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I started off by calling the number of numbers in my list "$n$". Since the integers are consecutive, I had $x + (x+1) + (x+2)...$ and so on. And since there were "$n$" numbers in my list, the last integer had to be $(x+n)$. This is where I got stuck. I didn't know how to proceed because I am not given the starting point of my integers, nor an ending point.

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    $\begingroup$ If there are $n$ numbers in your list and the first is $x = x + 0$, then the last is $x + n - 1$. $\endgroup$ – N. F. Taussig Jun 24 '17 at 10:37
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If your $n$ is odd, then the middle number has to be $50/n$. The odd divisors of $50$ are $1$ and $5$, which gives us two solutions $50=50$ and $8+9+10+11+12=50$

If $n$ is even, then $50/n$ is the half-integer between the middle two numbers. So $n$ has to be an even divisor of $100$, but not a divisor of $50$, so $n=4$ or $20$.

If $n=4$, then $50/4 = 12.5$ and we get $11+12+13+14=50.$

If $n=20$. then $50/20 = 2.5$ and we get $-7+-6+-5+\cdots +11+12 = 50.$

So there are 4 answers: $n=1, 4, 5, $ and $20$.

Edit: As Bill points out, I missed the divisors $25$ and $100$, which give two more answers: $50 = -10+-11+\cdots+14$ and $50 = -49 +-48+\cdots +50$.

Note that each solution with negative integers is related to an all-positive solution. From the solution $11+12+13+14=50$, we just prepend the terms $-10, -9, \ldots, 10$, which add to $0$, and we have another solution.

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    $\begingroup$ interesting the approach using the middle number! $\endgroup$ – G Cab Jun 24 '17 at 12:31
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    $\begingroup$ Great technique! n=25 and n=100 work aswell $\endgroup$ – Mike Jun 24 '17 at 13:47
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    $\begingroup$ Incomplete, e.g. the odd divisor $25\ \ $ $\endgroup$ – Bill Dubuque Jun 24 '17 at 14:01
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    $\begingroup$ @GCab If you think of it in terms of the "average" (mean), instead of the "middle" (median) number, then it's less surprising. The two only happen to be equal because the numbers are uniformly distributed. $\endgroup$ – Robin Saunders Jun 24 '17 at 15:24
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What may be helpful is to use the formula for the sum of an arithmetic progression: if you have a sequence whose first term is $a$ and each term is $d$ more than the rest, then the sum of the first $n$ terms is $na+\frac{n(n-1)}{2}d$. In this case, since we are looking at consecutive integers, $d=1$, and so you are trying to find $a$ and $n$ such that $na+\frac{n(n-1)}2=50$, or equivalently $n(2a+n-1)=100$.

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  • $\begingroup$ This is surely helpful. As there are only nine divisors of 100, you can try them all easily and look at which lead to an integer $a$. The nice thing is that (unlike when using a more sophisticated solution) you won't miss the negative first terms. $\endgroup$ – maaartinus Jun 25 '17 at 13:38
  • $\begingroup$ Yes, and you can make things slightly quicker using the fact that if $n$ is even then $2a+n-1$ is odd, and vice versa, so you can skip cases like $2\times 50$ where both factors are even. $\endgroup$ – Especially Lime Jun 25 '17 at 19:29
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$$\scriptsize\begin{align} 50 &=2\times 25 &&=25\boxed{+}25 &&=24.5\boxed{+}25.5 \text{ (AP but not integer AP)}\\ &=4\times 12.5 &&=12.5+12.5\boxed{+}12.5+12.5 &&=\color{red}{11+12\boxed{+}13+14}\\ &=5\times 10 &&=10+10+\boxed{10}+10+10 &&=\color{red}{8+9+\boxed{10}+11+12}\\ &=10\times 5 &&=\underbrace{\underbrace{5+5+\cdots+5}_{5}\boxed{+}\underbrace{5+5+\cdots +5}_{5} }_{10} &&=\underbrace{\underbrace{0.5+1.5+\cdots+4.5}_{25}\boxed{+}\underbrace{5.5+\cdots+9.5}_{25}}_{50}\\ & && && \quad \text{(AP but not integer AP)}\\ &=20\times 2.5 &&=\underbrace{\underbrace{2.5+2.5+\cdots+2.5}_{10}\boxed{+}\underbrace{2.5+\cdots +2.5}_{10} }_{20} &&=\color{red}{\underbrace{\underbrace{-7+(-6)+\cdots+2}_{10}\boxed{+}\underbrace{3+4+\cdots+12}_{10}}_{20}}\\ &=25\times 2 &&=\underbrace{\underbrace{2+2+\cdots+2}_{12}+\boxed{2}+\underbrace{2+\cdots +2}_{12} }_{25} &&=\color{red}{\underbrace{\underbrace{-10+(-9)+\cdots+1}_{12}+\boxed{2}+\underbrace{3+\cdots+13+14}_{12}}_{25}}\\ &=50\times 1 &&=\underbrace{\underbrace{1+1+\cdots+1}_{25}\boxed{+}\underbrace{1+\cdots +1}_{25} }_{50} &&=\underbrace{\underbrace{-23.5+(-22.5)+\cdots+0.5}_{25}\boxed{+}\underbrace{1.5+\cdots+25.5}_{25}}_{50}\\ & && && \quad \text{(AP but not integer AP)}\\ &=100\times 0.5 &&=\underbrace{0.5+0.5+\cdots+0.5}_{50}\boxed{+}\underbrace{0.5+0.5+\cdots+0.5}_{50} &&=\color{red}{\underbrace{\underbrace{-49+(-48)+\cdots+0}_{50}\boxed{+}\underbrace{1+2+\cdots +49+50}_{50} }_{100}} \end{align}$$


