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Numbers 5

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Numbers 5 [#permalink] New post 04 Nov 2009, 00:18
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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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Re: Numbers 5 [#permalink] New post 04 Nov 2009, 02:01
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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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Re: Numbers 5 [#permalink] New post 04 Nov 2009, 12:29
Please, can somebody elaborate on solution in details?
Thank you
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Re: Numbers 5 [#permalink] New post 04 Nov 2009, 13:52
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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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Re: Numbers 5 [#permalink] New post 08 Nov 2009, 18:39
1 it is.
taking the commons, it becomes easier.

thanks for the approach. +1 for you
Re: Numbers 5   [#permalink] 08 Nov 2009, 18:39
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