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## Comment on

What Must Be True?## Regardless of p, assuming q

q-1 or q-2: one of the two has to be even.

Thus, the result must be even.

Is this correct?

## That is correct, devroy.

That is correct, devroy. Great work.

## Hi, quick question; the

## I'm just plugging in 2 for an

I'm just plugging in 2 for an even value and 1 for an odd value, and then performing the necessary calculations.

HOWEVER, once I see that one of the pieces in the product equals zero, I immediately know that the final result will equal zero.

For example, when we plug in 2 for p and q in the first expression, we get 2(2-2)(2-1)

Once I recognize that 2-2 = 0, then we have 2(2-2)(2-1) = 2(ZERO)(2-1), which means the final product will be zero.

## understood. thanks :)

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