Data Sufficiency

Ian Stewart wrote:

knight247 wrote:One Important principal u may wanna keep in mind.

Suppose a number A*B divided by a number x leaves a certain remainder, that remainder is equal to the product of remainders produced when A is divided by x and B is divided by x individually.

Variation....If A divided by x leaves a certain remainder 'm' and B divided by x leaves a certain remainder 'n', then when A*B is divided by x the remainder is the product of m*n

These principles aren't quite right. Say you have two numbers, and the remainder is 5 when you divide the first number by 7, and the remainder is 2 when you divide the second number by 7. If you multiply these two numbers, the remainder will not be 5*2 = 10 when you divide the product by 7, since 10 is too large - when you divide anything by 7, the remainder must be between 0 and 6 inclusive. Instead you need, as the final step, to take the remainder when you divide 10 by 7; the remainder is 3 when you divide this product by 7.

So in the question in the original post, we're dividing by q. We can easily get a 'yes' answer to the question, using both statements, by letting r and s be 0, say, or 1. But if r*s is bigger than q, which can easily happen, then r*s can never be a remainder when you divide by q, since when you divide by q, the remainder must be smaller than q. So the answer is E.

chufus wrote:
Stewart is absolutely correct. I misinterpreted my solution right at the end. With the algebraic solution to this, the remainder comes to (rs)/q which means the remainder could still be anything. The only way remainder is rs is when q is 1 and that is not a precondition with the question. Do the algebra, it will actually show you very quickly what's wrong with assumptions defined by knight247

knight247 wrote: Variation....If A divided by x leaves a certain remainder 'm' and B divided by x leaves a certain remainder 'n', then when A*B is divided by x the remainder is equal to the remainder left when the product of m*n is divided by x

GmatMathPro wrote:

chufus wrote:
Stewart is absolutely correct. I misinterpreted my solution right at the end. With the algebraic solution to this, the remainder comes to (rs)/q which means the remainder could still be anything. The only way remainder is rs is when q is 1 and that is not a precondition with the question. Do the algebra, it will actually show you very quickly what's wrong with assumptions defined by knight247

Not exactly. The remainder is not rs/q. The remainder is the same as the remainder you would get if you divided rs by q. Make sure you see the difference.

Thus, it is not true that q has to be 1 for the remainder to be rs. The remainder will always be rs as long as rs is less than q. For example, 11 divided by 9 leaves a remainder of 2. 12 divided by 9 leaves a remainder of 3. 12*11=132, which leaves a remainder of 6 when divided by 9, which is the same as 2*3, the product of the remainders of 12 and 11 divided by 9. If we tried to use "the remainder is rs/q", we would get a remainder of 6/9 or 2/3, which clearly does not make sense.

knight247's principle can be salvaged with a slight modification:

knight247 wrote: Variation....If A divided by x leaves a certain remainder 'm' and B divided by x leaves a certain remainder 'n', then when A*B is divided by x the remainder is equal to the remainder left when the product of m*n is divided by x