2020 AMC 10A Problem 3

The tops and bottoms look almost the same, just written backwards. So try to put each top over its matching bottom.

1. When you multiply fractions, you can swap the bottoms around freely: it is all one big product. So rewrite the expression as
$$\frac{a-3}{3-a} \cdot \frac{b-4}{4-b} \cdot \frac{c-5}{5-c}.$$
2. Look at the first fraction. Try a number: if $a = 10$, it is $\frac{7}{-7} = -1$. In general $3 - a$ is the negative of $a - 3$, so the fraction is always $-1$. The problem says $a \ne 3$, so we never divide by zero.
3. The same thing happens with $b$ and with $c$. Each fraction is $-1$.
4. Multiply: $(-1)(-1)(-1) = -1$.

The answer is $\boxed{\textbf{(A)}\ -1}$.

Whenever you see $p - q$ and $q - p$ in the same expression, their ratio is $-1$. In a product you may re-pair tops and bottoms any way you like, so look for such pairs. Then count the minus signs: an odd number of them leaves the answer negative.

Plug in easy numbers. Take $a = 4$, $b = 5$, $c = 6$. The expression becomes
$$\frac{1}{-1} \cdot \frac{1}{-1} \cdot \frac{1}{-1} = -1.$$
Now check the choices with these numbers: (B) is $1$, (C) is $\frac{120}{60} = 2$, and (D) and (E) are $\pm\left(\frac{1}{120} - \frac{1}{60}\right)$, which are small fractions. Only (A) gives $-1$.