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  1. June 21st 2010, 01:37 PM #1
    Chris11
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    Number Theory Quicke

    This is a chalenge question. It has been approved by Mr.Fantastic





    Let Q(n)=\lfloor\frac{n}{\lfloor{n}^{1/2}\rfloor}\rfloor , where n is a natural number. Find the natural numbers a for which Q(a)>Q(a+1)



    Moderator edit: Approved Challenge question.
    Last edited by mr fantastic; June 21st 2010 at 01:42 PM. Reason: Approved.
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  2. June 21st 2010, 02:17 PM #2
    undefined
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    Spoiler:

    Let k\in\mathbb{Z},k>0 and consider the sets S=\{k^2,k^2+1,\dots, (k+1)^2-2\} and S'=\{k^2,k^2+1,\dots, (k+1)^2-1\}. It is clear that \forall a\in S':\lfloor\sqrt{a} \rfloor=k. Therefore \forall a\in S: Q(a)=\left\lfloor\frac{a}{\lfloor\sqrt{a}\rfloor}\ right\rfloor = \left\lfloor\frac{a}{k}\right\rfloor \le Q(a+1) = \left\lfloor\frac{a+1}{k}\right\rfloor.

    So the only candidates are of the form a=k^2-1 for some k\in\mathbb{Z},k>1. We know that \lfloor\sqrt{a} \rfloor=k-1 and \lfloor\sqrt{a+1} \rfloor=k. So we compare

    Q(a)=\left\lfloor\frac{a}{k-1}\right\rfloor=\left\lfloor\frac{k^2-1}{k-1}\right\rfloor=\left\lfloor k+1 \right\rfloor=k+1

    with

    Q(a+1)=\left\lfloor\frac{a+1}{k}\right\rfloor=\lef t\lfloor\frac{k^2}{k}\right\rfloor=k

    Therefore the complete solution set is \{a|a=k^2-1, k\in\mathbb{Z},k>1\}.
    Last edited by undefined; June 21st 2010 at 02:43 PM. Reason: typo
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  3. June 21st 2010, 03:04 PM #3
    Chris11
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    Good job!
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