Q:

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:

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:

$$\scriptsize\begin{align}
50
&=2\times 25
&&=25\boxed{+}25 
&&=24.5\boxed{+}25.5 \text{ (AP but not integer AP)}\\
&=4\times 12.5 
&&=12.5+12.5\boxed{+}12.5+12.5 
&&=\color{red}{11+12\boxed{+}13+14}\\
&=5\times 10 
&&=10+10+\boxed{10}+10+10   
&&=\color{red}{8+9+\boxed{10}+11+12}\\
&=10\times 5 
&&=\underbrace{\underbrace{5+5+\cdots+5}_{5}\boxed{+}\underbrace{5+5+\cdots +5}_{5} }_{10}
&&=\underbrace{\underbrace{0.5+1.5+\cdots+4.5}_{25}\boxed{+}\underbrace{5.5+\cdots+9.5}_{25}}_{50}\\
& && && \quad \text{(AP but not integer AP)}\\
&=20\times 2.5 
&&=\underbrace{\underbrace{2.5+2.5+\cdots+2.5}_{10}\boxed{+}\underbrace{2.5+\cdots +2.5}_{10} }_{20}
&&=\color{red}{\underbrace{\underbrace{-7+(-6)+\cdots+2}_{10}\boxed{+}\underbrace{3+4+\cdots+12}_{10}}_{20}}\\
&=25\times 2 
&&=\underbrace{\underbrace{2+2+\cdots+2}_{12}+\boxed{2}+\underbrace{2+\cdots +2}_{12} }_{25}
&&=\color{red}{\underbrace{\underbrace{-10+(-9)+\cdots+1}_{12}+\boxed{2}+\underbrace{3+\cdots+13+14}_{12}}_{25}}\\
&=50\times 1 
&&=\underbrace{\underbrace{1+1+\cdots+1}_{25}\boxed{+}\underbrace{1+\cdots +1}_{25} }_{50}
&&=\underbrace{\underbrace{-23.5+(-22.5)+\cdots+0.5}_{25}\boxed{+}\underbrace{1.5+\cdots+25.5}_{25}}_{50}\\
& && && \quad \text{(AP but not integer AP)}\\
&=100\times 0.5 
&&=\underbrace{0.5+0.5+\cdots+0.5}_{50}\boxed{+}\underbrace{0.5+0.5+\cdots+0.5}_{50}
&&=\color{red}{\underbrace{\underbrace{-49+(-48)+\cdots+0}_{50}\boxed{+}\underbrace{1+2+\cdots +49+50}_{50} }_{100}}
\end{align}$$

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.

