Iranian Mathematical Olympiad

Lemma. Suppose $n\ge 2$ is an integer, then all of the numbers are odd if and only if $n=2^k-1$ for an integer $k\ge 2$.

Proof. For an integer $t$, let $v(t)$ be the greatest integer $u$ such that $2^u\mid n$. We know that " If $p,q\in \mathbb{N}$ and $0\le p\le 2^q$ then $v(p)=v(2^q+p)=v(2^q-p)$" (1) Because if $p=2^{a}b$ with $a,b\in \mathbb{N}$ and $b$ an odd number then $a<q$ and $2^q\pm p=2^a(2^{q-a}\pm b)$ where $2^{q-a}\pm b$ are odd numbers. Now there is one and only one $m\in \mathbb{N}$ such that $2^m\le n<2^{m+1}$. Let $n=2^m+s$ with $0\le s<2^m$. Now consider the number . We have $$\left(\begin{array}{c}n\\ 2^m-1\end{array}\right)=\left(\begin{array}{c}{2}^m+s\\ 2^m-1\end{array}\right)=\left(\begin{array}{c}{2}^m+s\\ s+1\end{array}\right)=\frac{(2^m+s)(2^m+s-1)\cdots (2^m+1)(2^m)}{s(s-1)\cdots (1)(s+1)}.$$ By (1) we have $v(2^m+i)=v(i)$ where $1\le i\le s$, therefore $$v((2^m+s)\cdots (2^m+1))=v(s!),$$ and by assumption is odd therefore $v(2^m)=v(s+1)$ and consequently $2^m\mid s+1$ and $s+1\ge 1$. Hence we have $2^m-1\le s$ and therefore $s=2^m-1$ and $n=2^m+s=2^{m+1}-1$. Now if $n=2^k-1$ for some natural number $k$, for each $1\le c\le n$ we have $$\left(\begin{array}{c}{2}^k-1\\ c\end{array}\right)=\frac{(2^k-1)(2^k-2)\cdots (2^k-c)}{(1)(2)\cdots (c)}\text{and by (1) we know for}1\le l\le c$$ $$v(2^k-l)=v(l),\text{ so }\left(\begin{array}{c}{2}^k-1\\ c\end{array}\right)\text{ is odd for }0\le c\le n.\square$$

Now we return to main problem: Positive integer $n$ has the property of the problem if and only if all the numbers ($0\le i\le n$) have the same parity. It means that for every $0\le i\le n-1$, have different parities. So ($1\le i\le n-1$) is odd and as we know is also odd. Therefore by the lemma this is equivalence to $n+1=2^k-1$ for some integer $k\ge 2$ so $n=2^k-2$ where $k\ge 2$ is an integer. ​