Find the highest power of 2 dividing n

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Discussion Overview

The discussion revolves around determining the highest power of 2 that divides the factorial of a number, denoted as f(n). Participants explore different formulations and approaches to prove a proposed rule regarding f(n) and its relationship with powers of 2.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant presents a rule stating that if \(2^{k} = n\), then \(f(n) = 2^{k} - 1\); otherwise, \(n\) can be expressed as a sum of powers of 2, leading to a corresponding expression for \(f(n)\).
  • Another participant suggests using induction to show that \((2^k)!\) is divisible by \(2^{(2^k - 1)}\) but not by \(2^{(2^k)}\), referencing the factors of 2 in factorials.
  • A later reply questions the correlation of the proposed rule with the formula \(f(n) = \sum_{i = 1}^{b} \left[ \frac{n}{2^i} \right]\), where \(b\) is the highest power of 2 less than or equal to \(n\).
  • One participant expresses gratitude for the hint provided and indicates they are a beginner in number theory.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the validity of the proposed rule or its proof. Multiple approaches and formulations are discussed, but no definitive agreement is established.

Contextual Notes

The discussion includes various assumptions about the properties of factorials and powers of 2, and participants reference different methods of calculation without resolving the mathematical steps involved.

whodsow
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Hi.
I have a question.
I know the rule of the hightest power of 2 dividing n!, the highest power means if 2^{k}|n!, k is the greatest value. For example,
n the hightest power of 2 dividing n!
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Let f(n) is the highest power of 2 dividing n!, I think if 2^{k}=n, then f(n) = 2^{k}-1, else let n=2^{k_{1}}+2^{k_{2}}+...++2^{k_{r}}, then f(n)=f(2^{k_{1}})+f(2^{k_{2}})+...+f(2^{k_{r}})=2^{k_{1}}-1+2^{k_{2}}-1+...+2^{k_{r}}-1, more verification of n shows my rule is correct, but I can't prove that.
Help me to prove my rule. thanks.
 
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Split it into subproblems. Showing that (2^k)! is divisible by 2^(2^k - 1) but not by 2^(2^k) can be done by induction, starting with a base case of 1! being divisible by 1 but not by 2. For the inductive step, use that n! has a factor of 2 every 2, a factor of 4 every four, and so on, up to a factor of 2^k every 2^k (i.e. exactly once). This is a geometric series.
 
thank your hint, CRGreathouse.
I will try to prove it.
 
whodsow said:
Hi.
I have a question.

Let f(n) is the highest power of 2 dividing n!, I think if 2^{k}=n, then f(n) = 2^{k}-1, else let n=2^{k_{1}}+2^{k_{2}}+...++2^{k_{r}}, then f(n)=f(2^{k_{1}})+f(2^{k_{2}})+...+f(2^{k_{r}})=2^{k_{1}}-1+2^{k_{2}}-1+...+2^{k_{r}}-1, more verification of n shows my rule is correct, but I can't prove that.
Help me to prove my rule. thanks.
Did you try to corrolate your rule with the following
f(n) = \sum_{i = 1}^{b} \Left[ \frac{n}{2^i}\Right]
where the brackets stand for the foor function, and b is the highest power of 2 less than or equal to n
 
ramsey2879 said:
Did you try to corrolate your rule with the following
f(n) = \sum_{i = 1}^{b} \Left[ \frac{n}{2^i}\Right]
where the brackets stand for the foor function, and b is the highest power of 2 less than or equal to n

oh, thank you.
I am a beginner, just starting to learn number theorem, I did not know this formulation until you told me.
 

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