Math is Radical!

Algebra Level 4

$\sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006} = a\sqrt{2}+b\sqrt{3}+c\sqrt{5}$ where $a, b,$ and $c$ are positive integers. Find $abc$ .

1 solution

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May 12, 2015

Squaring both sides we get that:

$104\sqrt { 6 } +468\sqrt { 10 } +144\sqrt { 15 } +2006=2{ a }^{ 2 }+3{ b }^{ 2 }+5{ c }^{ 2 }+2\sqrt { 6 } ab+2\sqrt { 10 } ac+2\sqrt { 15 } bc$

Since $a,b,c$ are integers then

$2\sqrt { 6 } ab=104\sqrt { 6 } \Rightarrow ab=52$

$2\sqrt { 10 } ac=468\sqrt { 6 } \Rightarrow ac=234$

$2\sqrt { 15 } bc=144\sqrt { 15 } \Rightarrow bc=72$

From this it's easy to find the values of $a,b,c$ which are equal to $13,4,18$ respectively.

Therefore $abc=936$

It it not necessary to solve for the variables. Knowing that

$ab = 234$

$ac=234$

$bc=72$ ,

we can just say that $abacbc=52(234)(72) \Rightarrow (abc)^2=876096\Rightarrow abc = \boxed{936}$ .

Alex Delhumeau - 6 years, 1 month ago