### Instruction: For all real numbers $r$ and $s$, define the mathematical operation $\#$ such that the following conditions apply: $r\ \#\ 0 = r, r\ \#\ s = s\ \#\ r$, and $(r + 1)\ \#\ s = (r\ \#\ s) + s + 1$. What is the value of $11\ \#\ 5$?### Response: This problem involves a new operation $\#$ that has some properties that are similar to addition, but not exactly the same.
I wonder if I can use the properties to simplify the expression $11\ \#\ 5$ somehow.
The first property says that $r\ \#\ 0 = r$, so I can try to rewrite $11$ as something involving zero.
One way to do that is to use the fact that $11 = 10 + 1$, so $11\ \#\ 5 = (10 + 1)\ \#\ 5$.
The third property says that $(r + 1)\ \#\ s = (r\ \#\ s) + s + 1$, so I can apply that to the expression $(10 + 1)\ \#\ 5$.
That gives me $(10 + 1)\ \#\ 5 = (10\ \#\ 5) + 5 + 1$.
Now I have reduced the problem to finding $10\ \#\ 5$.
I can repeat the same process and write $10 = 9 + 1$, so $10\ \#\ 5 = (9 + 1)\ \#\ 5 = (9\ \#\ 5) + 5 + 1$.
Continuing this way, I get $9\ \#\ 5 = (8 + 1)\ \#\ 5 = (8\ \#\ 5) + 5 + 1$, and so on.
I can see a pattern emerging: each time I reduce the first argument of $\#$ by one, I add $5 + 1$ to the result.
So I can write $11\ \#\ 5 = (10\ \#\ 5) + 5 + 1 = (9\ \#\ 5) + 2(5 + 1) = (8\ \#\ 5) + 3(5 + 1) = \cdots = (0\ \#\ 5) + 11(5 + 1)$.
The second property says that $r\ \#\ s = s\ \#\ r$, so I can swap $0$ and $5$ and get $0\ \#\ 5 = 5\ \#\ 0$.
The first property says that $r\ \#\ 0 = r$, so I can simplify $5\ \#\ 0 = 5$.
Therefore, $11\ \#\ 5 = 5 + 11(5 + 1)$.
I can use the distributive property of multiplication over addition and get $11\ \#\ 5 = 5 + 55 + 11 = 71$.