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# M28-39

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 01:31
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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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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.

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16 Sep 2014, 22:23
Shouldn't x and y be interchanged in the answer explanation - $y =divisor*quotient+remainder= xq + r$ considering question asks when x is divided by y, i.e. x is dividend and y is divisor?
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19 Sep 2014, 02:10
Dienekes wrote:
Shouldn't x and y be interchanged in the answer explanation - $y =divisor*quotient+remainder= xq + r$ 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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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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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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21 Jan 2016, 11:13
I think this is a high-quality question and I agree with explanation.
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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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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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05 Jan 2017, 11:25
Good Question
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07 Apr 2017, 21:50
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?
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08 Apr 2017, 04:25
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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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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13 Jul 2017, 05:37
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.

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

Posted from my mobile device
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09 Oct 2018, 00:03
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.

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30 Nov 2018, 09:32
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.

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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# M28-39

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