The greatest common divisor  31C3,31C5…,31C29 is

The greatest common divisor  31C3,31C5,31C29 is

  1. A

    31

  2. B

    2

  3. C

    17

  4. D

    none of these

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    Solution:

    We have, for 3r29

    r!(31r)! 31Cr=31!

    As the prime 31 divides R.H.S. and 31 does not divide r! and

    (31r)! for βr29, we get 31 31Cr

    Also, Since  31C29=(31)(3)(5),31C3=(31)(29)(5) and 31C5=(31)(29)(7)

    No prime other than  31  can divide all the numbers.
    Thus greatest common divisors of the given numbers is  31 .

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