Note that \(27^{12} = (XY9)^6 = (ABCD1)^4\) will always have a unit digit of 1. So we only need to know the unit digit of \(3^{6q}\).

Also \(3^{6q} = 9^{3q}\) which can only have a unit digit of 9 or 1. This depends on whether q is even/odd, so we only need to know the parity of q for sufficiency.

Statement 1 Alone:

We don't care about the value of p, so this is insufficient.

Statement 2 Alone:

This tells us q is even, so this is sufficient.

Answer: B
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Hi Jerry,
How you got \(27^{12} = (XY9)^6\) ? My rational to get 1 as UD was that \(27^{12} = (7)^{12} = (7)^4 = 1\)
Thanks!

Hi lucasdachequi , thanks for your question!

We split the power of 12 into 2 and 6 first, then the expression would be \(27^{12} = (27^2)^6\). We know \(27^2\) ends with a digit of 9, so we will have something ending in 9 to the power of 6 (hence \(XY9^6\)). The rest follows with the same idea of splitting the powers. I think you split 12 into 3 and 4 and multiplied 27 by 3, but that's not how powers work! It would be \((27^3)^4\) in that case.

A handy trick for these types of questions is remembering every cycle of 4 has the same unit digit no matter which digit is last. Hence \(27^0, 27^4, 27^8, 27^{12}\) all have the same unit digit of 1!
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