Question

how many different shaped rectangles, with positive integer dimensions, have a perimeter equal to their area?

Answer 1

and what are the dimensions

Answer 2

So this? \[
nk = 2(n+k)
\]

Answer 3

yes

Answer 4

Are there any other restrictions on \(n\) and \(k\)? Seems like there are infinite solutions.

Answer 5

nope... only positive integers

Answer 6

thought I was a nice question of the day

and it's high school maths

Answer 7

k cannot be 2
if we write n as a function of k

Answer 8

i got \(6\) and \(3\)

Answer 9

\[nk=2(n+k) \\ nk-2n=2k \\ n(k-2)=2k \\ n=\frac{2k}{k-2} \\ k \neq 2 \\ \text{ pluggin \in k for 2 we get } 2n=2(n+2) \\ 2n=2n+4 \\0 \neq 4 \]

Answer 10

3 and 6 works

Answer 11

k = 2n/ (n - 2)

Answer 12

right so n cannot be 2 either

Answer 13

well 6 and 3 is correct... there is 1 more

Answer 14

i guess you need \(n-2|2n\)

Answer 15

i meant 4 and 4

Answer 16

k=4 . n=4 works

Answer 17

correct... that's it

Answer 18

why only those two are the solutions ?

Answer 19

\[n=\frac{2k}{k-2}=2+\frac{4}{k-2}\]
k-2 has to divide 4
--
3-2 divides 4
4-2 divides 4
5-2 doesn't
6-2 doesn't
7-2 exceeds 4
so there was only 2 right?

Answer 20

because that's what the maths says

\[2l + 2w = lw\]

then

\[w = \frac{2l}{l - 2}\]

then I did this

\[w = \frac{2l - 4}{l -2} + \frac{4}{l - 2}\]

which gives

\[w = 2 + \frac{4}{l - 2}\]

so remembering the dimensions are positive integer values... the only way \[\frac{4}{l -2}\] can be a positive integer is if (l - 2) = 1, 2 or 4

so for l - 2 to be positive and an integer then

l - 2 must have l = 3, 4 or 6

substitute the length values to find the widths...

and so only 2 solutions

hope it makes sense

Answer 21

Nice :)