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M28-39

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16 Sep 2014, 00:31
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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 $$q$$.
[Reveal] Spoiler: OA

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16 Sep 2014, 00:31
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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.

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16 Sep 2014, 21:23
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, 01:10
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, 08:11
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, 18:36
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, 10:13
I think this is a high-quality question and I agree with explanation.
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02 Jan 2017, 05:00
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, 05:07
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, 10:25
Good Question
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07 Apr 2017, 20:50
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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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08 Apr 2017, 03:25
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, 04:08
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, 04:37
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.

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08 Oct 2017, 04:02
This is a high quality question.

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Re: M28-39   [#permalink] 08 Oct 2017, 04:02
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