• jeff 👨‍💻@programming.dev
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    3 days ago

    What? No. The divisibility of the side lengths have nothing to do with this.

    The problem is what’s the smallest square that can contain 17 identical squares. If there were 16 squares it would be simply 4x4.

    • bitjunkie@lemmy.world
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      2 days ago

      And the next perfect divisor one that would hold all the ones in the OP pic would be 5x5. 25 > 17, last I checked.