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Manager
Joined: 19 Aug 2009
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Question Stats:
0% (00:00) correct
100% (02:33) wrong based on 0 sessions
Q.
What is the remainder when [2(8!) - 21(6!) ] divides [ 14(7!) + 14(13!) ]
1. 1
2. 7!
3. 8!
4. 9!
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Intern
Joined: 22 Sep 2009
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The numerator
reduces too 14* 7! * (1+8*9*10*11*12*13)
and denominator
7! * 13
the second term is divisible by 13
and the first term gives remainder 1
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Senior Manager
Joined: 18 Aug 2009
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Schools: UT at Austin, Indiana State University, UC at Berkeley
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Please, can somebody elaborate on solution in details? Thank you
_________________
Never give up,,,
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Senior Manager
Joined: 30 Nov 2008
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Sometimes Quant questions looks complicated but when you start putting on the paper, they turn out to be real simple.
Here is the approach -
Question is what is the remainder when [ 14(7!) + 14(13!) ] is divided by [2(8!) - 21(6!) ]
First simplfy the numerator part
14 * 7! * ( 13*12*11*10*9*8 + 1) (By taking 14 * 7! as the common factor. Don't try simplifying further until there is a need to.
Now the denominator -
6! * ( 2 * 8 * 7 - 21) ==> 7! *(16 - 3) = 7! * 13
Now Numerator / denominator is 14 * 7! * ( 13*12*11*10*9*8 + 1) / 7! * 13
7! cancel out. Left over is (14*13*12*11*10*9*8 / 13) + (14 / 13)
First part is divisible by 13 so remainder is zero
14 / 13 remainder is 1. Hence ans 1.
sometimes we need to mix and match the expressions to simplify the figures......
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Intern
Joined: 08 Nov 2009
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1 it is.
taking the commons, it becomes easier.
thanks for the approach. +1 for you
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