I was asked to make the simplest possible circuit to solve this problem: for an entry of 0000 to 1001 (0 to 9), detect the numbers that are divisible by 2 or 3. I decided to write the truth table with an output of 1 for functions that are NOT divisible by either 2 or 3, since that seemed like less trouble to me. I plotted the karnaugh map and solved as usual. However, I put an inverter on the output of the last gate since the question was asking to find the numbers that ARE divisible by 2 or 3. Is my solution still in the simplest possible form? If not, why not?