2000 AMC 10 Problem 15

1. Try an example first. Pick $a = 1$. The condition $ab = a - b$ becomes $b = 1 - b$, so $b = \tfrac12$. Then
$$\frac{a}{b} + \frac{b}{a} - ab = 2 + \tfrac12 - \tfrac12 = 2.$$
So $2$ is a possible value, which already answers the question. Now let us see why it is always $2$.
2. Put the two fractions together. Using the common denominator $ab$:
$$\frac{a}{b} + \frac{b}{a} = \frac{a^2 + b^2}{ab}.$$
3. Rewrite the top. Expanding $(a - b)^2 = a^2 - 2ab + b^2$ shows that $a^2 + b^2 = (a - b)^2 + 2ab$. We are told $a - b = ab$, so the top becomes $(ab)^2 + 2ab$.
4. Divide by $ab$.
$$\frac{(ab)^2 + 2ab}{ab} = ab + 2.$$
5. Subtract $ab$. The whole expression is $(ab + 2) - ab = 2$, for every pair that fits the condition.

The answer is $\boxed{\textbf{(E)}\ 2}$.

Expressions like $\frac ab + \frac ba$ can always be written using $a - b$ (or $a + b$) and $ab$. Once they are, the one given fact $ab = a - b$ lets everything cancel. When a problem asks for "a possible value" and gives you only one equation for two unknowns, the value is often the same for every valid pair, so a single example can find it.

Divide the condition $ab = a - b$ by $b$: you get $a = \frac ab - 1$, so $\frac ab = a + 1$. Divide it by $a$ instead: $b = 1 - \frac ba$, so $\frac ba = 1 - b$. Now
$$\frac ab + \frac ba - ab = (a + 1) + (1 - b) - (a - b) = 2.$$

Write the two fractions over a common denominator:
$$\frac{a}{b} + \frac{b}{a} = \frac{a^2 + b^2}{ab}.$$
Since $a^2 + b^2 = (a-b)^2 + 2ab$ and we are given $a - b = ab$, the numerator equals $(ab)^2 + 2ab$. Therefore
$$\frac{a^2+b^2}{ab} = \frac{(ab)^2 + 2ab}{ab} = ab + 2.$$
Subtracting $ab$ leaves $2$. So the expression equals $2$ for every pair satisfying the condition.

The answer is $\boxed{\textbf{(E)}\ 2}$.