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# PS#X,Y,A,B

Author Message
Manager
Joined: 08 Aug 2008
Posts: 226

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05 Dec 2008, 08:26
00:00

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Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

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x, y, a, and b are positive integers. When x is divided by y, the remainder is 6. When a is divided by b, the remainder is 9. Which of the following is NOT a possible value for y + b?

A)24
B)21
C)20
D)17
E)15

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Director
Joined: 14 Aug 2007
Posts: 695

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05 Dec 2008, 08:41
Lets try to find the least possible value of y+b

The least value y can have is 7 or else the remainder won't be 6.
Similarly to have a remainder of 9, b must be at least 10.

so minimum y+b is 17.

15 cant be the value.

E.
Current Student
Joined: 28 Dec 2004
Posts: 3292
Location: New York City
Schools: Wharton'11 HBS'12

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05 Dec 2008, 09:13
i am going with 20..

if you notice, the y+b can be written as 3*2N+3*3M or 3(2N+3m) SO BASICALLY any multiple of 3 is out.. 2N+3M=17 is possible if N=2 and M=1 ..

so 20 is the only choice
Senior Manager
Joined: 30 Nov 2008
Posts: 479
Schools: Fuqua

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05 Dec 2008, 11:38
I will go with E. Same explanation as alpha_plus_gamma provided.

20 is possible for y + b. Here is my example

x = 16, y = 10 leaves remainder 6
a = 19, b = 10 leaves remainder 9.

y + b = 20.

fresinha12 wrote:
i am going with 20..

if you notice, the y+b can be written as 3*2N+3*3M or 3(2N+3m) SO BASICALLY any multiple <SPAN id=google-navclient-hilite style="COLOR: black; BACKGROUND-COLOR: cyan">of</SPAN> 3 is out.. 2N+3M=17 is possible if N=2 and M=1 ..

so 20 is the only choice
SVP
Joined: 29 Aug 2007
Posts: 2427

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05 Dec 2008, 12:39
alpha_plus_gamma wrote:
Lets try to find the least possible value of y+b

The least value y can have is 7 or else the remainder won't be 6.
Similarly to have a remainder of 9, b must be at least 10.

so minimum y+b is 17.

15 cant be the value.

E.

Thats the approach. Agree with E.

In fact (y+b) cannot be 16 if so, then the reminders (9 and 6) wont be possible.
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Current Student
Joined: 28 Dec 2004
Posts: 3292
Location: New York City
Schools: Wharton'11 HBS'12

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05 Dec 2008, 14:17
i agreee with E..

I totally made a bad mistake, x/3 will never give you a remainder of 3, simply because multiples of 3 repeats every 3 numbers, which is less than 6 or 9

so based on that the smallest possible value for any number is 7 and 10..

17 it is..
Director
Joined: 29 Aug 2005
Posts: 818

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24 Jan 2009, 11:51
y>6
b>9
y+b>15
So, E.

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

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Re: PS#X,Y,A,B   [#permalink] 24 Jan 2009, 11:51
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