5
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By using $60$ matchsticks, $5\times5$ matchstick matrix is formed as below:

enter image description here

There are totally $55$ squares including all type of squares (such as $25$ $1\times1$ squares, $16$ $2\times2$ squares etc.)

What is the minimum number of matchsticks needed to be removed not to have a single square left in the shape above?

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4
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I did it by removing

$14$ matchsticks:
![enter image description here

This can probably be improved upon.

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  • $\begingroup$ you are fast as usual :) $\endgroup$ – Oray Aug 22 '17 at 19:53
  • $\begingroup$ I don't think this can be improved. You need to remove one match from the perimeter to break the 5x5 square. This also breaks one unit square. You need at least 12 more matches to break the remaining 24 unit squares, which turns them into dominoes. However, I don't think it is possible arrange those 12 dominoes without a pair of them creating a 2x2 square (with the leftover square on the perimeter). So you need a 14th match to break all squares. $\endgroup$ – Jaap Scherphuis Aug 22 '17 at 20:26

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