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.
number-theory elementary-number-theory discrete-mathematics divisibility 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.
number-theory elementary-number-theory discrete-mathematics divisibility integers
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 $k$.
â Jeremy Dover
Aug 28 at 16:48
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up vote
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down vote
favorite
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.
number-theory elementary-number-theory discrete-mathematics divisibility integers
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.
number-theory elementary-number-theory discrete-mathematics divisibility integers
asked Aug 28 at 16:41
Jimmy
446
446
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 $k$.
â Jeremy Dover
Aug 28 at 16:48
add a comment |Â
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 $k$.
â 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 $k$.
â 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 $k$.
â Jeremy Dover
Aug 28 at 16:48
add a comment |Â
1 Answer
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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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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
HINT: What is the smallest natural number $x$ that is divisible by $2^k$, where $x > 2^k$?
add a comment |Â
up vote
3
down vote
HINT: What is the smallest natural number $x$ that is divisible by $2^k$, where $x > 2^k$?
add a comment |Â
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 $x > 2^k$?
HINT: What is the smallest natural number $x$ that is divisible by $2^k$, where $x > 2^k$?
answered Aug 28 at 17:20
ArsenBerk
6,7351933
6,7351933
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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 $k$.
â Jeremy Dover
Aug 28 at 16:48