What is the area of the smallest rectangle that can fit 10 rectangles with areas 1 to 10, inclusive? Rectangles must have integer sides and cannot overlap.


2 Answers 2


This answer seems a bit too trivial to be the intended one, but:

The answer is 55.

The minimum area must be 1+2+3+4+5+6+7+8+9+10=55

This can be done by taking rectangles of dimensions 1$\times$1, 2$\times$1,..., 10$\times$1 and arranging them in a straight line to get a rectangle with dimensions 55$\times$1

  • 2
    $\begingroup$ Rot13(5k11 vf nyfb rkgerzryl fvzcyr (ebjf bs 10+1, 9+2, 8+3 rgp)) $\endgroup$
    – Jafe
    Jan 27 at 3:48
  • $\begingroup$ Oh that's right, then it's just a poorly framed question $\endgroup$ Jan 27 at 3:52
  • $\begingroup$ of course you are right. I completely stuffed up this question. It should have been a square... $\endgroup$ Jan 27 at 4:28

One example of smallest perimeter rectangle

Rectangle 8x7, with one empty 1x1 square, which is the nerest to an 8x8 square

enter image description here

  • $\begingroup$ that's very nice! This is the question I should have asked. $\endgroup$ Jan 27 at 12:04

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