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  • The greatest common divisor <img alt="\mathrm{{ }^{31} C_{3},{ }^{31} C_{5}, \ldots, { }^{31} C_{29}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%7B%20%7D%5E%7B31%7D%20C_%7B3

The greatest common divisor \mathrm{{ }^{31} C_{3},{ }^{31} C_{5}, \ldots, { }^{31} C_{29}} is

Option: 1

31


Option: 2

2


Option: 3

17


Option: 4

none of these


Answers (1)

best_answer

We have, for \mathrm{3 \leq r \leq 29}

\mathrm{r !(31-r) !\left({ }^{31} C_{r}\right)=31 !}
As the prime 31 divides R.H.S. and 31 does not divide \mathrm{r !} and \mathrm{(31-r) !} for \mathrm{3 \leq r \leq 29}, we get \mathrm{31 \mid\left({ }^{31} C_{r}\right)}.

Also, since \mathrm{{ }^{31} C_{29}=(31)(3)(5),{ }^{31} C_{3}=(31)(29)(5)\: and \: { }^{31} C_{5}=(31)(29)(7)}

no other prime can divide all the numbers.

\mathrm{\therefore } \mathrm{\quad g c d }  of the given numbers is 31 .

Posted by

Rakesh

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