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Math Help - AIME I Question

  1. #1
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    AIME I Question

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

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

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