Question

El número $\small \sqrt{104\sqrt{6} + 468\sqrt{10} + 144\sqrt{15} + 2006}$ puede escribirse como $a\sqrt{2} + b\sqrt{3} + c\sqrt{5},$ donde $a,$ $b,$ y $c$ son enteros positivos. Halla $a \cdot b \cdot c.$

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

Respuesta

Solución:

Elevar al cuadrado $a\sqrt{2} + b\sqrt{3} + c\sqrt{5}$ da $\begin{aligned} &2a^2 + 3b^2 + 5c^2 \\ &\quad {}+ 2ab\sqrt{6} + 2ac\sqrt{10} \\ &\quad {}+ 2bc\sqrt{15}. \end{aligned}$ Igualar coeficientes produce $2ab = 104,$ $2ac = 468,$ $2bc = 144,$ es decir $ab = 52,$ $ac = 234,$ $bc = 72,$ junto con $2a^2 + 3b^2 + 5c^2 = 2006.$

Entonces $\begin{aligned} (abc)^2 &= (ab)(ac)(bc) \\ &= 52 \cdot 234 \cdot 72 \\ &= 876096 = 936^2, \end{aligned}$ así que $abc = 936.$ Como comprobación, $a = \frac{abc}{bc} = 13,$ $b = \frac{abc}{ac} = 4,$ $c = \frac{abc}{ab} = 18,$ y $\begin{aligned} &2 \cdot 169 + 3 \cdot 16 + 5 \cdot 324 \\ &= 338 + 48 + 1620 \\ &= 2006, \end{aligned}$ como se requería.

Por lo tanto $a \cdot b \cdot c = 936.$

Squaring $a\sqrt{2} + b\sqrt{3} + c\sqrt{5}$ gives $\begin{aligned} &2a^2 + 3b^2 + 5c^2 \\ &\quad {}+ 2ab\sqrt{6} + 2ac\sqrt{10} \\ &\quad {}+ 2bc\sqrt{15}. \end{aligned}$ Matching coefficients yields $2ab = 104,$ $2ac = 468,$ $2bc = 144,$ that is $ab = 52,$ $ac = 234,$ $bc = 72,$ along with $2a^2 + 3b^2 + 5c^2 = 2006.$

Then $\begin{aligned} (abc)^2 &= (ab)(ac)(bc) \\ &= 52 \cdot 234 \cdot 72 \\ &= 876096 = 936^2, \end{aligned}$ so $abc = 936.$ As a check, $a = \frac{abc}{bc} = 13,$ $b = \frac{abc}{ac} = 4,$ $c = \frac{abc}{ab} = 18,$ and $\begin{aligned} &2 \cdot 169 + 3 \cdot 16 + 5 \cdot 324 \\ &= 338 + 48 + 1620 \\ &= 2006, \end{aligned}$ as required.

Therefore $a \cdot b \cdot c = 936.$

2006 AIME I Problema 5

Solución:

El Problema 5 en otros años