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Math Expert V
Joined: 02 Sep 2009
Posts: 55271
M28-39  [#permalink]

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Difficulty:   85% (hard)

Question Stats: 41% (01:28) correct 59% (01:29) wrong based on 211 sessions

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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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Math Expert V
Joined: 02 Sep 2009
Posts: 55271
Re M28-39  [#permalink]

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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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Shouldn't x and y be interchanged in the answer explanation - considering question asks when x is divided by y, i.e. x is dividend and y is divisor?
Math Expert V
Joined: 02 Sep 2009
Posts: 55271
Re: M28-39  [#permalink]

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Dienekes wrote:
Shouldn't x and y be interchanged in the answer explanation - considering question asks when x is divided by y, i.e. x is dividend and y is divisor?

Yes. I'll edit it to avoid confusion.
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GMAT 1: 610 Q46 V28 GMAT 2: 690 Q47 V38 GMAT 3: 710 Q49 V36 Re: M28-39  [#permalink]

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Yes. I'll edit it to avoid confusion.[/quote]

This is still incorrect. X and Y have not been switched.
Intern  Joined: 05 Aug 2015
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Re: M28-39  [#permalink]

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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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I think this is a high-quality question and I agree with explanation.
Manager  S
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Re: M28-39  [#permalink]

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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.
Math Expert V
Joined: 02 Sep 2009
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Re: M28-39  [#permalink]

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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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Re: M28-39  [#permalink]

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Good Question
Intern  S
Joined: 26 Feb 2017
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Re: M28-39  [#permalink]

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1
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?
Math Expert V
Joined: 02 Sep 2009
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Re: M28-39  [#permalink]

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vitorpteixeira wrote:
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?

______________
Edited. Thank you.
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Re M28-39  [#permalink]

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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.
Math Expert V
Joined: 02 Sep 2009
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Re: M28-39  [#permalink]

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1
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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Re: M28-39  [#permalink]

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This is a high quality question.

Posted from my mobile device
Director  V
Joined: 12 Feb 2015
Posts: 840
Re: M28-39  [#permalink]

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The key to answering this question is to read carefully. St (2) is insufficient by itelf if you read it carefully. You understand the importance of statement (1) once you realise that y needs to be a 2 digit number, i.e. greater than 9 so as to leave a definite remainder of 9.

Correct answer is C.
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Joined: 17 Jun 2017
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Re: M28-39  [#permalink]

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Bunuel wrote:
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

Hi ,
I'm still not able to understand this.

What will be dividend in this then ? If we are taking C both the statements ??? Re: M28-39   [#permalink] 30 Nov 2018, 09:32
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