Numbers 5

Question

Q.
What is the remainder when   divides  
1. 1
2. 7!
3. 8!
4. 9!

pradhan · Answer

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

gmatjon · Answer

Please, can somebody elaborate on solution in details?
Thank you

mrsmarthi · Answer

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   is divided by

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......

gmatcracker1 · Answer

1 it is. 
taking the commons, it becomes easier.

thanks for the approach. +1 for you

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