Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 ?

share|cite|improve this question
Relevant: Proof for formula for sum of sequence $1+2+3+\ldots+n$?. – user2468 Jul 30 '12 at 16:42
up vote 7 down vote accepted

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 $.

share|cite|improve this answer
is something wrong?? – Santosh Linkha Jul 30 '12 at 16:44
I am just curious how you got n<=13. I am trying to solve from $n^2+n-200$. – Rajeshwar Jul 30 '12 at 17:09
The quadratic gives $13.650...$ and $-14.6...$. So should I just consider 13 ? and why ? – Rajeshwar Jul 30 '12 at 17:13
@Rejeshwar $n=14$ yields $105$ thus $n\leq 13$ (strict monotonicity). – vanna Jul 30 '12 at 17:30
@Rajeshwar $13.650$ gives $100$ ... less than it gives less than 100. greatest possible natural number less than 13.650 is 13 – Santosh Linkha Jul 30 '12 at 17:31

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 $(14)(15)/2.$

share|cite|improve this answer

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

share|cite|improve this answer
$ \dfrac{14\cdot 15}{2}=\color{red}{105}$ – Quixotic Jul 30 '12 at 16:40

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

share|cite|improve this answer
The quadratic gives $13.650...$ and $-14.6...$. So should I just consider 13 ? and why ? – Rajeshwar Jul 30 '12 at 17:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.