Suppose that $a, b, c$, and $d$ are positive integers with $0<a<b<c<d<10$. What is the maximum possible value of $\frac{a-b}{c-d}$ ?
First, we note that $\frac{a-b}{c-d}=\frac{b-a}{d-c}$ and each of $b-a$ and $d-c$ is positive.

For $\frac{b-a}{d-c}$ to be as large as possible, we want $b-a$ to be as large as possible and $d-c$ to be as small as possible.

Since $d$ and $c$ are integers with $d>c$, then the smallest possible value of $d-c$ is 1 .

For $b-a$ to be as large as possible, we want $b$ to be as large as possible and $a$ to be as small as possible.

Since $a$ is an integer with $0<a$, then $1 \leq a$.

Since $b, c$ and $d$ are integers with $b<c<d<10$, then $b \leq 7$.

Therefore, the largest possible value of $b-a$ is $7-1$ which equals 6 .

Note that, in this case, it must be the case that $d=9$ and $c=8$ which does give $d-c=1$.

Thus, the largest possible value of $\frac{a-b}{c-d}$ is $\frac{6}{1}$ which equals 6 .

ANSWER: 6