A5. The sizes in degrees of the interior angles of a pentagon are consecutive even numbers. What is the size of the largest of these angles?
Solution
Let the interior angles of the pentagon be $2 n^{\circ},(2 n+2)^{\circ},(2 n+4)^{\circ},(2 n+6)^{\circ}$ and $(2 n+8)^{\circ}$. The interior angles of a pentagon sum to $540^{\circ}$. Therefore, $2 n+2 n+2+2 n+4+2 n+6+2 n+8=540$, which gives $10 n+20=540$ and $2 n=104$. Hence, the size of the largest angle is $(104+8)^{\circ}=112^{\circ}$.