The greatest common divisor ${ }^{31}{C}_3{,}^{31}{C}_5$$\dots {,}^{31}{C}_{29}$ is

We have, for $3\le r\le 29$

$r!(31-r)!\left({ }^{31}{C}_r\right)=31!$

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

$(31-r)!\text{ for }\beta \le r\le 29,\text{ we get }31\mid \left({ }^{31}{C}_r\right)$

Also, Since ${ }^{31}{C}_{29}=\left(31\right)\left(3\right)\left(5\right){,}^{31}{C}_3=\left(31\right)\left(29\right)\left(5\right){\text{ and }}^{31}{C}_5$$=\left(31\right)\left(29\right)\left(7\right)$

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