Sum of n different positive integers is less than 100. What is the greatest possible value for n? I am stumped on the following question:


The sum of n different positive integers is less than 100. What is the greatest possible value for n?
a) 10, b) 11, c) 12, d) 13, e) 14


The answer is d).
Any idea on how to solve it ?
 A: Sum of different numbers is least when it's consecutive numbers from beginning from $1$. The sum would be 
$$ {n(n+1) \over 2} \leq 100 $$
This inequality gives $  n \leq 13 $.
A: I would try to make the summands as small as possible. I.e. $1+2+3+\cdots+n=n(n+1)/2<100.$ Thus, compare $(13)( 14)/2$ with $(14)(15)/2.$
A: As 
\begin{equation}
1+2+\cdots +n = \dfrac{n(n+1)}{2}
\end{equation}
and $\dfrac{13\cdot 14}{2}=91$. We have $n=13.$ In fact, $\dfrac{14\cdot 15}{2}=105$.
A: Clearly, to fit the most numbers in to the sum, you will need to do $1+2+3+\cdots$.
You could manually work out how many numbers you can fit in, before you get to 100, or you could use the fact that:
$ 1+2+3+\cdots+n = \frac{n}{2}(n+1) $
So you want:
$\frac{n}{2}(n+1) < 100$.
$n(n+1) < 200$
Now, you could solve the quadratic, and find the valid regions, but I'm going to try it another (probably worse) way:
If you had $n(n+1) =200$, you can see that $\sqrt{200}$ is directly between $n$ and $n+1$.
So for $n(n+1) < 200$, the highest value of $n$ will always be either $\rm floor(\sqrt{200})$ or if that is too many, $\rm floor(\sqrt{200}) - 1$.
