2008 AMC 10A Problem 5

1. Look at the first few fractions: $\frac{8}{4}\cdot\frac{12}{8}\cdot\frac{16}{12}$. The $8$ on top of the first fraction matches the $8$ on the bottom of the second. The $12$ on top of the second matches the $12$ on the bottom of the third.
2. So these cancel. For the first three fractions,
$$\frac{8}{4}\cdot\frac{12}{8}\cdot\frac{16}{12} = \frac{16}{4}.$$
3. The same thing happens all the way along: every top number cancels with the bottom number of the fraction after it.
4. The only numbers with no partner are the bottom of the very first fraction, $4$, and the top of the very last fraction, $2008$. So the whole product is
$$\frac{2008}{4} = 502.$$

The answer is $\boxed{\textbf{(B)}\ 502}$.

When a long product has each top matching the next bottom, everything in the middle cancels and only the two ends are left. This is called a telescoping product, because it collapses like a folding telescope. Whenever neighbouring fractions share a number, check for this before doing any arithmetic.

Every number here is a multiple of $4$, so simplify each fraction first: $\frac{8}{4} = \frac{2}{1}$, $\frac{12}{8} = \frac{3}{2}$, $\frac{16}{12} = \frac{4}{3}$, and so on, up to $\frac{2008}{2004} = \frac{502}{501}$. The product $\frac{2}{1}\cdot\frac{3}{2}\cdot\frac{4}{3}\cdots\frac{502}{501}$ cancels down to $\frac{502}{1} = 502$.