### Instruction: 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.).### Response: 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 $5$.