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I am trying to figure out how to make a circuit that squares a 3-bit input using 2 3-bit binary adders and logic gates. Could someone point me in the right direction on how to design this? Thanks.

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If you insist using pure logic gates to make a 3 bit binary squarer (even such a thing?), HERE IT IS! 3-bit binary squarer

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  • \$\begingroup\$ If this answers your question, make sure to tick it (underneath the vote counter). If it doesn't answer your question, add a comment under here. \$\endgroup\$ – Bradman175 Apr 13 '16 at 9:35

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