Problem 1338

1. We are given the floor function $\lfloor x \rfloor$, which returns the greatest integer less than or equal to $x$. We need to find the maximum real number $z$ such that $\lfloor \frac{5}{z} \rfloor + \lfloor \frac{6}{z} \rfloor = 7$.

2. Let's denote $\lfloor \frac{5}{z} \rfloor$ by $a$ and $\lfloor \frac{6}{z} \rfloor$ by $b$. We are given that $a + b = 7$.

3. Since $a$ and $b$ are integers, we need to find values of $a$ and $b$ such that $a + b = 7$. The possible pairs $(a, b)$ are:
- $(3, 4)$
- $(4, 3)$
- $(2, 5)$
- $(5, 2)$
- $(1, 6)$
- $(6, 1)$
- $(0, 7)$
- $(7, 0)$

4. We need to check which of these pairs satisfy the conditions $\lfloor \frac{5}{z} \rfloor = a$ and $\lfloor \frac{6}{z} \rfloor = b$.

5. Let's start with the pair $(3, 4)$:
- $\lfloor \frac{5}{z} \rfloor = 3$ implies $3 \leq \frac{5}{z} < 4$.
- $\lfloor \frac{6}{z} \rfloor = 4$ implies $4 \leq \frac{6}{z} < 5$.

6. Solving these inequalities:
- $3 \leq \frac{5}{z} < 4$ implies $\frac{5}{4} < z \leq \frac{5}{3}$.
- $4 \leq \frac{6}{z} < 5$ implies $\frac{6}{5} < z \leq \frac{6}{4}$.

7. The intersection of these intervals is:
- $\frac{6}{5} < z \leq \frac{5}{3}$ and $\frac{5}{4} < z \leq \frac{6}{4}$.
- The common interval is $\frac{6}{5} < z \leq \frac{6}{4}$.

8. To find the maximum value of $z$, we take the upper bound of the interval:
- $z = \frac{6}{4} = 1.5$.

9. Therefore, the maximum value of $z$ is $1.5$. We need to find $20z$:
- $20z = 20 \times 1.5 = 30$.

The final answer is $\boxed{30}$.