Four "Lets" and three sets are perfect with three variables

Algebra Level 5

Let $X = \{ 3a^{2} + 27b^{2} + 5c^{2} - 18ab - 30c + 237 | a, b, c \in Z^{+} \}$ .

Let $x_{0} \in X$ be the lowest value of the set $X$ .

Let $Y = \{(a, b, c) | 3a^{2} + 27b^{2} + 5c^{2} - 18ab - 30c + 237 = x_{0}\}$

Let $Z = \{a + b + c | (a, b, c) \in Y\}$

Find the member in $Z$ which has the least value.

Clarification : $Z^{+}$ denote the set of positive integers

1 solution

Sanchayapol Lewgasamsarn
Apr 12, 2014

From $X$ :

$3a^{2} + 27b^{2} + 5c^{2} - 18ab - 30c + 237$

$= 3(a - 3b)^{2} + 5(c - 3)^{2} + 192 \geq 192$ .

So, $x_0 = 192$ .

From $Y$ :

$3a^{2} + 27b^{2} + 5c^{2} - 18ab - 30c + 237 = 192$

$3(a - 3b)^{2} + 5(c - 3)^{2} = 0$

This is only possible when

$a = 3b$ and $c = 3$ .

Since $b \in Z^{+}, (b)_{min} = 1$ .

So, $(a)_{min} = 3$ .

So, $(Z)_{min} = (a + b + c)_{min} = (3 + 1 + 3) = \boxed{7}$