Algebra with Sets 1

Note that $A\cap C = \{1,5\}$ if and only if $a-b=5$ . Furthermore, $A\cup B = \{1,2,4,5,6,a^2+b\}$ . Thus $a^2+b \in \{1,2,4,5,6\}$ . Together with our initial conclusion, we get $a^2 + a - 5 \in \{1,2,4,5,6\}$ . Thus we have five possible equations, each yielding two possible values for $a$ (it is straightforward to check that all the quadratics indeed yield solutions). Since adding a constant to a quadratic necessarily changes the roots, these values must all be distinct. And each value of $a$ corresponds to precisely one value of $b$ .

Therefore there are $5\times 2 = \boxed{10}$ solutions.