When you make the table, you'll realize that all of the numbers after being multiplied are either \( 0 \text{ (mod 6)}\) or \( 3 \text{ (mod 6)}\), so actually none of the numbers are \( 1 \text{ (mod 6)}\) The reason why this doesn't work is because \(3\) is a factor of \(6\)! And actually, you can remake this problem with any even number and you'll find that there are a lot of exceptions.

BUT, what you said is actually **really useful!!** Number theory often involves prime numbers, so everything you figured out will save you a lot of time when actually doing problems. Way to go for understanding everything!