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M16-35

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Re: M16-35 [#permalink]

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New post 29 May 2017, 04:37
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AAmulya wrote:
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


How did you come up with that conclusion? It's not true. You could test it yourself.

Say x = 5, a = 3, and b = 2. All conditions are satisfied:

\(x=5\) is divided by \(a=3\), the remainder is \(b=2\).
\(x=5\) is divided by \(b=2\), the remainder is \(a-2=1\).

But x + b = 5 + 2 = 7, which is NOT divisible by a = 3.

Hope it's clear.
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Re M16-35 [#permalink]

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New post 26 Jun 2017, 18:09
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 19 Sep 2017, 07:44
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 14 Jan 2018, 10:23
hensap wrote:
Key takeway that really helps *thanks Bunuel for showing this* is remainder must be less than divisor


Thank you very much for pointing it out so clearly
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Re: M16-35 [#permalink]

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New post 24 Jan 2018, 18:14
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



How do we know we can use the same co-efficient for the quotients?
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Re: M16-35 [#permalink]

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New post 24 Jan 2018, 21:14
Mco100 wrote:
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



How do we know we can use the same co-efficient for the quotients?


This is addressed here: https://gmatclub.com/forum/m16-184104.html#p1705676
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Re M16-35 [#permalink]

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New post 22 May 2018, 13:41
I think this is a high-quality question and I don't agree with the explanation. I don't completely agree with the explanation.
In the first line, the question says "when x is divided by a, the remainder is b" so using the basic division laws, x will be divisible by a when b is added (x + b will be divisible by a)
So the option "x+b is divisible by a" is also correct.

Let me know if I am thinking in the wrong direction :)
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Re: M16-35 [#permalink]

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New post 22 May 2018, 21:58
PrayasT wrote:
I think this is a high-quality question and I don't agree with the explanation. I don't completely agree with the explanation.
In the first line, the question says "when x is divided by a, the remainder is b" so using the basic division laws, x will be divisible by a when b is added (x + b will be divisible by a)
So the option "x+b is divisible by a" is also correct.

Let me know if I am thinking in the wrong direction :)


That's not true.

For example, x = 1, a = 5 and b = 1.

x = 1 divided by a = 5 gives the remainder of b = 1 but x + b = 2 is NOT divisible by a = 5.
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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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Re: M16-35 [#permalink]

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New post 29 May 2018, 09:14
It would have been correct had the option said x-b is divisible by a. You need to subtract the remainder from the dividend in order to make it divisible by the quotient.
Re: M16-35   [#permalink] 29 May 2018, 09:14
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