# Design a combinational circuit that has an output of 1 if the binary value is even? [closed]

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?

• Binary value of what is even? Do you realize that the low bit of a binary number indicates its oddness? – Olin Lathrop Jan 2 '18 at 12:08
• Sorry for not being clear enough. The output is 1 when the binary value of the inputs is an even number. – Crazy Jan 2 '18 at 12:17
• 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... – user105652 Jan 5 '18 at 2:49
• It is z right? because $$2^n$$ is an even number for any positive integer. $$z.(2^0)=1 \iff z=1$$ – Crazy Jan 5 '18 at 6:35

• 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. – Crazy Jan 2 '18 at 12:30