Find the largest real number $x$ such that
\[\frac{\lfloor x \rfloor}{x} = \frac{9}{10}.\]
We can write $9x = 10 \lfloor x \rfloor.$  Since $x = \lfloor x \rfloor + \{x\},$
\[9 \lfloor x \rfloor + 9 \{x\} = 10 \lfloor x \rfloor.\]Then $9 \{x\} = \lfloor x \rfloor.$ Since $\{x\} < 1,$ $\lfloor x \rfloor = 9 \{x\} < 9.$   Thus, $\lfloor x \rfloor \le 8.$

If $\lfloor x \rfloor = 8,$ then $\{x\} = \frac{8}{9},$ so the largest possible value of $x$ is $8 + \frac{8}{9} = \boxed{\frac{80}{9}}.$
\frac{80}{9}
Level 4
Intermediate Algebra
Problem: Find the largest real number $x$ such that
\[\frac{\lfloor x \rfloor}{x} = \frac{9}{10}.\]
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the equation \(\frac{\lfloor x \rfloor}{x} = \frac{9}{10}\), we start by letting \(\lfloor x \rfloor = n\), where \(n\) is an integer. This gives us the equation:

\[
\frac{n}{x} = \frac{9}{10}.
\]

Rearranging this equation leads to:

\[
10n = 9x \quad \Rightarrow \quad x = \frac{10n}{9}.
\]

Since \(n = \lfloor x \rfloor\), it follows that \(n \leq x < n + 1\). Substituting \(x\) into this inequality gives:

\[
n \leq \frac{10n}{9} < n + 1.
\]

We will analyze each part of this compound inequality separately.

1. For the left side of the inequality \(n \leq \frac{10n}{9}\):

Multiplying both sides by 9 (which is positive) yields:

\[
9n \leq 10n \quad \Rightarrow \quad 0 \leq n.
\]

This indicates that \(n\) must be a non-negative integer.

2. For the right side of the inequality \(\frac{10n}{9} < n + 1\):

Multiplying both sides by 9 gives:

\[
10n < 9(n + 1) \quad \Rightarrow \quad 10n < 9n + 9 \quad \Rightarrow \quad n < 9.
\]

Thus, \(n\) can take integer values from 0 to 8, inclusive.

Next, we will find the corresponding values of \(x\) for each integer \(n\):

- If \(n = 0\), then \(x = \frac{10 \cdot 0}{9} = 0\).
- If \(n = 1\), then \(x = \frac{10 \cdot 1}{9} = \frac{10}{9}\).
- If \(n = 2\), then \(x = \frac{10 \cdot 2}{9} = \frac{20}{9}\).
- If \(n = 3\), then \(x = \frac{10 \cdot 3}{9} = \frac{30}{9} = \frac{10}{3}\).
- If \(n = 4\), then \(x = \frac{10 \cdot 4}{9} = \frac{40}{9}\).
- If \(n = 5\), then \(x = \frac{10 \cdot 5}{9} = \frac{50}{9}\).
- If \(n = 6\), then \(x = \frac{10 \cdot 6}{9} = \frac{60}{9} = \frac{20}{3}\).
- If \(n = 7\), then \(x = \frac{10 \cdot 7}{9} = \frac{70}{9}\).
- If \(n = 8\), then \(x = \frac{10 \cdot 8}{9} = \frac{80}{9}\).

Among these values, the largest \(x\) occurs when \(n = 8\), yielding:

\[
x = \frac{80}{9}.
\]

This value satisfies both the original equation and the condition \(n = \lfloor x \rfloor\). Therefore, the final answer is \(\frac{80}{9}\).<|END_OF_LATENT|> The final answer is $\frac{80}{9}$.