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  • Let . The least positive integer <img alt="\mathrm{k}" src="https://entrancecorner.oncodec

Let \mathrm{n=2015}. The least positive integer \mathrm{k} for which

\mathrm{k\left(n^{2}\right)\left(n^{2}-1^{2}\right)\left(n^{2}-2^{2}\right)\left(n^{3}-3^{2}\right) \ldots\left(n^{2}-\right.\left.(n-1)^{2}\right)=r !}
for some positive integer \mathrm{r} is

Option: 1

2014


Option: 2

2013


Option: 3

1


Option: 4

2


Answers (1)

best_answer

We can rewrite the given expression as
\mathrm{\begin{array}{ll} & k\left(n^{2}\right)(n-1)(n+1)(n-2)(n+2)(n-3) \\ & (n+3) \ldots(n+n-1)(n-n+1)=r ! \\ \Rightarrow \quad & k n(1)(2) \ldots(n-1) n(n+1)(n+2) \ldots \\ \Rightarrow \quad & (2 n-1)=r ! \\ & k n(2 n-1) !=r ! \end{array}}

\mathrm{\therefore \quad}   To convert L.H.S. to a factorial, we shall require,\mathrm{k= 2} which will convert it into \mathrm{(2 n)!}

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Sayak

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