2
$\begingroup$

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...

$\endgroup$
3
$\begingroup$

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...

$\endgroup$
  • 2
    $\begingroup$ Most questions look easy after you've figured out the solution :P $\endgroup$ – Simply Beautiful Art Sep 16 '19 at 21:42
3
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ How to show that $i$ is smaller than n? $\endgroup$ – yuanming luo Sep 16 '19 at 21:33
  • $\begingroup$ is that true for all equations like this? or was it just luck that it worked by taking the floor of the equation after setting i = 0? $\endgroup$ – rhavelka Sep 16 '19 at 21:34
  • 1
    $\begingroup$ @rhvelka It is true because of $\frac{n(n+1)}{2}$ is increasing function, i is an integer. So what you get is bigger than the answer you want. Then take the floor. Becuase $\frac{n(n+1)}{2}$ is always an integer, so $1986-\frac{n(n+1)}{2}$. The only question is the uniqueness of $i$ $\endgroup$ – yuanming luo Sep 16 '19 at 21:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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