Consider the binomial coefficients $\binom{n}{k}=\frac{n!}{k!(n-k)!}\ (k=1,2,\ldots n-1)$. Determine all positive integers $n$ for which $\binom{n}{1},\binom{n}{2},\ldots ,\binom{n}{n-1}$ are all even numbers.
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Tags: ARML, binomial coefficients
TZF
03.01.2011 20:33
This topic was touched upon in one of the ARML Penn State talks (it was either 2006 or 2007). I wish they'd keep transcripts of the annual talks. In any case, check out this article (PDF).
yashagarwal199528
04.01.2011 21:10
all n=(2^r) would follow the result
Alex00
23.02.2016 22:42
Yes, It's true.
Ankoganit
24.02.2016 06:11
Answer: $n\in\{2^k|k\in\mathbb{N}\}$. Proof: Suppose $n$ is not a power of two. We shall constuct a number $0<m<n$ such that $\binom n m$ is odd. Express $n$ in binary digits. Clearly the leftmost digit is $1$, and since it is not a power of two, it has another non-zero digit (i.e. $1$), say at position $j$. Now take a number $m$ as follows: in binary representation, it has $1$ at position $j$, and $0$ at all other positions. From Lucas' theorem, it is easily seen that $\binom n m$ is indeed odd, completing our proof.