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# M16-35

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Math Expert
Joined: 02 Sep 2009
Posts: 42281

Kudos [?]: 132984 [0], given: 12400

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16 Sep 2014, 01:00
Expert's post
25
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BOOKMARKED
00:00

Difficulty:

85% (hard)

Question Stats:

60% (01:44) correct 40% (02:15) wrong based on 128 sessions

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If $$x$$, $$a$$, and $$b$$ are positive integers such that when $$x$$ is divided by $$a$$, the remainder is $$b$$ and when $$x$$ is divided by $$b$$, the remainder is $$a-2$$, then which of the following must be true?

A. $$a$$ is even
B. $$x+b$$ is divisible by $$a$$
C. $$x-1$$ is divisible by $$a$$
D. $$b=a-1$$
E. $$a+2=b+1$$
[Reveal] Spoiler: OA

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Math Expert
Joined: 02 Sep 2009
Posts: 42281

Kudos [?]: 132984 [4], given: 12400

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16 Sep 2014, 01:00
4
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Expert's post
5
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Official Solution:

If $$x$$, $$a$$, and $$b$$ are positive integers such that when $$x$$ is divided by $$a$$, the remainder is $$b$$ and when $$x$$ is divided by $$b$$, the remainder is $$a-2$$, then which of the following must be true?

A. $$a$$ is even
B. $$x+b$$ is divisible by $$a$$
C. $$x-1$$ is divisible by $$a$$
D. $$b=a-1$$
E. $$a+2=b+1$$

When $$x$$ is divided by $$a$$, the remainder is $$b$$: $$x=aq+b$$ and $$remainder=b \lt a=divisor$$ (remainder must be less than divisor);

When $$x$$ is divided by $$b$$, the remainder is $$a-2$$: $$x=bp+(a-2)$$ and $$remainder=(a-2) \lt b=divisor$$.

So we have that: $$a-2 \lt b \lt a$$, as $$a$$ and $$b$$ are integers, then it must be true that $$b=a-1$$ (there is only one integer between $$a-2$$ and $$a$$, which is $$a-1$$ and we are told that this integer is $$b$$, hence $$b=a-1$$).

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Manager
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19 Feb 2016, 15:49
1
KUDOS
Slightly different approach:

x=aq+b
x=bq+a-2

aq+b=bq+a-2
factor out q's
re-arranging gives: 2b=2a-2
Divide by 2
b=a-1
Ans D

Kudos [?]: 28 [1], given: 3

Manager
Joined: 12 Nov 2015
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Location: Uruguay
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08 Mar 2016, 14:41
Would it work to pick numbers?

Kudos [?]: 25 [0], given: 126

Math Expert
Joined: 02 Sep 2009
Posts: 42281

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09 Mar 2016, 11:25
Avigano wrote:
Would it work to pick numbers?

Considering 3 variables and constraints, picking numbers won't be the best way for this problem.
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28 Mar 2016, 18:19
I think this is a high-quality question and I agree with explanation.

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Manager
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20 Jun 2016, 16:55
I think this is a high-quality question and I agree with explanation.
_________________

Appreciate any KUDOS given !

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05 Jul 2016, 01:02
Slightly different approach:

x=aq+b
x=bq+a-2

aq+b=bq+a-2
factor out q's
re-arranging gives: 2b=2a-2
Divide by 2
b=a-1
Ans D

Is this correct? I find it extremely confusing.

Kudos [?]: [0], given: 0

Math Expert
Joined: 02 Sep 2009
Posts: 42281

Kudos [?]: 132984 [1], given: 12400

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05 Jul 2016, 06:48
1
KUDOS
Expert's post
gmatprepeugene2014 wrote:
Slightly different approach:

x=aq+b
x=bq+a-2

aq+b=bq+a-2
factor out q's
re-arranging gives: 2b=2a-2
Divide by 2
b=a-1
Ans D

Is this correct? I find it extremely confusing.

When you say "factor out q's" where does q go? It cannot just disappear.

But more importantly, the quotient should not be the same. If you check the solution above you'll see that it's x=aq+b in one case and x=bp+(a-2). We don't know whether q = p.
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07 Jul 2016, 04:35
Hi ,

Please check this and correct me if i am wrong

X= aq + b,

x= bp + (a-2).

so substituting values

a=4,q=1,b=2,x=6,p=3, a-2=0.

