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 $\frac{502}{1} = 502$.

The general factor is $\frac{4n+4}{4n}$: its numerator $4(n+1)$ is exactly the denominator of the next factor. So in the full product every numerator cancels against the following denominator.

What survives is the very last numerator and the very first denominator:
$$\frac{8}{4}\cdot\frac{12}{8}\cdot\frac{16}{12}\cdots\frac{2008}{2004} = \frac{2008}{4} = 502.$$

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