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 Post subject: Number Theory Challenge
PostPosted: Tue, 3 Aug 2010 22:22:39 UTC 
Prove that \mbox{log n} \geq k \cdot \mbox{log 2}, where n is a natural number and k the number of distinct primes that divide n.


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PostPosted: Tue, 3 Aug 2010 22:29:23 UTC 
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PostPosted: Wed, 4 Aug 2010 02:13:07 UTC 
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Casdan1 wrote:
Prove that \mbox{log n} \geq k \cdot \mbox{log 2}, where n is a natural number and k the number of distinct primes that divide n.

\log 35=\log5+\log7\geq 2\log 2


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PostPosted: Wed, 4 Aug 2010 06:49:27 UTC 
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Denis wrote:
Be this Shadow in the sunlight? 8)


No, but I can see the thoughts behind that. I actually got bunny from another poster on this board a long time ago, myself.

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PostPosted: Wed, 4 Aug 2010 16:28:06 UTC 
Matt wrote:
Casdan1 wrote:
Prove that \mbox{log n} \geq k \cdot \mbox{log 2}, where n is a natural number and k the number of distinct primes that divide n.

\log 35=\log5+\log7\geq 2\log 2



let n be a natural number greater than 1.

let \mbox{p_{1}, p_{2}, ... , p_{k}} be the prime factors of n. and let
p_{1}^{\alpha_1}, p_{2}^{\alpha_{2}}, ... , p_{k}^{\alpha_{k}} be the highest powers of those primes that divide n.

Therefore n =  p_{1}^{\alpha_1}p_{2}^{\alpha_{2}}\cdot\cdot\cdot p_{k}^{\alpha_{k}}

Therfore since none of the primes are less than 2, and each \alpha_{i} \geq 1

Then n\geq 2^{\alpha_{1} + \alpha_{2} + \ldots + \alpha_{k}}\geq 2^{k}.

Hence \mbox{ log n \geq k\cdot log 2}


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PostPosted: Wed, 4 Aug 2010 18:06:50 UTC 
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Casdan1 wrote:
let n be a natural number greater than 1.

let \mbox{p_{1}, p_{2}, ... , p_{k}} be the prime factors of n. and let
p_{1}^{\alpha_1}, p_{2}^{\alpha_{2}}, ... , p_{k}^{\alpha_{k}} be the highest powers of those primes that divide n.

Therefore n =  p_{1}^{\alpha_1}p_{2}^{\alpha_{2}}\cdot\cdot\cdot p_{k}^{\alpha_{k}}

Therfore since none of the primes are less than 2, and each \alpha_{i} \geq 1

Then n\geq 2^{\alpha_{1} + \alpha_{2} + \ldots + \alpha_{k}}\geq 2^{k}.

Hence \mbox{ log n \geq k\cdot log 2}

Looks great!


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