For positive integers $a$, $b$ and $c$, if $ab + bc = 13$ and $bc + ca = 45$, what is the minimum possible value of $abc$?
Answer and solution
Correct answer: 36
Solution
$b(a + c) = 13$. Since 13 is prime and $a + c \ge 2$: $b = 1,\; a + c = 13$.
$c(a + b) = 45 \Rightarrow c(a + 1) = 45 \Rightarrow c(14 - c) = 45$
$c^2 - 14c + 45 = 0 \Rightarrow c = 5$ or $9$
$c = 5, a = 8$: $abc = 40$; $\;c = 9, a = 4$: $abc = 36$
Minimum $abc = 36$
$c(a + b) = 45 \Rightarrow c(a + 1) = 45 \Rightarrow c(14 - c) = 45$
$c^2 - 14c + 45 = 0 \Rightarrow c = 5$ or $9$
$c = 5, a = 8$: $abc = 40$; $\;c = 9, a = 4$: $abc = 36$
Minimum $abc = 36$