In more detail:

$$\frac {50}n=m$$ If $n$ is even $(n=2p)$, then we want $m=a+0.5 \;\;(a\in \mathbb Z)$ $\cdots$ Condition $(1)$

  • The AP would comprise $p$ consecutive integers on either side of $m$.

If $n$ is odd $(n=2p+1)$, then we want $m\in \mathbb Z$ $\cdots$ Condition $(2)$

  • The AP would comprise $p$ consecutive integers on either side of $m$, as well as $m$ itself.

Try different values of $n$ (excluding the trivial case $n=1$):

  • If $n=2$, then $m=25$, hence Condition ($1$) not satisfied.

  • If $n=3$, then $m=16\frac 23$, hence Condition ($2$) not satisfied.

  • If $n=4$, then $m=12.5$, hence Condition $(1)$ satisfied, so the AP is $\lbrace (m-1.5), (m-0.5), (m+0.5), (m+1.5)\rbrace$, i.e. $\color{red}{11,12,13,14}$ ($AP1$).
    Adding negative terms and corresponding positive terms which cancel out gives: $\color{blue}{-10,-9,-8\cdots, 0\cdots, 8,9,10,}\color{red}{11,12,13,14}$ ($AP 1'$)

  • If $n=5$, then $m=10 $, hence Condition $(2)$ satisfied, so the AP is $\lbrace (m-2),(m-1),m,(m+1),(m+2)\rbrace$, i.e. $\color{red}{ 8,9,10,11,12}$ ($AP 2$).
    Adding negative terms and corresponding positive terms which cancel out gives: $\color{blue}{-7,-6,-5\cdots, 0\cdots, 5,6,7,}\color{red}{8,9,10,11,12}$ ($AP 2'$)

  • If $n=6, 7,8,9$, then $m\notin\mathbb Z$ and $m\neq a+0.5 $, hence neither Conditions $(1)$ or $(2)$ satisfied.

  • If $n=10$, then $m=5$, hence Condition $(2)$ not satisfied - not possible.

  • If $n=11,12,\cdots, 24$, then $m\notin\mathbb Z$ and $m\neq a+0.5$, hence neither Conditions $(1)$ or $(2)$ satisfied.

  • If $n=25$, then $m=2$, hence Condition $(2)$ satisfied, so AP is $\color{red}{\underbrace{-10,-9,\cdots 0,1}_{12\text{ terms}},2,\underbrace{3,\cdots 9,10,11,12,13,14}_{12 \text{ terms}}}$ (same as $AP 1'$)

  • If $n=26,27,\cdots, 49$, then $m\notin\mathbb Z$ and $m\neq a+0.5 $, hence neither Conditions $(1)$ or $(2)$ satisfied.

  • If $n=100$, then $m=0.5$, hence Condition $(1)$ satisfied, so AP is $\color{red}{\underbrace{-49,-48,\cdots -2,-1,0}_{50\text{ terms}},\underbrace{1,2,3,\cdots 48,49,50}_{50 \text{ terms}}}$ ($AP 3$)

  • For higher values of $n$, $0<m<0.5$ - not possible.


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Without reading other answers... this should tell you how an old computer programmer thinks, versus a real mathematician.

First the obvious answer is the single integer 50. However, if negative numbers are allowed, then we can scoop up the sequence from -49 to 50 for 100 consecutive numbers. This is the longest possible sequence.

If n is the starting number of a sequence and s is the number of consecutive numbers, then we end up with

50 = (n+0) + (n+1) + (n+2) +... + (n+s-1)
50 = ns + (s-1)(s)/2

This is helpful, maybe, as ns pretty much puts a box around the solution set. Since we know the nature of the rightmost term, we can tell that s is 10 or less.

Consider the transformation if we multiply both sides by 2/s:

100/s = 2n + s - 1

The right side will always be an integer. Therefore, s must always divide 100 evenly. From above we know that s is between 1 and 10, so the only possible values for s are 1, 2, 4, 5, and 10.

