Find the largest solution to \[\lfloor x \rfloor = 5 + 100 \{ x \},\]where $\{x\} = x - \lfloor x \rfloor.$
This problem involves the floor function $\lfloor x \rfloor$, which rounds down $x$ to the nearest integer, and the fractional part function $\{x\}$, which is the difference between $x$ and $\lfloor x \rfloor$. For example, if $x = 3.7$, then $\lfloor x \rfloor = 3$ and $\{x\} = 0.7$.
[*) To solve for $x$, I need to get rid of the floor and fractional part functions somehow. I know that $\{x\}$ is always between $0$ and $1$, so $100 \{x\}$ is between $0$ and $100$. Therefore, the right-hand side of the equation is between $5$ and $105$. The left-hand side of the equation is an integer, so it can only take discrete values in that interval. That means there are only a finite number of possible solutions for $x$. I can try to find them by testing different values of $\lfloor x \rfloor$ and seeing if they satisfy the equation. For example, if $\lfloor x \rfloor = 6$, then the equation becomes $6 = 5 + 100 \{x\}$, which implies that $\{x\} = 0.01$. This gives me one solution: $x = 6 + 0.01 = 6.01$. If $\lfloor x \rfloor = 7$, then the equation becomes $7 = 5 + 100 \{x\}$, which implies that $\{x\} = 0.02$. This gives me another solution: $x = 7 + 0.02 = 7.02$. I can keep doing this until I reach the upper bound of the right-hand side. If $\lfloor x \rfloor = 105$, then the equation becomes $105 = 5 + 100 \{x\}$, which implies that $\{x\} = 1$. However, this is not possible, because $\{x\}$ can never be exactly $1$. Therefore, the largest possible value of $\lfloor x \rfloor$ is $104$. If $\lfloor x \rfloor = 104$, then the equation becomes $104 = 5 + 100 \{x\}$, which implies that $\{x\} = 0.99$. This gives me the largest solution: $x = 104 + 0.99 = 104.99$.

