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If x and y are positive integer, what is the remainder when

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If x and y are positive integer, what is the remainder when [#permalink] New post 02 Nov 2009, 07:15
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If x and y are positive integer, what is the remainder when x is divided by y ?

(1) When x is divided by 2x, the remainder is 4.
(2) When x + y is divided by y, the remainder is 4.

OA: B
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Re: Tough DS: please help [#permalink] New post 02 Nov 2009, 09:00
gmat620 wrote:
If x and y are positive integer, what is the remainder when x is divided by y ?

(1) When x is divided by 2x, the remainder is 4.
(2) When x + y is divided by y, the remainder is 4.

OA: B


x=yn+r. Question is r=?

(1) This statement is clearly insufficient, no info about y.
Maybe there is a typo? If it were: "When x is divided by 2y, the remainder is 4." It would make more sense, though still would be insufficient.

But still let's consider this case too:
x=2yk+4. x=20 y=8 --> x/2y=20/16 remainder 4 and x/y=20/8 remainder also 4, BUT x=10 y=3 --> x/2y=10/6 remainder 4 and x/y=10/3 remainder 1. Two different remainders: not sufficient.

(2) x+y=yp+4 --> x=y(p-1)+4 --> this statement directly gives a remainder of 4 upon dividing x by y. Sufficient.

Answer: B.
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Re: Tough DS: please help [#permalink] New post 02 Nov 2009, 11:32
Bunuel wrote:
gmat620 wrote:
If x and y are positive integer, what is the remainder when x is divided by y ?

(1) When x is divided by 2x, the remainder is 4.
(2) When x + y is divided by y, the remainder is 4.

OA: B


x=yn+r. Question is r=?

(1) This statement is clearly insufficient, no info about y.
Maybe there is a typo? If it were: "When x is divided by 2y, the remainder is 4." It would make more sense, though still would be insufficient.

But still let's consider this case too:
x=2yk+4. x=20 y=8 --> x/2y=20/16 remainder 4 and x/y=20/8 remainder also 4, BUT x=10 y=3 --> x/2y=10/6 remainder 4 and x/y=10/3 remainder 1. Two different remainders: not sufficient.

(2) x+y=yp+4 --> x=y(p-1)+4 --> this statement directly gives a remainder of 4 upon dividing x by y. Sufficient.

Answer: B.


Hi Bunuel!,

Look at the red coloured equation above.

x= 2yk + 4
=> x= y(2k) + 4
=> x divided by y has a remainder of 4. Is my understanding correct?

2nd query: For first equation (the red one mentioned above) x= 2yk + 4, you solved it by plugging in the numbers while for second correct equation x= y(p-1) + 4 you didn't solve it by plugging the numbers. Why did you not plug the numbers in equation 2? I mean how did you find out without counter checking that equation 2 will be true for all positive integers???

Thanks in Advance!!!
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Re: Tough DS: please help [#permalink] New post 02 Nov 2009, 12:13
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Hussain15 wrote:
Bunuel wrote:
gmat620 wrote:
If x and y are positive integer, what is the remainder when x is divided by y ?

(1) When x is divided by 2x, the remainder is 4.
(2) When x + y is divided by y, the remainder is 4.

OA: B


x=yn+r. Question is r=?

(1) This statement is clearly insufficient, no info about y.
Maybe there is a typo? If it were: "When x is divided by 2y, the remainder is 4." It would make more sense, though still would be insufficient.

But still let's consider this case too:
x=2yk+4. x=20 y=8 --> x/2y=20/16 remainder 4 and x/y=20/8 remainder also 4, BUT x=10 y=3 --> x/2y=10/6 remainder 4 and x/y=10/3 remainder 1. Two different remainders: not sufficient.

(2) x+y=yp+4 --> x=y(p-1)+4 --> this statement directly gives a remainder of 4 upon dividing x by y. Sufficient.

Answer: B.


Hi Bunuel!,

Look at the red coloured equation above.

x= 2yk + 4
=> x= y(2k) + 4
=> x divided by y has a remainder of 4. Is my understanding correct?

2nd query: For first equation (the red one mentioned above) x= 2yk + 4, you solved it by plugging in the numbers while for second correct equation x= y(p-1) + 4 you didn't solve it by plugging the numbers. Why did you not plug the numbers in equation 2? I mean how did you find out without counter checking that equation 2 will be true for all positive integers???

Thanks in Advance!!!


That's a good 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 (coefficient) can be any integer.

Look at equation A, the coefficient is 2k, 2k is even. It can be rephrased as x divided by y will give the remainder of 4 IF coefficient is even. But what about the cases when coefficient is odd? We don't know that so we must check to determine this.

As for B. Coefficient 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.

Hope it's clear.
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PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. NEW!!!

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set. NEW!!!


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VP
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Status: The last round
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GMAT 1: 680 Q48 V34
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Kudos [?]: 383 [1] , given: 156

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Re: Tough DS: please help [#permalink] New post 02 Nov 2009, 13:16
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Thanks boss!!!

You will score +50 in your test definitely. :)
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Re: Tough DS: please help   [#permalink] 02 Nov 2009, 13:16
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