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If x and y are different positive integers, which of the following COU
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29 Oct 2018, 09:58

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2

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A

B

C

D

E

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85% (hard)

Question Stats:

32% (03:26) correct 68% (01:31) wrong based on 59 sessions

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If x and y are different positive integers, which of the following COULD be true:

i) When x is divided by y, the remainder is 2x ii) When x is divided by 2y, the remainder is y iii) When (2x + y) is divided by (x + y), the remainder is y

A) i only B) ii only C) iii only D) i & ii only E) ii & iii only

ASIDE: Many Integer Properties questions can be solved by identifying values that satisfy some given conditions. This question is intended to strengthen that skill.

If x and y are different positive integers, which of the following COU
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29 Oct 2018, 11:16

1

If x and y are different positive integers, which of the following COULD be true:

i) When x is divided by y, the remainder is 2x Remainder cannot be > than the dividend ... So not possible

ii) When x is divided by 2y, the remainder is y when x =3y or x=5 y or x=7y and so on, this will be true

iii) When (2x + y) is divided by (x + y), the remainder is y 2x+y=x+x+y will leave a remainder of x when divided by x+y.. But it is given as y, this means x=y .. But it is given that x and y are distinct so not possible

If x and y are different positive integers, which of the following COU
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Updated on: 03 Nov 2018, 08:07

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GMATPrepNow wrote:

If x and y are different positive integers, which of the following COULD be true:

i) When x is divided by y, the remainder is 2x ii) When x is divided by 2y, the remainder is y iii) When (2x + y) is divided by (x + y), the remainder is y

A) i only B) ii only C) iii only D) i & ii only E) ii & iii only

ASIDE: Many Integer Properties questions can be solved by identifying values that satisfy some given conditions. This question is intended to strengthen that skill.

i) When x is divided by y, the remainder is 2x The remainder cannot be greater than the dividend (the number we're dividing) For example, it CANNOT be the case that 17 divided by 5 leaves a remainder of 34 Statement i can never be true Check the answer choices. . . . ELIMINATE A and D

ii) When x is divided by 2y, the remainder is y Nice rule: If N divided by D leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc. For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc.

So, some possible values of x are: y, y + 2y, y + 4y, y + 6y, . . . etc Let's TEST the 1st option: x = y. No good. The question tells us that x and y are different

So, let's TEST the 2nd option: x = y + 2y = 3y So, how about x = 15 and y = 5 When we plug those values into statement ii, we get: When 15 is divided by 10, the remainder is 5 PERFECT! Statement ii CAN be true. Check the answer choices. . . . ELIMINATE C

iii) When (2x + y) is divided by (x + y), the remainder is y When we apply the above rule, we get.... Some possible values of (2x + y) are: y, y + (x + y), y + 2(x + y), y + 3(x + y), . . . etc

Let's TEST the 1st option: (2x + y) = y. Solve to get x = 0 No good. We're told x is POSITIVE

Let's TEST the 2nd option: (2x + y) = y + (x + y) Solve to get: x = y No good. The question tells us that x and y are different

Let's TEST the 3rd option: (2x + y) = y + 2(x + y) Solve to get: y = 0 No good. We're told y is POSITIVE

Let's TEST the 4th option: (2x + y) = y + 3(x + y) Solve to get: x = -3y No good. If y is POSITIVE, then that means x is NEGATIVE, but we're told x is POSITIVE

Let's TEST the 5th option: (2x + y) = y + 4(x + y) Solve to get: 2x = -4y No good. If y is POSITIVE, then that means x is NEGATIVE, but we're told x is POSITIVE

At this point, we should recognize that, if we keep going, we'll keep running into the same problem where either x or y is NEGATIVE (which contradicts the given information. So, statement iii can never be true

Answer: B

Cheers, Brent

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Originally posted by GMATPrepNow on 03 Nov 2018, 08:03.
Last edited by GMATPrepNow on 03 Nov 2018, 08:07, edited 1 time in total.

