How many distinct rectangles are there with integer side lengths such that the numerical value of area of the rectangle in square units is equal to $5$ times the numerical value of the perimeter in units? (Two rectangles are considered to be distinct if they are not congruent.)
To find the rectangles that satisfy the given condition, I need to relate the area and the perimeter of a rectangle in terms of its side lengths. Let $x$ and $y$ be the side lengths of the rectangle. Then the area is $xy$ and the perimeter is $2(x+y)$. The condition states that $xy = 5 \cdot 2(x+y)$. Simplifying, I get $xy - 10x - 10y = 0$. This is a quadratic equation in $x$ and $y$. To solve it, I can try to factor it into two linear terms. I notice that $xy - 10x - 10y + 100 = (x-10)(y-10)$. So, adding $100$ to both sides, I get $(x-10)(y-10) = 100$. Now, I need to find all the pairs of factors of $100$ that are positive integers, since $x$ and $y$ have to be positive side lengths. The factors of $100$ are $1, 2, 4, 5, 10, 20, 25, 50,$ and $100$. The pairs of factors that multiply to $100$ are $(1,100), (2,50), (4,25), (5,20),$ and $(10,10)$. These correspond to the pairs of side lengths $(11,110), (12,60), (14,35), (15,30),$ and $(20,20)$ for the rectangle. These are all the possible rectangles that satisfy the condition, and they are all distinct. Therefore, the answer is $5$. 