Theorem regarding greatest common divisor of certain Binomial coefficients. Recently my friend asked following question-
find the greatest common divisor of all binomial coefficient for a given n so the problem is in mathematical form
$\gcd\left(\binom{2n}1,\binom{2n}2,\binom{2n}3,...,\binom{2n}{2n-1}\right)$. Unfortunately I forget the theorem regarding this something called Lucas theorem. Please suggest me the appropriate answer.
NOTE - is this problem makes any easy solution for  $\gcd\left(\binom{2n}1,\binom{2n}3,\binom{2n}5,...,\binom{2n}{2n-1}\right)$
 A: It is a known fact (called Kummer's theorem) that the multiplicity of a prime number $p$ as factor in the binomial coefficient $\binom{a+b}a$ equals the number of carries produced in the addition of $a$ and $b$ in the base$~p$ number system. For instance to count the number of factors $2$ in $\binom{73}{27}$, note first that $73=27+46$, write $27=11011_2$ and $46=101110_2$; then computing $11011_2+101110_2=1001001_2=73$ gives carries in positions $1,2,3,4,5$ (counting from the right starting at $0$), so the number of factors is $5$; indeed $74604711829408425056/2^5=2331397244669013283$ is (integer and) odd.
Here the value $m=2n=a+b$ is fixed (or maybe you meant $n$; odd numbers are no different than even ones for this problem, except that $m>1$ is needed for the statement to make sense), and $a,b$ can be any positive integers. Given the base$~b$ representation $R$ of$~m$, one can always find $a,b$ that add up to $m$ without producing any carries, unless $R$ contains a single digit$~1$ and all other digits are$~0$. If there is more than one nozero digit one can split off one such digit (as in $13010=13000+10$), and if there is a single nonzero digit greater than$~1$, one can split off a unit from that digit (as in $30000=10000+20000$).
So the only case where $p$ divides all $\binom ma$ with $0<a<m$ is when $m$ is a power of$~p$. Moreover, when this is the case, say $m=p^k$ (with $k>0$ to avoid $m=1$) we can write $p^k=p^{k-1}+(p-1)p^{k-1}$, which involves only a single carry (in the leftmost position).
Concluding the $\gcd$ of the question, which involves any prime number to a power which is the minimum of its multiplicities in any $\binom ma$ with $0<a<m$, equals $1$ unless $m$ is a (positive) power of a prime number$~p$ (obviously unique), in which case the $\gcd$ is $p$.
