We have had many questions here about the divisibility of $\binom{n}{k}$, many of them dealing with divisibility by powers of primes, or expressions involving the $\textrm{gcd}(n,k)$ (I originally gave several more examples but took TheSimpliFire's advice to shorten the list, and many other examples can still be found by looking at this question's edit history):
- How can we prove that a binomial coefficient (n choose k) is divisible by the ratio of n and the gcd(n,m)?
- Proving prime $p$ divides $\binom{p}{k}$ for $k\in\{1,\ldots,p-1\}$
The topic has also lead to some interesting research papers recently:
- In 2014: Some divisibility properties of binomial and $q$-binomial coefficients.
- In 2017: Divisibility of binomial coefficients and generation of alternating groups.
- In 2019: On the divisibility of binomial coefficients.
There's also many associated theorems:
However I've come across a related problem which is expressed completely in the title, and is remarkably not covered by the above extensive body of literature. In MathJax, if $s>0$ is an integer (let's also make it the largest one for which $2^s$ divides $n$), under what conditions of $k$ do we have the following:
$$ 2^s \mid n \implies 2^s \left\vert \binom{n}{k} \right. . \tag{1} $$
Since $\binom{n}{k} = \frac{n}{k} \binom{n-1}{k-1}$ and $2^s \mid n$ we are left with determining when $\frac{q}{k} \binom{n-1}{k-1}$ is an integer ($q$ being the result when $n$ is divided by $2^s$).
The implication works for a remarkable number of cases, but not always (for example if $\binom{n}{k}$ is odd).
from Hot Weekly Questions - Mathematics Stack Exchange
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