I am struggling with a concept on remainder problems regarding the quotient about what form it can or cannot take. For example lets look at one example problem:
What is the remainder when B is divided by 6 if B is a positive integer?
(1) When B is divided by 18, the remainder is 3
(2) When B is divided by 12, the remainder is 9
Answer::
What is the remainder when B is divided by 6 if B is a positive integer?
(1) When B is divided by 18, the remainder is 3 --> B=18q+3=6*(3q)+3, thus B divided by 6 yields the remainder of 3. Sufficient.
(2) When B is divided by 12, the remainder is 9 --> B=12p+9=6*(2p+1)+3, thus B divided by 6 yields the remainder of 3. Sufficient.
Answer: D.
Notice how in (2) the quiotent=2p+1, is an ODD number. Yet based on this logic we were able to deduce that anytime we divided B by 6, the remainder would always be 3. Easy enough. No issues so far.
Lets look at the second example highlighting the main issue:
If x and y are positive integers, what is the remainder when x is divided by y?
(1) When x is divided by 2y, the remainder is 4
(2) When x + y is divided by y, the remainder is 4
Positive integer x divided by positive integer y yields remainder of r can be expressed as x=yq+r. Question is r=?
(1) When x is divided by 2y, the remainder is 4. If x=20 and y=8 (satisfies the given statemnet as 20 divided by 2*8=16 yields reminder of 4), then x divided by y yields r=4 (20 divided by 8 yields remainder of 4) BUT if x=10 and y=3 (satisfies the given statemnet as 10 divided by 2*3=6 yields reminder of 4), then x divided by y yields r=1 (10 divided by 3 yields remainder of 1). Two different answers. Not sufficient.
(2) When x + y is divided by y, the remainder is 4 --> x+y=yp+4 --> x=y(p-1)+4 (x is 4 more than multiple of y)--> this statement directly tells us that x divided by y yields remainder of 4. Sufficient.
Answer: B.
Hope it's clear.
Bunuel also elaborated on the question with the following response to another forum members question:
A. x=y(2k)+4, k any integer >=0.
B. x=y(p-1)+4, p any integer >=0.
Why A is not sufficient to determine the remainder and B is? Why did I use number plugging to show this in the first case and didn't in the second?
If we are told that x divided by y gives a remainder of 4, means x=yp+4 where p is integer >=0. We don't know x and y so p (quotient) can be any integer.
Look at equation A, the quotient is 2k, 2k is even. It can be rephrased as x divided by y will give the remainder of 4 IF quotient is even. But what about the cases when quotient is odd? We don't know that so we must check to determine this.
As for B. Quotient here is (p-1), which for integer values of p can give us ANY value: any even as well as any odd. So basically x=y(p-1)+4 is the same as x=yp+4. No need for double checking.
Can someone explain why the same logic doesn't work on both the questions?? In the first problemthe quotient was odd yet we were able to deduce that B divided by 6 will always yield a remainder of 3. Yet in the second problem. The quotient was 2k, even quotient , but we were forbidden to make any more conclusions because it conditioned the scenarios such that the quotient was even. Hence we were unable to concluded that 4 was the remainder in the equation x=y(2k)+4, k any integer >=0.
What am I not understanding??![]()
What is the remainder when B is divided by 6 if B is a positive integer?
(1) When B is divided by 18, the remainder is 3
(2) When B is divided by 12, the remainder is 9
Answer::
What is the remainder when B is divided by 6 if B is a positive integer?
(1) When B is divided by 18, the remainder is 3 --> B=18q+3=6*(3q)+3, thus B divided by 6 yields the remainder of 3. Sufficient.
(2) When B is divided by 12, the remainder is 9 --> B=12p+9=6*(2p+1)+3, thus B divided by 6 yields the remainder of 3. Sufficient.
Answer: D.
Notice how in (2) the quiotent=2p+1, is an ODD number. Yet based on this logic we were able to deduce that anytime we divided B by 6, the remainder would always be 3. Easy enough. No issues so far.
Lets look at the second example highlighting the main issue:
If x and y are positive integers, what is the remainder when x is divided by y?
(1) When x is divided by 2y, the remainder is 4
(2) When x + y is divided by y, the remainder is 4
Positive integer x divided by positive integer y yields remainder of r can be expressed as x=yq+r. Question is r=?
(1) When x is divided by 2y, the remainder is 4. If x=20 and y=8 (satisfies the given statemnet as 20 divided by 2*8=16 yields reminder of 4), then x divided by y yields r=4 (20 divided by 8 yields remainder of 4) BUT if x=10 and y=3 (satisfies the given statemnet as 10 divided by 2*3=6 yields reminder of 4), then x divided by y yields r=1 (10 divided by 3 yields remainder of 1). Two different answers. Not sufficient.
(2) When x + y is divided by y, the remainder is 4 --> x+y=yp+4 --> x=y(p-1)+4 (x is 4 more than multiple of y)--> this statement directly tells us that x divided by y yields remainder of 4. Sufficient.
Answer: B.
Hope it's clear.
Bunuel also elaborated on the question with the following response to another forum members question:
A. x=y(2k)+4, k any integer >=0.
B. x=y(p-1)+4, p any integer >=0.
Why A is not sufficient to determine the remainder and B is? Why did I use number plugging to show this in the first case and didn't in the second?
If we are told that x divided by y gives a remainder of 4, means x=yp+4 where p is integer >=0. We don't know x and y so p (quotient) can be any integer.
Look at equation A, the quotient is 2k, 2k is even. It can be rephrased as x divided by y will give the remainder of 4 IF quotient is even. But what about the cases when quotient is odd? We don't know that so we must check to determine this.
As for B. Quotient here is (p-1), which for integer values of p can give us ANY value: any even as well as any odd. So basically x=y(p-1)+4 is the same as x=yp+4. No need for double checking.
Can someone explain why the same logic doesn't work on both the questions?? In the first problemthe quotient was odd yet we were able to deduce that B divided by 6 will always yield a remainder of 3. Yet in the second problem. The quotient was 2k, even quotient , but we were forbidden to make any more conclusions because it conditioned the scenarios such that the quotient was even. Hence we were unable to concluded that 4 was the remainder in the equation x=y(2k)+4, k any integer >=0.
What am I not understanding??