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M28-39

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M28-39 [#permalink]

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If \(x\) and \(y\) are both positive integers and \(x \gt y\), what the remainder when \(x\) is divided by \(y\)?


(1) \(y\) is a two-digit prime number.

(2) \(x=qy+9\), for some positive integer \(q\).

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Re M28-39 [#permalink]

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New post 16 Sep 2014, 01:31
Official Solution:


If \(x\) and \(y\) are positive integers, there exist unique integers \(q\) and \(r\), called the quotient and remainder, respectively, such that \(x =divisor*quotient+remainder= yq + r\) and \(0 \leq r \lt y\).

(1) \(y\) is a two-digit prime number. Clearly insufficient since we know nothinf about \(x\).

(2) \(x=qy+9\), for some positive integer \(q\). It's tempting to say that this statement is sufficient and \(r=9\), since given equation is very similar to \(x =divisor*quotient+remainder= yq + r\) . But we don't know whether \(y \gt 9\)?: remainder must be less than divisor.

For example:

If \(x=10\) and \(y=1\) then \(10=1*1+9\), then the remainder upon division 10 by 1 is zero.

If \(x=11\) and \(y=2\) then \(11=1*2+9\), then the remainder upon division 11 by 2 is one.

Not sufficient.

(1)+(2) From (2) we have that \(x=qy+9\) and from (1) that \(y\) is more than 9 (since it's a two-digit number), so we have direct formula of remainder, as given above. Sufficient.


Answer: C
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Re: M28-39 [#permalink]

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New post 16 Sep 2014, 22:23
Shouldn't x and y be interchanged in the answer explanation - Image considering question asks when x is divided by y, i.e. x is dividend and y is divisor?
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Re: M28-39 [#permalink]

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Re: M28-39 [#permalink]

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New post 28 Dec 2015, 09:11
Yes. I'll edit it to avoid confusion.[/quote]

This is still incorrect. X and Y have not been switched.
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Re: M28-39 [#permalink]

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New post 18 Jan 2016, 19:36
msalvi wrote:
Yes. I'll edit it to avoid confusion.


This is still incorrect. X and Y have not been switched.[/quote]

yup, very confusing -- still not changed. thanks.
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Re: M28-39 [#permalink]

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New post 21 Jan 2016, 11:13
I think this is a high-quality question and I agree with explanation.
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Re: M28-39 [#permalink]

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New post 02 Jan 2017, 06:00
Hi All, I don't understand the explanation.Taking statement 1 and 2 together and setting q to 1 we get x=y+9, which gives multiple remainders. Please advise, thanks.
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Re: M28-39 [#permalink]

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New post 02 Jan 2017, 06:07
Cez005 wrote:
Hi All, I don't understand the explanation.Taking statement 1 and 2 together and setting q to 1 we get x=y+9, which gives multiple remainders. Please advise, thanks.


No it won't. Plug there two-digit prime number for y and see what you get.
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Collection of Questions:
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Re: M28-39 [#permalink]

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New post 05 Jan 2017, 11:25
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Re: M28-39 [#permalink]

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New post 07 Apr 2017, 21:50
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Shouldn't x and y be interchanged in the answer explanation - Image considering question asks when x is divided by y, i.e. x is dividend and y is divisor?
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Re: M28-39 [#permalink]

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Re M28-39 [#permalink]

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New post 13 Jul 2017, 05:08
I think this is a poor-quality question and I don't agree with the explanation. When you say "We don't know whether y>9?" Could you
please elaborate. I have never seen that when divisor is 2 and dividend=11, the quotient can be 1.
The definition of a remainder as stated is r<y.
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Re: M28-39 [#permalink]

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New post 13 Jul 2017, 05:37
vil1 wrote:
I think this is a poor-quality question and I don't agree with the explanation. When you say "We don't know whether y>9?" Could you
please elaborate. I have never seen that when divisor is 2 and dividend=11, the quotient can be 1.
The definition of a remainder as stated is r<y.


In (2) we are simply given that \(x=qy+9\). We are not given that when "x is divided by y the remainder is 9", we are just given that \(x=qy+9\). x, q, and y can be any set of positive integers satisfying this equation.

Next, consider the examples given in the solution:
If \(x=10\) and \(y=1\) then \(10=1*1+9\). In this case the remainder upon division x = 10 by y = 1 is zero.

If \(x=11\) and \(y=2\) then \(11=1*2+9\). In this case then the remainder upon division x = 11 by y = 2 is one.

Two different answers. Not sufficient.
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Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

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. ,11 Mixed Questions, 12 Fresh Meat

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., 11 New DS set.


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Re: M28-39 [#permalink]

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New post 08 Oct 2017, 05:02
This is a high quality question.

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Re: M28-39   [#permalink] 08 Oct 2017, 05:02
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