Remainder

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by Ozlemg » Fri Jun 24, 2011 8:22 am
my answer is C

50/26: reminder is 24
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by Brent@GMATPrepNow » Fri Jun 24, 2011 8:38 am
MBA.Aspirant wrote:The maximun remainder when 50 is divided by a positive integer less than 50 is

a.1 b.2 c.24 d.26 e.29
We know that when N is divided by some positive integer d, then the remainder r is such that: 0 < r < d

To find the maximum remainder, we can start at E and work our way to A.

E) If the remainder (r) is 29, then the d must be greater than 29. But if d is greater than 29, we can see that the remainder will be less than 21 when we divide 50 by 29.
For example, if d=30, then we get a remainder of 20 when we divide 50 by 30.
If d=31, then we get a remainder of 19 when we divide 50 by 31. etc.
So, if the remainder were somehow 29, we see that the remainder would then have to be less than 21.
Since we get a contradiction here, we can see that r cannot be 29, so eliminate E

D) If the remainder (r) is 26, then the d must be greater than 26. But if d is greater than 26, we can see that the remainder will be less than 24 when we divide 50 by 26. Since we get a contradiction here, we can see that r cannot be 26, so eliminate D

C) If the remainder (r) is 24, then the d must be greater than 24.
If d=25, the remainder is 0.
If d=26, the remainder is 24.
Voila!
The answer is C

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Brent
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by MBA.Aspirant » Mon Jun 27, 2011 12:03 am
Thanks Brent for your help.

the way I did it now is:

50 = qx + 24

qx = 26

since 26*2 = 52 this must be the max remainder you can get from an integer < 50.

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by casperkamal » Tue Jun 28, 2011 3:19 am
In my opinion elimination is the best way to approach this problem....
Kamal