Sum of n different positive integers is less than 100. What is the greatest possible value for n?

Question

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?

<p>a) 10, b) 11, c) 12, d) 13, e) 14</p>

The answer is d).

Any idea on how to solve it ?

Answer 1

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 $.

Answer 2

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 $(14)(15)/2.$

Answer 3

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, $\dfrac{14\cdot 15}{2}=105$.

Answer 4

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, $\rm floor(\sqrt{200}) - 1$.