St1: Clearly INSUFF
If p-q = 10
pq-(p+q) can be written as q^2+8q-10
or (q^2 - 1) + (8q-9)
8q-9 will give a remainder of 3 when divided by 4
now lets see q^2 - 1
q^2 - 1 = (q-1) (q+1)
since q is odd both (q-1) and (q+1) are divisible by 2 hence it is divisible by 4.
Final remainder = 3: SUFF
Why is (1) clearly insufficient? Shouldn't you know me better by now?
Answer should be D.
p = 10x +1
q = 10y+1
where x and y are positive integers
Now pq-(p+q) can be written as
(10x+1) (10y+1) - 10x-1-10y-1
100xy +10x+10y +1 - 10x-1-10y-1
100xy - 1 so the remainder will always be 3: SUFF
A little out of practice...