Natural number n w/ over 2017 divisors, w/ sqrt[n] <= d < 1,01 sqrt[n]

[VanBuren]

Prove that there exists a natural number n, which has over 2017 divisors d satisfying:

$\displaystyle √n ≤ d < 1,01 √n$

To be honest, I don't really know where to start, again, but I assume that:

$\displaystyle n = 2^a + 3^b + 5^c$, where $\displaystyle a+b+c>2016$ (because 1 is also a divisor)

 

[Ishuda]

Prove that there exists a natural number n, which has over 2017 divisors d satisfying:

$\displaystyle √n ≤ d < 1,01 √n$

To be honest, I don't really know where to start, again, but I assume that:

$\displaystyle n = 2^a + 3^b + 5^c$, where $\displaystyle a+b+c>2016$ (because 1 is also a divisor)

Let p_k be the k^th prime and suppose
u = $\displaystyle p_1^2\, *\, p_2^2\, *\, p_3^2\, ...*\, p_m^2\, =\, \Pi_{k=1}^{k=m}\,\,\, p_k^2$
and
s = $\displaystyle \sqrt{u}$
How many prime divisors would s have? How many prime divisors would 1.01 s have? How does this tell you how to pick an n?

 

[VanBuren]

Let p_k be the k^th prime and suppose
u = $\displaystyle p_1^2\, *\, p_2^2\, *\, p_3^2\, ...*\, p_m^2\, =\, \Pi_{k=1}^{k=m}\,\,\, p_k^2$
and
s = $\displaystyle \sqrt{u}$
How many prime divisors would s have? How many prime divisors would 1.01 s have? How does this tell you how to pick an n?

Well, I think s has $\displaystyle 2^m$ divisors, but $\displaystyle 1.01*s$?
N would be $\displaystyle \Pi_{k=1}^{k=m}\,\,\, p_k^2$ , where $\displaystyle m>2017$?

 

[Ishuda]

Well, I think s has $\displaystyle 2^m$ divisors, but $\displaystyle 1.01*s$?
N would be $\displaystyle \Pi_{k=1}^{k=m}\,\,\, p_k^2$ , where $\displaystyle m>2017$?

Sorry, what I should have said is how many divisors does n have, not how many prime divisors. Also, not that p1, p2, p3, ... are the 1st, 2nd, 3rd, ... primes but rather a set of primes such that p1 < p2 < p3 < ...

Now, more ramblings [I'm not sure we are going any where useful]. Let m=0 [i.e. n=1]. The number of divisors is 1=3⁰.
[1]

Let m=1 [i.e. n = p₁²]. The number of divisors is 3 = 3¹
[1*[1], p₁*[1], p₁²*[1]] = [1, p₁, p₁²]

Let m=2. The number of divisors is 9=3²
[1*[1, p₁, p₁²], p₂*[1, p₁, p₁²], p₂²*[1, p₁, p₁²]]
= [1, p₁, p₁², p₂, p₁ p₂, p_1² p₂, p₂², p₁ p₂², p₁² p₂²]

Let m=3 The number of divisors is 27=3³
[1*[1, p₁, p₁², p₂, p₁ p₂, p_1² p₂, p₂², p₁ p₂², p₁² p₂²], p₃*[1, p₁, p₁², p₂, p₁ p₂, p_1² p₂, p₂², p₁ p₂², p₁² p₂²], p₃²*[1, p₁, p₁², p₂, p₁ p₂, p_1² p₂, p₂², p₁ p₂², p₁² p₂²]]
=[...]

Now generalize that, i.e.
$\displaystyle s=\sqrt{n} = \Pi_{k=1}^m\, p_k^q$
The number of divisors of s is q^m. If we let q=p_k, then obviously d is greater than or equal to s. Can we find an m so that p_k^m< 1.01 s. What about m < log(1.01 s)/log(p_k)

Is that enough to play around with for a while?

BTW: There should be some easier way to do this. Anybody?


EDIT: Sorry, that should have been (q+1)^k I think

 

[pka]

Prove that there exists a natural number n, which has over 2017 divisors d satisfying:
$\displaystyle √n ≤ d < 1,01 √n$
To be honest, I don't really know where to start, again, but I assume that:
$\displaystyle n = 2^a + 3^b + 5^c$, where $\displaystyle a+b+c>2016$ (because 1 is also a divisor)

I admit that I do not understand your notation. But this may help you.
$\displaystyle 2^4*5^2*7^6*11=517655600=N$ has $\displaystyle (5)(3)(7)(2)=210$ divisors.

Every divisor of $\displaystyle N$ have the form $\displaystyle 2^a\cdot 5^b\cdot 7^c \cdot 11^d$ where $\displaystyle 0\le a\le 4~,~0\le b\le 2~,~ 0\le c\le 6~,~0\le d\le 1$

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