Number Theory: Divisibility on a set of consecutive integers

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Consider the set S of integers 1,2,...,n. Let 2$^k$ be the integer in S that is the highest power of 2. Prove that 2$^k$ is not a divisor of any other integer in S.
I don't want a complete proof here. I just want to know how to approach this problem, so that I can write a formal proof.







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    Suppose that $2^k$ does divide some $m ne 2^k$ in this range. Then $m=2^kx$ for some integer $x ge 2$. Show that this contradicts the definition of $k$.
    – Jeremy Dover
    Aug 28 at 16:48














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Consider the set S of integers 1,2,...,n. Let 2$^k$ be the integer in S that is the highest power of 2. Prove that 2$^k$ is not a divisor of any other integer in S.
I don't want a complete proof here. I just want to know how to approach this problem, so that I can write a formal proof.







share|cite|improve this question
















  • 1




    Suppose that $2^k$ does divide some $m ne 2^k$ in this range. Then $m=2^kx$ for some integer $x ge 2$. Show that this contradicts the definition of $k$.
    – Jeremy Dover
    Aug 28 at 16:48












up vote
1
down vote

favorite
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up vote
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down vote

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Consider the set S of integers 1,2,...,n. Let 2$^k$ be the integer in S that is the highest power of 2. Prove that 2$^k$ is not a divisor of any other integer in S.
I don't want a complete proof here. I just want to know how to approach this problem, so that I can write a formal proof.







share|cite|improve this question












Consider the set S of integers 1,2,...,n. Let 2$^k$ be the integer in S that is the highest power of 2. Prove that 2$^k$ is not a divisor of any other integer in S.
I don't want a complete proof here. I just want to know how to approach this problem, so that I can write a formal proof.









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asked Aug 28 at 16:41









Jimmy

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    Suppose that $2^k$ does divide some $m ne 2^k$ in this range. Then $m=2^kx$ for some integer $x ge 2$. Show that this contradicts the definition of $k$.
    – Jeremy Dover
    Aug 28 at 16:48












  • 1




    Suppose that $2^k$ does divide some $m ne 2^k$ in this range. Then $m=2^kx$ for some integer $x ge 2$. Show that this contradicts the definition of $k$.
    – Jeremy Dover
    Aug 28 at 16:48







1




1




Suppose that $2^k$ does divide some $m ne 2^k$ in this range. Then $m=2^kx$ for some integer $x ge 2$. Show that this contradicts the definition of $k$.
– Jeremy Dover
Aug 28 at 16:48




Suppose that $2^k$ does divide some $m ne 2^k$ in this range. Then $m=2^kx$ for some integer $x ge 2$. Show that this contradicts the definition of $k$.
– Jeremy Dover
Aug 28 at 16:48










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HINT: What is the smallest natural number $x$ that is divisible by $2^k$, where $x > 2^k$?






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    1 Answer
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    up vote
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    HINT: What is the smallest natural number $x$ that is divisible by $2^k$, where $x > 2^k$?






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      up vote
      3
      down vote













      HINT: What is the smallest natural number $x$ that is divisible by $2^k$, where $x > 2^k$?






      share|cite|improve this answer






















        up vote
        3
        down vote










        up vote
        3
        down vote









        HINT: What is the smallest natural number $x$ that is divisible by $2^k$, where $x > 2^k$?






        share|cite|improve this answer












        HINT: What is the smallest natural number $x$ that is divisible by $2^k$, where $x > 2^k$?







        share|cite|improve this answer












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        answered Aug 28 at 17:20









        ArsenBerk

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