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There are three variables x,y,z respectively.

\begin{array}{|c|c|c|c|} \hline x & y & z & F\\ \hline 0 & 0 & 0 & 1 \\ \hline 0 & 0 & 1 & 0 \\ \hline 0 & 1 & 0 & 1 \\ \hline 0 & 1 & 1 & 0 \\ \hline 1 & 0 & 0 & 1 \\ \hline 1 & 0 & 1 & 0 \\ \hline 1 & 1 & 0 & 1 \\ \hline 1 & 1 & 1 & 0 \\ \hline \end{array}

Is my table correct for even binary digit? My book is suggesting.

\begin{array}{|c|c|c|c|} \hline x & y & z & F\\ \hline 0 & 0 & 0 & 0 \\ \hline 0 & 0 & 1 & 1 \\ \hline 0 & 1 & 0 & 0 \\ \hline 0 & 1 & 1 & 1 \\ \hline 1 & 0 & 0 & 0 \\ \hline 1 & 0 & 1 & 1 \\ \hline 1 & 1 & 0 & 0 \\ \hline 1 & 1 & 1 & 1 \\ \hline \end{array}

I am quite confused?

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    \$\begingroup\$ Binary value of what is even? Do you realize that the low bit of a binary number indicates its oddness? \$\endgroup\$ – Olin Lathrop Jan 2 '18 at 12:08
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    \$\begingroup\$ Sorry for not being clear enough. The output is 1 when the binary value of the inputs is an even number. \$\endgroup\$ – Crazy Jan 2 '18 at 12:17
  • \$\begingroup\$ I think you missed the boat, as well as Olin's comment. Look a x,y and z and think about which one determines an odd or even output... \$\endgroup\$ – user105652 Jan 5 '18 at 2:49
  • \$\begingroup\$ It is z right? because $$2^n$$ is an even number for any positive integer. $$z.(2^0)=1 \iff z=1$$ \$\endgroup\$ – Crazy Jan 5 '18 at 6:35
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You are apparently given three binary signals, X, Y, and Z. These three together are taken as a 3-bit binary value, with X the MSB and Z the LSB. Your assignment is to make a circuit that produces a 1 when this input number is even.

Stop and actually think about it. What does evenness mean? Since you only have 23 = 8 possible input cases, write them all down. For each, show the decimal value of the number, and whether it is even or not. Then show the output your circuit is supposed to produce.

Do you see a simple pattern? Look closely at the Z input, and compare that to your desired output.

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    \$\begingroup\$ Oh! Since $$z$$ is the least significant digit and always has a value of $1(2^0)$. It implies ODD. Whereas $$x=(2^2)$$ and $$y=(2^1)$$. X and Y is always even no matter what. So when $$z=1$$ we have the ODD input. \$\endgroup\$ – Crazy Jan 2 '18 at 12:30

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