The sum of consecutive integers is $50$. How many integers are there? 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. 
 A: 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.
A: 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.
A: 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 $n(2a+n-1)=100$.
A: 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 $S_n$
A: 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.
A: 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.
A: 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.
