(p q)^(r s) is worked out

[sanju09]

What will be the digit at the unit’s place, when (p q)^(r s) is worked out?

(1) p^2 + q^2 = 41 and r s = 4 t for some positive integer t.

(2) p q = 20.



MBM

[ketkoag]

imo : c
coz if i take only B then its (20)^(rs), now rs can be 0 or non zero. so the unit's place differs in both the cases.
but when we take both the statements then pq = 20 and rs is a positive integer so that means that unit's place would alwways be 0.

hence C.. please correct me if i am missing something.

[Brent@GMATPrepNow]

What will be the digit at the unit’s place, when (p q)^(r s) is worked out?

(1) p^2 + q^2 = 41 and r s = 4 t for some positive integer t.

(2) p q = 20.
MBM

Can we assume that p, q, r and s are all integers?
If so, then (1) is sufficient since p^2 + q^2 = 41 tells us that (p,q) is either (4,5), (-4,-5), (4,-5), (5,-4), (-5,-4) and so on.
This means that pq = 20 or -20

"r s = 4 t for some positive integer t" tells us that rs is positive.

So, 20 (or -20) to the power of a positive integer will have zero as its units digit.

The answer is A (if p,q,r and s are integers).

[sanju09]

Same question here Brent! Can we assume?