kevincan wrote:

ps_dahiya wrote:

B

St1: Clearly INSUFF

St2:

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?

Oops...

Answer should be D.

St1:

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...