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M16-35

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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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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 \(b=a-1\)).


Answer: D
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Re: M16-35 [#permalink]

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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

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Re: M16-35 [#permalink]

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New post 08 Mar 2016, 14:41
Would it work to pick numbers?

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Re M16-35 [#permalink]

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I think this is a high-quality question and I agree with explanation.

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New post 20 Jun 2016, 16:55
I think this is a high-quality question and I agree with explanation.
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Re: M16-35 [#permalink]

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New post 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. :(

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Re: M16-35 [#permalink]

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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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Re: M16-35 [#permalink]

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New post 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.

Please explain.

Thanks
John

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New post 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.

Please explain.

Thanks
John


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

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Re M16-35 [#permalink]

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I think this is a high-quality question and I agree with explanation.

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Re: M16-35 [#permalink]

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New post 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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Re: M16-35 [#permalink]

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New post 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

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Re: M16-35 [#permalink]

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New post 27 Nov 2016, 01:40
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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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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

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New post 30 Mar 2017, 04:30
brunel : This is a very good question, thanks for a lucid explanation!!!

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Re: M16-35 [#permalink]

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New post 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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Re: M16-35 [#permalink]

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New post 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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Re M16-35 [#permalink]

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New post 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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