Find a + b + c + bc.

Question

My math trainer friend ask my help to look for the shortest possible solution for this problem:

Let $a$, $b$ and $c$ be positive integers such that $$\left \{\begin{matrix}a + b + ab = 15 \\ b + c + bc = 99 \\ c + a + ca = 399\end{matrix}\right. $$

Find $a + b + c + bc.$

I tried elimination but it took us for about 3 min to do it. This question was intended for 15 seconds only.

Answer 1

Adding $1$ to each equation, we get that $$\left \{\begin{matrix}a + b + ab +1 = 16 \\ b + c + bc+1 = 100 \\ c + a + ca+1 = 400\end{matrix}\right. $$ Factoring, we get $$\left \{\begin{matrix}(b+1)(a+1) = 16 \\ (b+1)(c+1) = 100 \\ (c+1)(a+1) = 400\end{matrix}\right. $$ Dividing equation 2 from equation 3 and multiplying the result with equation 1, we get that $(a+1)^2=64$, meaning $a+1=8$. Equation one gives us $b+1=2$, and equation three gets us $c+1=50$, getting that $$a+b+c+bc=7+1+49+49=106$$

Answer 2

$(a+1)(b+1) = 16 $
$(b+1)(c+1) = 100$
$(c+1)(a+1) = 400$
Multiply all $3$ equations to get value of $(a+1)(b+1)(c+1) = 800$ then
$c+1 = 50,\quad a+1 = 8,\quad b+1 = 2$ , required is $a+(99) = 106$

Answer 3

Also elimination leads to the result (as the OP wanted), namely we obtain exactly two solutions over any field of characteristic zero: $$ (a,b,c)=(-9,-3,-51),(7,1,49). $$ Then $a+b+c+bc=7+1+49=106$, since only the positive solution was asked.