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I don't know if its possible to find out how many pieces are the boarder?

Ruben
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Edwin Barroso
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3 Answers3

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Suppose that you have a rectangle on a grid of squares. Assume the rectangle is $3n$ squares wide and $n$ squares tall. The area would then be $3n^2$. The perimeter would be $8n-4$. For your problem $1000\approx 3 n^2$, so $n\approx \sqrt{333}\approx 18$. Thus I would guess the perimeter would have about $8(18)-4=140$ pieces.

irchans
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  • You cannot have 18 pieces on one side, because that is not a divisor of 1000. – Martin R Jun 26 '19 at 14:40
  • @MartinR You are, of course, absolutely correct. I was just doing an approximation. You might get a better approximation by setting $n=\sqrt{1000/3}\approx 18.2574$ in which case $8n-4\approx 142$ , but $n$ is not an integer. I am thinking that you can often get better approximations by violating assumptions. – irchans Jun 26 '19 at 14:47
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Well if the pieces are different sizes, the best you can do is approximate without any information about the distribution of sizes or other information. Thus, we will assume they are (about) the same size.

To start, we know there are 1000 pieces, and the area of the puzzle is $99\text{ cm} * 33\text{ cm} = 3267 \text{ cm}^2$, or $3.267\text{ cm}^2$ per piece if they're the same size. Assuming the piece are square, this gives $\sqrt{3.267\text{ cm}^2}=1.8\text{ cm}$ to a side. Thus we have $\frac{99\text{ cm}}{1.8\text{ cm}}=55$ pieces on the long side, and $\frac{3\text{ cm}}{1.8\text{ cm}}\approx 18$ pieces on the short side.

So, if your puzzle pieces are all approximately square and approximately the same size, you can expect the puzzle to be 55pcs by 18pcs, for a total perimeter of 142 pieces.

Vedvart1
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    Of course $55 \times 18 \ne 1000$. If it is a rectangular grid, it is likely to be either $50 \times 20$ or $40 \times 25$. – Robert Israel Jun 26 '19 at 14:41
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If the puzzle has exactly $1000$ pieces, and the pieces are rectangularly gridded (i.e. there are identifiable rows and columns of pieces); and if each piece is roughly the same width and height; then it seems reasonable to think that the puzzle is $50$ by $20$ (the factorization of $1000$ closest to a $3:1$ width to length ratio).

This would mean that there are $136$ edge pieces.

paw88789
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  • Thank you guys the answers helped and it's my first time using this site so thank you even if I didn't put the right tag! – Edwin Barroso Jun 28 '19 at 09:44