Veritas Prep Official Solution:
Simplifying the given expression, the question really provides that 2(xyz)
ends in a 2. And finding the pattern for exponential units digits, one can see that the cycle goes:
\(2^1=2\)
\(2^2=4\)
\(2^3=8\)
\(2^4=16(ends−in−6)\)
\(2^5=32(ends−in−2)\)
And because the ending digit of 2 is the first repeating digit, it is known that every fourth exponent ends in 6, and that every 4th-plus-one exponent will yield a units digit of 2. So the exponent xyz must be odd, meaning that the each of x, y, and z must be odd, for if any one term was even the entire term would be even.
That eliminates A and C (and guarantees B). D is wrong because, whatever the odd value (it's 17) that yields 2x=131072
, y and z could each be 1 leaving all the value to come from x, so the values must not yield a prime product. And, similarly, E is incorrect because both y and z could be -1, still allowing for xyz to be 17 (which you don't need to know - you can just know that all the positive value could come from one term).