How to find the positive integer solution?

Question

The question askes the sum of $1$ to $n$ and a number $i$ which $1\le i \le n$ is equal to 1986

Which means $$\frac{n(n+1)}{2}=1986-i$$ Then how to find the solution algebraically? I get $i=33,n=62$ by listing them...

Answer 1

Sorry, It is a stupid question.

We get $$\frac{n(n+1)}{2}<1986, \frac{n(n+1)}{2}>1986-n$$

Which result the only integer $n=62$ satisfies the equation...

Answer 2

Find $n$ as the floor of the value you get with the quadratic formula and $i=0$. Then $i$ is just what's left.

Answer 3

You can estimate $(n+\frac 12)^2=n^2+n+\frac 14 \lt 2\times 1986=3972\approx 3969=63^2$ so $n\lt 63$.

On the other hand $61\times 62 =3782=3972-190$ would be too small. So $n\gt 61$. There is only one possible integer answer.

You really only need the first of these which shows you that $62$ will be close.