MHB Creating terms that have the value 0

[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 !

 

[awkward]

Hi Ilijnnasil,

Welcome to MHB.

Let's say an integer $n \ge 4$ is "acceptable" if there are numbers $a_4, a_5, a_6, \dots , a_n$ where $a_i \in \{-1,1\}$ and $\sum_{i=4}^n a_i \ i = 0$. You have already shown that 7 is acceptable. Another acceptable number is 8, because $1 \cdot 4 + 1 \cdot 5 + 1 \cdot 6 - 1 \cdot 7 - 1 \cdot 8 = 0$.

If $n$ is acceptable, then $n+4$ is also acceptable, because then

$\sum_{i=4}^n a_i \ i + 1 \cdot (n+1) -1 \cdot (n+2) - 1 \cdot (n+3) + 1 \cdot (n+4) = 0$.

So we know all numbers of the form $4n+3$ and $4n+4$ are acceptable, where $n = 1,2,3, \dots $.

Can there be any other acceptable numbers? Well, notice that
$$S(n) = \sum_{i=4}^n i = \frac{n(n+1)}{2} - 6$$
If $n$ is acceptable, then the terms with +1 attached must sum to
$$\frac{1}{2} S_n = \frac{n(n+1)}{4} - 3$$
which must be an integer, so $n(n+1)$ must be a multiple of 4. Now any integer is congruent to one of 0, 1, 2, or 3 modulo 4, and we see that $n(n+1) \equiv 0 \pmod 4$ when n is 0 or 3, and $n(n+1) \equiv 2 \pmod 4$ when n is 1 or 2. So if $n$ is acceptable, we must have $n \equiv 0 \pmod 4$ or $n \equiv 3 \pmod 4$.

This rules out any numbers but $4n+3$ and $4n+4$ for $n = 1,2,3, \dots$,
so that is the complete list of acceptable numbers.