1
$\begingroup$

This question is purely out of curiosity. My little brother got a question for homework to find a rectangle where the Area = Outline. Both sides must also be integers, obviously. He found the square 4x4 which was sufficient. Now, using trial and error where $ y = \frac{2x}{x-2} $ I found that when x = 3 it works. We get y = 6, x = 3. However I would like all values of y that are integers, so in order to do this shouldnt I solve for x, where $ 2x = 0\mod\ (x-2) $ ? And if so, how would I do it?

$\endgroup$
2
$\begingroup$

(You need that $x-2$ divides $2x$, this is $2x \equiv 0 \pmod{x-2}$.)

Since $2x = 2(x-2) + 4$, this simplifies to $4 \equiv 0 \pmod{x-2}$. In other words you need $x-2 \mid 4$.

Yielding $x=6$, $x=4$, $x=3$.

$\endgroup$
  • $\begingroup$ Sorry, that was an accident. My main question, is how would I solve for that? $\endgroup$ – Gregory Peck Oct 23 '14 at 20:10
  • $\begingroup$ Okay. I updated. $\endgroup$ – quid Oct 23 '14 at 20:11
  • 1
    $\begingroup$ Excellent and simple! should have though of that ! $\endgroup$ – Gregory Peck Oct 23 '14 at 20:14
0
$\begingroup$

Since the area is $xy$ and the outline is $2(x+y)$, $xy = 2(x+y)$ or $xy-2(x+y) = 0$, or $xy-2(x+y)+4 = 4$, or $(x-2)(y-2) = 4$.

If $x$ and $y$ are integers, since $4$ can only be factored as $2\cdot 2, 1 \cdot 4$, and $4\cdot 1$, the only possible values for $(x, y)$ are $(4, 4), (3, 6)$, and $(6, 3)$.

$\endgroup$

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.