Results 1 to 2 of 2

Thread: Divisible integers

  1. #1
    Oct 2010

    Divisible integers

    Thank you for your help!!

    Here's the question:
    Find all positive integers n divisible by all positive integers m where m is less than or equal to n^(1/2)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Nov 2009
    First note that the only $\displaystyle n<36$ which fill the condition are

    $\displaystyle n = 1, 2, 3, 4, 6, 8, 12, 24$. We'll see there are no others.

    Suppose $\displaystyle n$ fills the conditions, and let $\displaystyle t = n-\lfloor \sqrt n \rfloor^2$.

    We can suppose now that $\displaystyle \lfloor \sqrt n \rfloor \ge 6$ and then arrive at a contradiction.

    $\displaystyle \lfloor \sqrt n \rfloor \mid n$, so there exists a positive integer $\displaystyle s$ such that

    $\displaystyle s\lfloor \sqrt n \rfloor = n = \lfloor \sqrt n \rfloor^2+t$

    Therefore $\displaystyle \lfloor \sqrt n \rfloor\mid t$ and we can write $\displaystyle n-\lfloor \sqrt n \rfloor^2 = t = d\lfloor \sqrt n \rfloor$

    That is, $\displaystyle n = \lfloor \sqrt n \rfloor^2+d\lfloor \sqrt n \rfloor$.

    Write $\displaystyle \lfloor \sqrt n \rfloor = \sqrt{n}-\epsilon,\:\: 0\le\epsilon<1$. Plugging in the previous equality yields:

    $\displaystyle d = \epsilon\frac{2\sqrt{n}-\epsilon}{\sqrt{n}-\epsilon} < \frac{2\sqrt{n}-\epsilon}{\sqrt{n}-\epsilon} = 1+\frac{\sqrt{n}}{\sqrt{n}-\epsilon} < 1+ \frac{\sqrt{n}}{\sqrt{n}-2}$

    And since $\displaystyle \sqrt{n}\ge 6$ we have $\displaystyle \frac{\sqrt{n}}{\sqrt{n}-2}<2$, so $\displaystyle d<1+2 = 3 \Rightarrow d\in\{0,1,2\}$

    Thus $\displaystyle n\le h^2+2h$ where we denote $\displaystyle h=\lfloor \sqrt n \rfloor$

    We have $\displaystyle h\mid n$ and $\displaystyle h-1\mid n$ and these numbers

    are coprime so $\displaystyle h(h-1)\mid n$. Furthermore, $\displaystyle h-2\mid n$

    and is coprime to $\displaystyle h-1$, and $\displaystyle \gcd(h-2,h)\le2$,

    so $\displaystyle h(h-1)(h-2) \mid 2n$, and thus $\displaystyle h(h-1)(h-2) \le 2n \le 2h^2+4h$

    Therefore $\displaystyle h^2-3h+2 \le 2h+4 \Rightarrow h^2-5h -2 \le 0 \Rightarrow h < 6$,

    contradicting our assumption that $\displaystyle h=\lfloor\sqrt{n}\rfloor \ge 6 $ (see head of the post).
    Last edited by Unbeatable0; Oct 18th 2010 at 09:41 AM. Reason: typo
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: Feb 20th 2013, 09:32 AM
  2. Replies: 3
    Last Post: Jun 1st 2010, 01:41 AM
  3. Replies: 3
    Last Post: Feb 3rd 2010, 09:30 AM
  4. Replies: 5
    Last Post: Jan 1st 2010, 01:59 AM
  5. Replies: 2
    Last Post: Sep 22nd 2007, 02:03 PM

Search Tags

/mathhelpforum @mathhelpforum