I have a square cloth with side length $x$ cm, and I am going to cut it into at least $n$ squares with side length $1$ cm for my customer, and also you cannot cut the cloth to thinner pieces (reminded by @risky mysteries). I cannot glue any bits of cloth together. What is the minimum value of $x$?

Problem inspired from a math test problem in my school (about a month ago).

Thanks to @Jaap Scherphuis, I now know this is an unsolved problem. So of course I still haven't solved it. You can use a computer!

  • $\begingroup$ Is folding valid? $\endgroup$ Jul 28, 2020 at 4:04
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    $\begingroup$ @riskymysteries Yes, but you can't say a $1$ by $1.1$ rectangle is a $1$ by $1$ square with folding. You need to cut it. $\endgroup$ Jul 28, 2020 at 4:09
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    $\begingroup$ Just to make this clear, so is this $x$ should be defined in term of $n$ a.k.a $x = f(n)$? $\endgroup$
    – athin
    Jul 28, 2020 at 5:07
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    $\begingroup$ Isn't this an open problem in mathematics? en.wikipedia.org/wiki/Square_packing_in_a_square $\endgroup$ Jul 28, 2020 at 5:51
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    $\begingroup$ This isn't clear: the problem is inspired from a math test problem, but has not been solved by the OP. So there is no definitive answer? This should be closed IMO. $\endgroup$
    – Earlien
    Jul 28, 2020 at 11:14

4 Answers 4



(the ceiling of the square root of n)


If you need n squares which are 1cm wide, you essentially require a piece of cloth with an area of , i.e. a square of length √n.
But given you can't glue pieces together we take the ceiling of √n to ensure we have enough pieces whose length are 1cm


Clearly, $x \geq \sqrt n$, otherwise your original square would have less area than the n smaller squares.

In most cases, you will need $x = \lceil \sqrt n \rceil$, as that will be the smallest square that allows you to cut the necessary pieces.

However, for $n=5$,

You can make do with $x = 2\sqrt2$:
Square with diagonal smaller squares
This wastes 3 units, which is less than would be wasted for a size 3 square.

This does not continue to work for higher $n$. For example, for $x=3\sqrt2$, you could fit 13 squares in a similar pattern, but the slightly smaller $x = 4$ would have room for 16.

So, in conclusion,

you need $x = \lceil \sqrt n \rceil$, except for $n=5$, where $x=2\sqrt 2$ will work.


Is the answer



The cloth can be cut into 2 pieces like this


  • $\begingroup$ No, you cannot cut it to thinner pieces. I will add that. $\endgroup$ Jul 28, 2020 at 3:29
  • $\begingroup$ @CulverKwan Make sure to acknowledge the edit! $\endgroup$ Jul 28, 2020 at 3:30
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    $\begingroup$ Ok, I have acknowledge it! $\endgroup$ Jul 28, 2020 at 3:35

In both the square, area will be the same.

suppose big square has area, A = x*x

all small square total area will be A = n * 1 *1

so, x*x = n * 1 * 1

x = √n, for all x>=n

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    $\begingroup$ How can you cut for example 2 whole squares from a piece of cloth of size $\sqrt{2}\times\sqrt{2}$? $\endgroup$ Jul 28, 2020 at 6:52
  • $\begingroup$ i think solution will work for all x>=n $\endgroup$
    – Amit Huda
    Jul 29, 2020 at 8:41
  • $\begingroup$ You already said that $x=\sqrt{n}$, so how can it be that $x\ge n$? $\endgroup$ Jul 29, 2020 at 8:58

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