Remainder with 5 integers

[karthikpandian19]

Let r s m n and q be positive integers. Is rs the remainder when mn is divided by q?

(1) m divided by q leaves a remainder of r

(2) n divided by q leaves a remainder of s

[knight247]

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

(1) m = q(z) + r ....Where z is the quotient
This statement is insufficient as it gives no info about n and s
(2) n = q (x) + s ....Where x is the quotient
This statement is insufficient as it gives no info about m and r


Using the principal mentioned above, If m divided by q leaves a remainder of r and n divided by q leaves a remainder of s, then m*n divided by q will leave a remainder of r*s. Hence C

[ArunangsuSahu]

Ans is (E)

(A) not Sufficient
(B) not Sufficient

Let's combine both
Case I:
m=7 q=5 implies r=2
n=9 q=5 implies s=4

So mn=63, rs=8, q=5 but mn%q=3. So rs not equal to Remainder of mn divided by 5 but will be rqual if rs is dived by 5 as rs > 5

Case II:
m=4 q=3 r=1
n=2 q=3 r=2
So mn=8, rs=2, q=3 and mn%3=2 so it conforms

But we don't have a definite solution

So E is the answer

So mn=

[chufus]

Just saw the flaw in my approach so removed the solution....... It was wrong and don't want to misguide anyone...

[Ian Stewart]

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]

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.

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

[GmatMathPro]

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:

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

[ArunangsuSahu]

See my answer in the 2nd post

[karthikpandian19]

OA is E

[chufus]

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:

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

Yea bang on.. I guess that is what i meant but just conveyed it in the wrong manner. Thanks for pointing the subtle difference out.