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Re M1635 [#permalink]
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16 Sep 2014, 01:00
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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 \(a2\), then which of the following must be true? A. \(a\) is even B. \(x+b\) is divisible by \(a\) C. \(x1\) is divisible by \(a\) D. \(b=a1\) 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 \(a2\): \(x=bp+(a2)\) and \(remainder=(a2) \lt b=divisor\). So we have that: \(a2 \lt b \lt a\), as \(a\) and \(b\) are integers, then it must be true that \(b=a1\) (there is only one integer between \(a2\) and \(a\), which is \(a1\) and we are told that this integer is \(b\), hence \(b=a1\)). Answer: D
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Re: M1635 [#permalink]
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19 Feb 2016, 15:49
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Slightly different approach:
x=aq+b x=bq+a2
aq+b=bq+a2 factor out q's rearranging gives: 2b=2a2 Divide by 2 b=a1 Ans D



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Re: M1635 [#permalink]
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08 Mar 2016, 14:41
Would it work to pick numbers?



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09 Mar 2016, 11:25



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28 Mar 2016, 18:19
I think this is a highquality question and I agree with explanation.



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20 Jun 2016, 16:55
I think this is a highquality question and I agree with explanation.
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Re: M1635 [#permalink]
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05 Jul 2016, 01:02
Slightly different approach: x=aq+b x=bq+a2 aq+b=bq+a2 factor out q's rearranging gives: 2b=2a2 Divide by 2 b=a1 Ans D Is this correct? I find it extremely confusing.



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Re: M1635 [#permalink]
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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 + (a2).
so substituting values
a=4,q=1,b=2,x=6,p=3, a2=0.
6 = 4(1) + 2x =aq + b
6 = 2(3) + (22)x = bp + (a2)
but D doesnt satisfy the above b = a1 . b is not equal to a1.
Please explain.
Thanks John



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Re: M1635 [#permalink]
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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 + (a2).
so substituting values
a=4,q=1,b=2,x=6,p=3, a2=0.
6 = 4(1) + 2x =aq + b
6 = 2(3) + (22)x = bp + (a2)
but D doesnt satisfy the above b = a1 . b is not equal to a1.
Please explain.
Thanks John If a2=0, then a=2 but you consider a=4 in the first case.
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Re: M1635 [#permalink]
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07 Jul 2016, 05:03
hahaha crazy me.tnx buneul.



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11 Sep 2016, 04:41
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I think this is a highquality question and I agree with explanation.



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Re: M1635 [#permalink]
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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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Re: M1635 [#permalink]
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26 Nov 2016, 13:23
Bunuel wrote: gmatprepeugene2014 wrote: Slightly different approach: x=aq+b x=bq+a2 aq+b=bq+a2 factor out q'srearranging gives: 2b=2a2 Divide by 2 b=a1 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=a q+b in one case and x=b p+(a2). 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+a2 factor out q's rearranging gives: 2b=2a2



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Re: M1635 [#permalink]
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27 Nov 2016, 01:40
vtomar20 wrote: Bunuel wrote: gmatprepeugene2014 wrote: Slightly different approach: x=aq+b x=bq+a2 aq+b=bq+a2 factor out q'srearranging gives: 2b=2a2 Divide by 2 b=a1 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=a q+b in one case and x=b p+(a2). 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+a2 factor out q's rearranging gives: 2b=2a2 This is an incorrect method by gmatprepeugene2014, which is pointed out in my post.
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Re: M1635 [#permalink]
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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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Re: M1635 [#permalink]
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06 Apr 2017, 21:54
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Key takeway that really helps *thanks Bunuel for showing this* is remainder must be less than divisor



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Re: M1635 [#permalink]
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25 May 2017, 02:44
DJ1986 wrote: Slightly different approach:
x=aq+b x=bq+a2
aq+b=bq+a2 factor out q's rearranging gives: 2b=2a2 Divide by 2 b=a1 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 M1635 [#permalink]
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29 May 2017, 04:06
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I think this is a highquality 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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