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Q.

 Let n=2015 The least positive integer k for which kn2n212n222n332n2(n1)2=r!  for some positive integer r is

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An Intiative by Sri Chaitanya

a

2014 

b

2013

c

1

d

2

answer is D.

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Detailed Solution

We can rewrite the given expression as 

kn2(n1)(n+1)(n2)(n+2)(n3)(n+3)…….(n+n1)(nn+1)=r!

kn(1)(2)(n1)n(n+1)(n+2)(2n1)=r!kn(2n1)!=r!

 To convert L.H.S. to a factorial, we shall require

k = 2 which will convert it into (2n)!

 

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 Let n=2015 The least positive integer k for which kn2n2−12n2−22n3−32…n2−(n−1)2=r!  for some positive integer r is