For positive integers $a$, $b$ and $c$, if $ab + bc = 13$ and $bc + ca = 45$, what is the minimum possible value of $a + b + c$?
Answer and solution
Correct answer: 14
Solution
$b(a+c)=13$. Since 13 is prime and $a,b,c$ are positive integers with $a+c \ge 2$: $b=1,\ a+c=13$.
$c(a+b)=45 \Rightarrow c(a+1)=45$. Substituting $c=13-a$: $(13-a)(a+1)=45 \Rightarrow a^2-12a+32=0 \Rightarrow (a-4)(a-8)=0$.
Solutions: $(a,b,c)=(4,1,9)$ or $(8,1,5)$. Both give $a+b+c=14$.
$c(a+b)=45 \Rightarrow c(a+1)=45$. Substituting $c=13-a$: $(13-a)(a+1)=45 \Rightarrow a^2-12a+32=0 \Rightarrow (a-4)(a-8)=0$.
Solutions: $(a,b,c)=(4,1,9)$ or $(8,1,5)$. Both give $a+b+c=14$.