Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

enter image description hereI have a square of length $2n^2$, could it be possible to fill it by small small squares triplets? well, I am not able to guess how to proceed. please help

share|cite|improve this question
How do you define "small small" squares triplets? It doesn't immediately suggest a clear idea to me. – hardmath Oct 2 '12 at 11:27
okay imagine a chess board and hope you know how the path of horse is ;) I mean that kind of triplet box – Un Chien Andalou Oct 2 '12 at 11:28
But there are four squares involved in a knight's move... – Dennis Gulko Oct 2 '12 at 11:33
ooops sorry then omit one box – Un Chien Andalou Oct 2 '12 at 11:35
up vote 2 down vote accepted

With length $2n^2$, the area is $4n^4$. Your triplet has three boxes, so a necessary condition is that $n$ is divisible by $3$.

To show that $n$ divisible by 3 is sufficient: join two triplets together to form a $2\times 3$ rectangle. If $n$ is divisible by $3$, we can line $2n^2/3$ rectangles end to end to form a row that is $2\times 2n^2$ in size. Then stacking $n^2$ of them you get the square.

share|cite|improve this answer
"that is $2\times 2n^2$ in size" how? and what do u mean by size here? could you tell me just? – Un Chien Andalou Oct 2 '12 at 11:57
Eh.. $3 \times (2n^2 / 3) = 2n^2$? So if you line up a bunch ($2n^2 /3$ many) of boxes that are 2 units tall and 3 units wide, the entire row will be 2 units tall and $2n^2$ units wide. – Willie Wong Oct 2 '12 at 11:59
got it.. thank you – Un Chien Andalou Oct 2 '12 at 12:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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