Re: If x and y are different positive integers, which of the following COU
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03 Nov 2018, 08:06

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chetan2u wrote:

If x and y are different positive integers, which of the following COULD be true:

i) When x is divided by y, the remainder is 2x Remainder cannot be > than the dividend ... So not possible

ii) When x is divided by 2y, the remainder is y when x =3y or x=5 y or x=7y and so on, this will be true

iii) When (2x + y) is divided by (x + y), the remainder is y 2x+y=x+x+y will leave a remainder of x when divided by x+y.. But it is given as y, this means x=y .. But it is given that x and y are distinct so not possible

I thought I'd mention that testing only one pair of values for x and y isn't enough to say statement iii cannot be true. If that were the case, we could also conclude that statement ii is not true, because if we test the first possible value of x, we get x = y, which isn't allowed.

If x and y are different positive integers, which of the following COU
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03 Nov 2018, 09:31

GMATPrepNow wrote:

chetan2u wrote:

If x and y are different positive integers, which of the following COULD be true:

i) When x is divided by y, the remainder is 2x Remainder cannot be > than the dividend ... So not possible

ii) When x is divided by 2y, the remainder is y when x =3y or x=5 y or x=7y and so on, this will be true

iii) When (2x + y) is divided by (x + y), the remainder is y 2x+y=x+x+y will leave a remainder of x when divided by x+y.. But it is given as y, this means x=y .. But it is given that x and y are distinct so not possible

I thought I'd mention that testing only one pair of values for x and y isn't enough to say statement iii cannot be true. If that were the case, we could also conclude that statement ii is not true, because if we test the first possible value of x, we get x = y, which isn't allowed.

I am not saying that iii is not possible because of one example, rather I have not taken any examples. But iii is possible only when x=y and I have derived it. I have just rewritten 2x+y as X+X+y. When I divide this by X+y....(X+X+y)/(X+y)=X/(X+y) + (X+y)/(X+y), so clearly remainder is x as X+y /X+y will leave 0 as remainder... But choice iii mentions y as a remainder in this case, so we derived X and it is given en it is equal to y... So x=y, but both are distinct so not possible.

So I didn't take any examples, but proved it that X has to be equal to y for iii to be true
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Re: If x and y are different positive integers, which of the following COU
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03 Nov 2018, 09:53

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chetan2u wrote:

I am not saying that iii is not possible because of one example, rather I have not taken any examples. But iii is possible only when x=y and I have derived it. I have just rewritten 2x+y as X+X+y. When I divide this by X+y....(X+X+y)/(X+y)=X/(X+y) + (X+y)/(X+y), so clearly remainder is x as X+y /X+y will leave 0 as remainder... But choice iii mentions y as a remainder in this case, so we derived X and it is given en it is equal to y... So x=y, but both are distinct so not possible.

So I didn't take any examples, but proved it that X has to be equal to y for iii to be true

You're absolutely right! I misread your solution. Sorry about that.

Re: If x and y are different positive integers, which of the following COU
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03 Nov 2018, 17:11

GMATPrepNow wrote:

If x and y are different positive integers, which of the following COULD be true:

i) When x is divided by y, the remainder is 2x ii) When x is divided by 2y, the remainder is y iii) When (2x + y) is divided by (x + y), the remainder is y

A) i only B) ii only C) iii only D) i & ii only E) ii & iii only

Since the remainder can’t be greater than the dividend or the divisor, we see that statement (i) is not true.

If x = 9 and y = 3, we see that the remainder is 3 when 9 is divided by 6. So statement (ii) could be true.

Since (2x + y)/(x + y) = (x + y + x)/(x + y) = (x + y)/(x + y) + x/(x + y) = 1 + x/(x + y), we see that the remainder must be x. However, since x and y are different positive integers, then the remainder can’t be y. We see that statement (iii) is not true.

Answer: B
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