Number Theory Challenge

Guest · **Posted:** 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.

Denis · **Posted:** Tue, 3 Aug 2010 22:29:23 UTC

Be this Shadow in the sunlight?

Matt · **Joined:** Wed, 1 Oct 2003 04:45:43 UTC **Posts:** 9633

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.

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

Shadow · **Posted:** Wed, 4 Aug 2010 06:49:27 UTC

Denis wrote:

Be this Shadow in the sunlight?

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

Guest · **Posted:** 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 $\mbox{ log n \geq k\cdot log 2}$

Matt · **Joined:** Wed, 1 Oct 2003 04:45:43 UTC **Posts:** 9633

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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