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.

1 Answer 1


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

  • $\begingroup$ Can you elaborate this? I think what you are trying to say is that if two numbers are divisible by highest power of 2 then it isn’t the highest power and we can find a new highest power of 2. @lordsharktheunknown $\endgroup$ Commented Aug 28, 2018 at 17:15
  • $\begingroup$ That's basically it. @Jimmy $\endgroup$ Commented Aug 28, 2018 at 17:16

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