Sum of certain consecutive numbers is $1000$.

Question

Question:

The sum of a certain number (say $n$) of consecutive positive integers is $1000$. Find these integers.

I have no idea how to approach the problem. I did try the following but did not arrive anywhere: I said that $$1+2+3+....+44=990$$ Then I subtracted numbers from $1$ to $9$ and added $45$. Then subtracted $10$ and added $46$ and continued the process. But arrived nowhere.

Thanks for the help!!

P.S. I do know that $n=1$ such that the "numbers" belong to the set $S=\{1000\}$ is trivial solution but looking for others.

Edit : I am asking for the integers and not the number of ways it can be done in.

Answer 1

The sum you're looking at is $$ a+(a+1)+\dots+(a+n-1)=na+(1+2+\dots+(n-1))=na+\frac{n(n-1)}{2} $$ so you get the equation $$ n^2+(2a-1)n-2000=0 $$ An integer solution has to be a divisor of $2000$, so of the form $n=2^x5^y$, with $0\le x\le 4$ and $0\le y\le 3$. The condition is then that $$ 2a-1=\frac{2000-n^2}{n}=\frac{2000}{n}-n $$ This number must be odd. If $n$ is even (that is, $x>0$), we need that $2000/n$ is odd, so $x=4$. If $n$ is odd (that is, $x=0$), any value of $y$ is good.

Thus we have $n\in\{1,5,25,125,16,80,400,2000\}$.

If you want $a>0$, then $n^2<2000$ and the choices are reduced to $n\in\{1,5,25,16\}$, corresponding to \begin{array}{cc} n & a \\ \hline 1 & 1000 \\ 5 & 198 \\ 25 & 28 \\ 16 & 55 \end{array}

Answer 2

You could think as follows:

Consecutive numbers: $$ \ldots, r-2, r-1, r, r+1, r+2, \ldots $$

If you sum these up you will get

$$ n\cdot r $$

Where $n$ is an odd number

So we want $n\cdot r= 1000$

Try for example $n=5$

hope this helps :)

Edit: read the comments below

Answer 3

The sum of $n$ consecutive numbers is $n$ times the average. I.e.

$$n\frac{i+i+n-1}2=\frac{n(2i+n-1)}2.$$

We can look for $n$ among the factors of $2000$, using

$$i=\frac12\left(\frac{2000}n-n+1\right).$$

The ratio $$\frac{2000}n$$ must have the opposite parity of $n$.