The greatest integer contained in (3+1)6 is 

The greatest integer contained in (3+1)6 is 

  1. A

    208

  2. B

    416

  3. C

    415

  4. D

    207

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    Solution:

    Let R=(3+1)6 and R[R]=f

    Put F=(31)6 , then 0<F<1 and 

    R+F=(3+1)6+(31)6

    =233+6C232+6C4(3)+1=2[27+135+45+1]=416

     f+F=416[R] is an integer 

    Now, show that f+F=1

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