# Take the positive integers 1,2,3... and add them all up, since I missed one of them, I'm going to get 2001, so what I missed?

Using arithmetic series summation formula, I got $$S_n=\frac{n(n+1)}{2}$$

Let $$S_n-k=2001$$, $$S_{63}=2016$$, I got $$k=2016-2001=15$$

But I think my process is a little loose, what should I do?

• "But I think my process is a little loose" -- I'm not sure what you mean here. What precisely is your question? Commented Apr 15, 2023 at 7:08
• If by "a little loose" you mean "not rigorous," then notice that you never show that $63$ is unique. It's not difficult—you can just point out that $n \geq 64$ always means that the difference $S_n-2001$ is greater than $n$—but that's one extra thing you can do. Commented Apr 15, 2023 at 17:19

## 2 Answers

I don't know how you chose $$S_{63}$$, If it is a MCQ exam, your method might be the quickest, nonetheless this is how I would solve it $$2001 + n \ge \frac{n(n+1)}{2} > 2001 \\ 0 \ge n^2-n-4002 \ \ \ \ and \ \ \ \ n^2+n-4002 > 0$$

Solving this you obtain $$n= 63$$

• You are right! I was so stupid that I tried all the $S_n$ one by one. Commented Apr 15, 2023 at 9:11

Let $$k$$ be the missing number. Then, $$\frac{n(n+1)}{2} - 2001 = k$$.

To tighten your process, you need to convert this lower and upper limit for $$n$$ and explore all feasible solution in this range. We have

$$k = \frac{n(n+1)}{2} - 2001 \ge 1$$

and

$$k = \frac{n(n+1)}{2} - 2001 \le n$$

Solving these two quadratic inequalities we get $$62.7 \le n \le 63.76$$. Hence the only feasible value is $$n = 63$$ and we get $$k = 15$$.

• To convince yourself $n=63$ is the only possibility, you can also note that $S_{62}$ is only $1953$, and for lower $n$ it gets only worse, so $n\ge 63$. Then note that $S_{64}$ is $2080$, and the difference to the desired $2001$ is $79$, exceeding the last term $64$ in the sum, and for greater $n$ it just becomes worse, so $n\le 63$. Commented Apr 15, 2023 at 7:23