That reduces the problem to 5 single-degree-of-freedom equations. Let's do it.

100/1 = 2n, n = 50, seq is (50)
100/2 = 2n + 1, no solution
100/4 = 2n + 4 - 1, 25-3 = 2n, n = 11, seq is (11, 12, 13, 14)
100/5 = 2n + 5 - 1, 20 = 2n + 4, 16=2n, n = 8, seq is (8, 9, 10, 11, 12)
100/10 = 2n + 10 - 1, 10 = 2n +9, 1 = 2n, no solution

So those are all the solutions where n and s both > 0.

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Using the standard summation of an arithmetic progression formula:
$S_n=\frac{n(2a_1+(n-1)d)}{2}$
here since $d=1$
$2S_n=n(2a_1+n-1)$
$2S_n=n(a_1+(a_1+n-1))$
Here $a_1+n-1$ is just the last term.
$2S_n=n(a_1+a_n)$
Rewrite as

$n=\frac{2S_n}{a_1+a_n}$

If n is even:

$n=\frac{2S_n}{a_1+a_n+a_{n/2}-a_{n/2}}$


$n=\frac{2S_n}{2a_{n/2}}$

Since $a_1+a_{n/2}+a_n-a_{n/2}=2a_{n/2}$
hence

$n=\frac{S_n}{a_{n/2}}$

and since $n\in Z$ , so $a_{n/2}$ is a divisor of $S_n$
once you get the required $n, a=a_{n/2}-n/2$
Similarly for odd $S_n$ you get

$n=\frac{S_n}{a_{(n-1)/2}-1}$


where $a_{(n-1)/2}-1$ is a divisior of $S_n$

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The sum of consecutive numbers $1\dots n$ up to $n$ starting from 1 is $\frac{n(n+1)}{2}$. Since you want only a partial sum from, say, $m+1$ to $n$, you can just subtract to get the partial sum:

$$ \frac{n(n+1)}{2} - \frac{m(m+1)}{2} = \frac{n^2+n-m^2-m}{2} = 50 $$

Multiplying by 2 gets you

$$ 100 = n^2+n-m^2-m = (n+m)(n-m) + n-m = (n+m+1)(n-m) $$

Now the trick: $n+m$ and $n-m$ are either both even or both odd. This means that in the last formula, one term must be even, the other one must be odd. Factorization of $100 = 5\cdot 5 \cdot 2 \cdot 2$ means that there are very limited solutions. For instance, one term is 25, the other is 4 (there is one other nontrivial combination, see below). Since $n+m+1$ is larger than $n-m$, in the choice shown here, you must have \begin{align} n+m+1 &= 25\\ n-m &=4 \end{align} Solving this will give you the desired numbers.

Addendum You get the combination by observing that you get all acceptable combinations of the factorisation via: \begin{align} 100 &= 1\cdot (5\cdot5\cdot2\cdot2)\\ &= 5\cdot (5\cdot2\cdot2)\\ &= (5\cdot 5)\cdot(2\cdot2) \end{align} where you have to stop as all the following factorisations will contain only even factors.

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Let tere be $n\geq1$ numbers, the smallest of them being $p\in{\mathbb Z}$. We then want $$p+(p+1)+\ldots+\bigl(p+(n-1)\bigr)=50\ ,$$ which amounts to $np+{(n-1)n\over2}=50$, or $$n(n+2p-1)=100\ .$$ Going with $n$ through the $9$ divisors of $100$ we obtain the following table: $$\matrix{n:&1&2&4&5&10&20&25&50&100 \cr p:&50&&11&8&&-7&-10&&-49\cr}$$ When $n\in\{2,10,50\}$ solving $n+2p-1={100\over n}$ for $p$ does not lead to an integer $p$. It follows that there are $6$ solutions in all.

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  • $\begingroup$ How did you get from your first equation to the second? $\endgroup$ – Tony Ennis Jun 25 '17 at 21:26
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First, welcome to Mathematics stack exchange. Second, if you start like $x+(x+1)+...+(x+n)$, you will end nowhere. $50$ is a concrete and small number so what can you try is to use that fact, sum of two consecutive numbers it is not, due to sum of two consecutive numbers can not be divided with $2$ in $\mathbb{Z}$, sum of three consecutive numbers is divisible with $3$, so it is not sum of three numbers. Four consecutive numbers- now this is on first look possible(remainder mod $4$ is $2$) but each one of them has to be around $12$, so first options that you have is $11+12+13+14=23+27= 50$, hence the solution. Maybe there are even other solution, but I feel free to interpret your question as required to see one possible solution.

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    $\begingroup$ There are other solutions. The simplest is $50$. $\endgroup$ – N. F. Taussig Jun 24 '17 at 10:42

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