AIME I Question

**Jameson** · March 12th 2006, 12:38 PM

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?

**ThePerfectHacker** · March 12th 2006, 01:12 PM

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,
$abc=936$

**ThePerfectHacker** · March 12th 2006, 02:06 PM

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 $b=z$
--------------

**Jameson** · March 12th 2006, 02:14 PM

Thanks for the help

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