6 = 4(1) + 2---------------------x =aq + b

6 = 2(3) + (2-2)------------------x = bp + (a-2)

but D doesnt satisfy the above b = a-1 . b is not equal to a-1.

Thanks
John

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Math Expert
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Kudos [?]: 132984 [0], given: 12400

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07 Jul 2016, 04:44
sidjohn wrote:
Hi ,

Please check this and correct me if i am wrong

X= aq + b,

x= bp + (a-2).

so substituting values

a=4,q=1,b=2,x=6,p=3, a-2=0.

6 = 4(1) + 2---------------------x =aq + b

6 = 2(3) + (2-2)------------------x = bp + (a-2)

but D doesnt satisfy the above b = a-1 . b is not equal to a-1.

Thanks
John

If a-2=0, then a=2 but you consider a=4 in the first case.
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07 Jul 2016, 05:03
hahaha crazy me.tnx buneul.

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11 Sep 2016, 04:41
1
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I think this is a high-quality question and I agree with explanation.

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28 Sep 2016, 09:59
Avigano wrote:
Would it work to pick numbers?

I used numbers x=5 a=3 and b=2. Worked like a charm. But I also admit I may have got lucky.

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26 Nov 2016, 13:23
Bunuel wrote:
gmatprepeugene2014 wrote:
Slightly different approach:

x=aq+b
x=bq+a-2

aq+b=bq+a-2
factor out q's
re-arranging gives: 2b=2a-2
Divide by 2
b=a-1
Ans D

Is this correct? I find it extremely confusing.

When you say "factor out q's" where does q go? It cannot just disappear.

But more importantly, the quotient should not be the same. If you check the solution above you'll see that it's x=aq+b in one case and x=bp+(a-2). We don't know whether q = p.

Hi Bunuel, I am extremely sorry but I didnt understand how we got rid of the quotients and got the below equation. Can you please help me understand it

aq+b=bq+a-2
factor out q's
re-arranging gives: 2b=2a-2

Kudos [?]: [0], given: 0

Math Expert
Joined: 02 Sep 2009
Posts: 42281

Kudos [?]: 132984 [1], given: 12400

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27 Nov 2016, 01:40
1
KUDOS
Expert's post
vtomar20 wrote:
Bunuel wrote:
gmatprepeugene2014 wrote:
Slightly different approach:

x=aq+b
x=bq+a-2

aq+b=bq+a-2
factor out q's
re-arranging gives: 2b=2a-2
Divide by 2
b=a-1
Ans D

Is this correct? I find it extremely confusing.

When you say "factor out q's" where does q go? It cannot just disappear.

But more importantly, the quotient should not be the same. If you check the solution above you'll see that it's x=aq+b in one case and x=bp+(a-2). We don't know whether q = p.

Hi Bunuel, I am extremely sorry but I didnt understand how we got rid of the quotients and got the below equation. Can you please help me understand it

aq+b=bq+a-2
factor out q's
re-arranging gives: 2b=2a-2

This is an incorrect method by gmatprepeugene2014, which is pointed out in my post.
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30 Mar 2017, 04:30
brunel : This is a very good question, thanks for a lucid explanation!!!

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06 Apr 2017, 21:54
Key takeway that really helps *thanks Bunuel for showing this* is remainder must be less than divisor

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25 May 2017, 02:44
DJ1986 wrote:
Slightly different approach:

x=aq+b
x=bq+a-2

aq+b=bq+a-2
factor out q's
re-arranging gives: 2b=2a-2
Divide by 2
b=a-1
Ans D

There is a problem with approach. Its nowhere mentioned that the quotient "q" is same in both the cases as assumed here. This makes this approach incorrect.

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29 May 2017, 04:06
I think this is a high-quality question and I don't agree with the explanation. Question says " when x is divided by a, the remainder is b" so that would mean that x+b is divisible by a.
So the option "x+b is divisible by a" is also correct

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Re M16-35   [#permalink] 29 May 2017, 04:06

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# M16-35

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