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I really need help because I need to come up with a circuit with 3 inputs (input A, input B and input C) and a truth table where in every case except for when input A and input C are 1 the output is 1. So basically if input A and input C are both 1, the output is 0. In all the other cases, the output is 1. I'm also wondering if that's even possible to make.

Table 1. Truth table for OP to edit.

A  B  C   Out
-------------
0  0  0   1
1  0  0   1
0  1  0   1
1  0  1   0
0  1  1   1
1  1  0   1
0  0  1   1
1  1  1   1
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    \$\begingroup\$ If the output doesn't depend on input B, you only need a two input logic gate. You can just have the B input connected to nothing inside your circuit. \$\endgroup\$ – The Photon Feb 28 '20 at 16:29
  • \$\begingroup\$ Try starting with the truth table first. Once you have a fully filled out truth table, the circuit may be easier to think about. \$\endgroup\$ – John M Feb 28 '20 at 16:38
  • \$\begingroup\$ @Maria: I've started the truth table for you. Hit the edit link below your question and complete it, filling in the output column. \$\endgroup\$ – Transistor Feb 28 '20 at 16:39
  • \$\begingroup\$ I have added the homework tag to help potential folks trying to answer avoid giving away the complete solution. \$\endgroup\$ – Michael Karas Feb 28 '20 at 16:50
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    \$\begingroup\$ 101 and 111 both result in output = 0? B = don't care? Then 2 input NAND. I don't understand where you are having difficulty. \$\endgroup\$ – CrossRoads Feb 28 '20 at 17:47
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schematic

simulate this circuit – Schematic created using CircuitLab

schematic

simulate this circuit

NAND

A  C  Out
0  0  1
0  1  1
1  0  1
1  1  0

OR 
A/C  B  Out
0    0  0    So A = C = 1, B = 0, Output = 0
0    1  1
1    0  1
1    1  1
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The answer seems simple. Just a 3 input NAND gate with the B input inverted.

Why does this work? Because it means that in any case except for that 1 0 1 case, it will be on.

Simulatorio: https://simulator.io/board/KeuHkS0cwt/1

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