A 4-bit combinatorial circuit that outputs 1 when integer greater than 7 and is odd number I need to draw a combinatorial circuit that when an integer is greater than 7 and is an odd number, the output will be 1. There are 4 inputs representing 4 bits and 1 output wire. The output wire is only 1 if an integer is greater than 7 and is an odd number.

My attempt: If I segment the 4-bit input as r x y z where if the input was 8, the 4-bits would be 1 0 0 1. I know that r and z must have an AND gate but cannot figure out a way to get x y to work. I drew a truth table and found that  


1 0 0 1
1 0 1 1
1 1 0 1
1 1 1 1 


must all equal to 1 to satisfy the condition placed.
Any help is appreciated.
You've got the answer!

The number is eight or higher if the first bit is set. The number is odd if the last bit is set. So the answer is simply those two bits ANDed together. The two middle bits don't matter, you don't need to bother with them.