+7) = -4 + 5 + 6 - 7 = 0.

Note that your example can be written as
(4-5) - (6-7) = 0
so, as HallsofIvy said, you have two groups which add to the same thing. Thus if you always start with 4, n = 4m - 1 and m is greater than 1, you always have at least two solutions:
(4-5) - (6-7) + (8-9) - (10-11) +...+{[(4m-4)-(4m-3)] - [(4m-2)-(4m-1)]}
and the negative of that.

 

[Ishuda]

Oh, and you always have at least two combinations for n=4m, m greater than 1. To prove consider the 4m-1 term
[-(4 + 5) + (6+7)] + [- (8 +9) + (10 + 11)] + ....+{-[(4m-4)+(4m-3)] + [(4m-2)+(4m-1)]} = 4m