Vtyjkyuk

Prove that in any block of consecutive positive integers there is a unique integer divisible by a higher power of $2$ than any of the others.

1. I tried first doing this using the Fundamental theorem of Arithmetic but I was stuck in proving that the integer divisible by higher power of $2$ is unique. Can someone help me with this?


2. Then I tried to prove it by defining a set $S=2^k-n,2^k-n+1,ldots,2^k-1,2^k,2^k+1,ldots,2^k+n$, $n<2^k$ because the set must contain all positive integers. But I realized that this is not a general proof and applicable only if set is symmetric and has $2^k$. Can someone help me extend this proof for general case if possible?


3. Also can someone help me come to another proof of this statement. I think this problem uses results of this
   Number Theory: Divisibility on a set of consecutive integers.

Suppose there are two of them: $2^r a$ and $2^r b$ say with $a$ and $b$ odd and
$a<b$.
Then $bge a+2$, and the list of consecutive numbers also contains
$2^r(a+1)$. But $a+1$ is even.