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very hard ds question

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very hard ds question [#permalink] New post 14 Dec 2007, 23:25
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Question Stats:

100% (02:56) correct 0% (00:00) wrong based on 2 sessions
if m and n are both 2-digit numbers, and m-n=11x, is x an integer?
1. the tens digit and the units digit of m are the same
2. m+n is a multiple of 11.

Please help explain how you have derived your answer. Kaplan did a bad job explaining.


OA ******************





***************
is C
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Re: very hard ds question [#permalink] New post 15 Dec 2007, 00:15
aliensoybean wrote:
if m and n are both 2-digit numbers, and m-n=11x, is x an integer?
1. the tens digit and the units digit of m are the same
2. m+n is a multiple of 11.

Please help explain how you have derived your answer. Kaplan did a bad job explaining.





Since m - n = 11x, is x an integer?

1. m = aa means m is divisible by 11, but n = ? do not know.
2. m+n = 11x, do not m and n each alone.

togather: since m is aa, and m+n = 11x, then n must be also divisible by 11. so suff.

C.


Please help explain how you have derived your answer. Kaplan did a bad job explaining. [/quote]
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 [#permalink] New post 15 Dec 2007, 00:23
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C

1. we don't know n.INSUFF

2. m+n=11k. There are two patterns for reminders of m and n: (0,0) and (r,11-r), where 0<r<11.

So, for (0,0) m-n=11s, for (r,11-r): (m-n)(mod 11)=r-11+r=2r but 2r<>0 and 2r<>11. ISUFF.

1&2. from 1. m=AA=A*10+A=11*A. So, from 2 we have only one pattern: (0,0). SUFF.
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 [#permalink] New post 15 Dec 2007, 13:34
Tiger,
1. m = aa means m is divisible by 11, but n = ? do not know.

If m = aa, doesn't that force n to be a mutiple of 11 since the original equation says that m-n is a multiple of 11. Ex: m=55-n=11X, n = bb, so is a multiple of 11, which means X is an integer
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Re: very hard ds question [#permalink] New post 15 Dec 2007, 22:26
aliensoybean wrote:
if m and n are both 2-digit numbers, and m-n=11x, is x an integer?
1. the tens digit and the units digit of m are the same
2. m+n is a multiple of 11.

Please help explain how you have derived your answer. Kaplan did a bad job explaining.


OA ******************





***************
is C


Ive stopped using Kaplan for this reason. While the problems can be decent practice, the explanations are terrible. Its like someone hurried up to write them so he/she could hurry up to go to party or somethin. Just poorly done.

Anyway:

2 is easier so i started w/ it.

2: (m+n)/11 --> Could be 12+10 22/11 12-10 =2. So X is not an integer. b/c 11x=2 --> x=2/11

Then we could have 66+33 --> 99/11 66-33 =33 X=3 so x is an integer.

1: Doesnt help us at all.


Together: the only way for m+n to be divisble by 11 is to have both integers divisible by 11. This means that their sum or difference is ALSO divisible by 11.

So C.
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 [#permalink] New post 15 Dec 2007, 23:06
aliensoybean wrote:
Tiger,
1. m = aa means m is divisible by 11, but n = ? do not know.

If m = aa, doesn't that force n to be a mutiple of 11 since the original equation says that m-n is a multiple of 11. Ex: m=55-n=11X, n = bb, so is a multiple of 11, which means X is an integer


m - n = 11x, but we do not know whether x is an integer. in fact thats the question.

suppose, m = 44 and n = 14, then m - n = 30
so 30 = 11x
x = 30/11, which is not an integer

if n = 22, then m-n = 44-22=22,
22 = 11x
x = 2, which is an integer.

therefore, 1 is not suff.
  [#permalink] 15 Dec 2007, 23:06
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