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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.

AryanSonwatikar
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    The number of odd factors of 1000 is the number of possible sets. Try to find a set with 5 consecutive numbers. – Hw Chu Feb 18 '19 at 12:25
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    To begin with, find a formula to compute $a+(a+1)+(a+2)+\ldots+(a+n-1)$ – Hagen von Eitzen Feb 18 '19 at 12:28
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    $198+199+200+201+202$ is one – Eleven-Eleven Feb 18 '19 at 12:29
  • @HagenvonEitzen it's $na + \frac {n (n-1)}{2} $ but that has not led me anywhere. – AryanSonwatikar Feb 18 '19 at 12:31
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    First of all, $$ 1+2+3+ \ldots +n = \frac{n(n+1)}{2} $$ (proof omitted). Let's say $s(n) = \frac{1}{n}(n+1)$. Now we are interested in $$ s(b) - s(a) = 1000 $$ A quick numerical search gives the solutions $(a,b) = (27,52), (54,70), (197,202), (999,1000)$ ... – Matti P. Feb 18 '19 at 12:31
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    @MattiP. I think that's "(a,b]" for these examples (and yes, that's all of them). – Marco13 Feb 18 '19 at 12:34
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    The process that you described could have led you to a result pretty soon. Always add the next number, and remove the smallest numbers until the sum is <= 1000. After adding 52, you had found the first solution. (That's the "computer scientist approach", however. Mathematicians might tackle this differently...) – Marco13 Feb 18 '19 at 12:36
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    @HwChu Since $1000$ is divisible by $8$ but not $16$ there is also a solution with $2\times 8=16$ consecutive integers. The first and last of these need to sum to $\frac {1000}{8}=125$ and will therefore be centred around $\frac {125}2$. Matti P has found this with $55+ \dots + 70$. – Mark Bennet Feb 18 '19 at 12:46
  • Possible duplicate of In how many ways can a number be expressed as a sum of consecutive numbers? – user Feb 18 '19 at 13:30
  • @MarkBennet I am still pretty sure that my arguments is true (there is a proof). that solution from 55 to 70 corresponds to the odd factor 125. Consider a set of consecutive integers of 125 terms centered at $1000/125=8$, the set will be from $8-62=-54$ to $8+62=70$. Note that numbers between $-54$ and $54$ cancelled each other pairwise, so you end up with the set from 55 to 70. The link from the comment above this comment gives ideas from another perspective. – Hw Chu Feb 19 '19 at 15:19
  • @HwChu Apologies, I misunderstood what you were saying - that is a neat way of looking at things which I hadn't seen before. – Mark Bennet Feb 19 '19 at 17:40

3 Answers3

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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

Vinyl_cape_jawa
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    There is also a solution with an even number of summands. Suppose $a=2^bc$ where $c$ is odd, then $2^{b+1}$ integers centred on $\frac c2$ will sum to $a$. Some of these may be negative. So we have $2=-1+0+1+2$, but there is a solution here where all are positive. To be more general, if $c=de$ then $2^{b+1}d$ summands centred on $\frac e2$ will also work. Here that would be, for example, $80$ summands centred around $12.5$ and some of these would be negative (or $400$ summands centred around $2.5$, or $2000$ summands centred around $0.5$) – Mark Bennet Feb 18 '19 at 12:56
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    @MarkBennet True, but my answer aimed at giving some hints how to tackle problems with "consecutive integers", rather than being a complete answer. I rarely give solutions on this site since I believe MS is about learning – Vinyl_cape_jawa Feb 18 '19 at 13:01
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    I was just conscious that the "easy" case is with an odd number of summands and that one of the comments to the original question has missed the even possibility out altogether. I tried to honour the spirit of your answer by not giving all the details for the case which actually works. – Mark Bennet Feb 18 '19 at 13:11
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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}

egreg
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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 $n$.