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Question

The least positive integer k for which the value k×n2(n212)(n222)....(n2(n1)2) turns into a factorial of some positive integer is


  1. 2

  2. 1

  3. 4

  4. 8


Solution

The correct option is A

2


k×n2(n1)(n+1)(n2)(n+2)....(n+n1)(nn+1)=r!

k×n.1.2.3.4....(n1)n(n+1).....(2n1)=r!

k×n(2n1)!=r!

If k = 2 then L.H.S = (2n)! = r!

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