$$ \left\textbackslash{\begin{array}{l} a(b+1)=1443=3 \times 13 \times 37=37 \times 39 \textbackslash\textbackslash{} b(a+1)=1444=38^{2} \end{array}\right. $$  Compare and $a=37, b=38 .$ So $10 a+b=370+38=408$. 
Let $a$ and $b$ be positive integers such that $a+a b=1443$ and $a b+b=1444$. Find $10 a+b$. 
a. $$407$$ 
b. $$408$$ 
c. $$418$$ 
d. $$419$$ 
e. $$428$$ 