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M16-35

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Re: M16-35  [#permalink]

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New post 29 May 2017, 03:37
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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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New post 26 Jun 2017, 17:09
I think this is a high-quality question and I agree with explanation.
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New post 19 Sep 2017, 06:44
I think this is a high-quality question and I agree with explanation.
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New post 14 Jan 2018, 09:23
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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New post 24 Jan 2018, 17:14
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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New post 24 Jan 2018, 20:14
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Re M16-35  [#permalink]

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New post 22 May 2018, 12:41
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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New post 22 May 2018, 20:58
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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Re: M16-35  [#permalink]

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New post 29 May 2018, 08:14
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.
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New post 12 Jul 2018, 19:41
I think this is a high-quality question and I agree with explanation.
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New post 25 Jul 2018, 03:39
If u add the remainder to the dividend, it does not become perfectly divisible.
Eg: If u have 12/5...... The Quotient is 2 and Remainder is 2. If u add 2 to 12 i.e 14/5 is still not divisible. Hence that analogy is incorrect.
If u subtract the remainder from the Dividend i.e 12-2 then the result will be perfectly divisible.
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New post 09 Nov 2018, 05:55
dude option b is correct isn't it? When x is divided by a, b is the remainder. So that means x+b will be divisible by a!
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New post 09 Nov 2018, 06:07
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New post 23 Jul 2019, 12:28
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 Aug 2019, 13:11
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


Incorrect approach stated.

Example: 10/3 ---> quotient is 3 and remainder is 1

10+1 (x+b) is not divisible by 3 (a).
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New post 03 Sep 2019, 15:52
I think this is a high-quality question and I agree with explanation.
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New post 04 Oct 2019, 21:41
Since b is remainder, x-b is divisible by a, not x+b. I hope you got this point.
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New post 04 Oct 2019, 23:11
Here is my approach to this question

Posted from my mobile device
>> !!!

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New post 17 Nov 2019, 21:19
Can I please get some links to similar problems?
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New post 17 Nov 2019, 23:47
ashita5678 wrote:
Can I please get some links to similar problems?


6. Remainders



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Re: M16-35   [#permalink] 17 Nov 2019, 23:47

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