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Math Help - Divisible integers

  1. #1
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    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)
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  2. #2
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    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; October 18th 2010 at 09:41 AM. Reason: typo
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