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Thread: AIME I Question

  1. #1
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    AIME I Question

    The number $\displaystyle \sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}$ can be written as $\displaystyle a\sqrt{2}+b\sqrt{3}+c\sqrt{5}$, where a,b, and c are positive integers. Find a*b*c.

    Any hints?
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  2. #2
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    Quote Originally Posted by Jameson
    The number $\displaystyle \sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}$ can be written as $\displaystyle a\sqrt{2}+b\sqrt{3}+c\sqrt{5}$, where a,b, and c are positive integers. Find a*b*c.

    Any hints?
    $\displaystyle 104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006=(a\sqrt {2}+b\sqrt{3}+c\sqrt{5})^2$
    Open parantheses,
    $\displaystyle 104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006$=$\displaystyle 2a^2+3b^2+5c^2+2ab\sqrt{6}+2ac\sqrt{10}+2bc\sqrt{1 5}$
    Thus, we have,
    $\displaystyle 2006=2a^2+3b^2+5c^2$ (1)
    $\displaystyle 144=2bc$ (2)
    $\displaystyle 468=2ac$ (3)
    $\displaystyle 104=2ab$ (4)
    Now multiply (2)(3)(4) to get,
    $\displaystyle 7008768=8a^2b^2c^2=8(abc)^2$
    Thus,
    $\displaystyle abc=936$
    Last edited by ThePerfectHacker; Mar 12th 2006 at 01:49 PM.
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  3. #3
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    The reason being that,
    $\displaystyle a_0+a_1\sqrt{r_1}+...+a_n\sqrt{r_n}=+b_0+b_1\sqrt{ r_1}+...+b_n\sqrt{r_n}$ and all $\displaystyle r_j \not= s^2$ thus, they are not squares. Then, $\displaystyle a_j=b_j$.

    Here is an outline of the proof, to keep thing simple let us do it with only two terms,
    $\displaystyle a\sqrt{x}+b=y\sqrt{x}+z$ and $\displaystyle a,b,x,y,z$ are positive integers with $\displaystyle x$ a non-square.
    Now if $\displaystyle a>y$ then, $\displaystyle a=y+y',y'>0$
    $\displaystyle y\sqrt{x}+y'\sqrt{x}+b=y\sqrt{x}+z$
    Thus,
    $\displaystyle y'\sqrt{x}=z-b$ Impossible because LHS is irrational.
    If you have, $\displaystyle a<y$ you get the same thing.
    Thus, $\displaystyle a=y$ which implies that $\displaystyle b=z$
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    Thanks for the help
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