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 n<36 which fill the condition are

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

    contradicting our assumption that h=\lfloor\sqrt{n}\rfloor \ge 6 (see head of the post).
    Last edited by Unbeatable0; Oct 18th 2010 at 10:41 AM. Reason: typo
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: Feb 20th 2013, 10:32 AM
  2. Replies: 3
    Last Post: Jun 1st 2010, 02:41 AM
  3. Replies: 3
    Last Post: Feb 3rd 2010, 10:30 AM
  4. Replies: 5
    Last Post: Jan 1st 2010, 02:59 AM
  5. Replies: 2
    Last Post: Sep 22nd 2007, 03:03 PM

Search Tags

/mathhelpforum @mathhelpforum