2002 AMC 10A Problem 1

Numbers like $10^{2000}$ are far too big to work out, so the plan is to pull out a common factor and let it cancel.

1. Try a small version first. Replace the exponents $2000, 2001, 2002$ by $0, 1, 2$:
$$\frac{10^0 + 10^2}{10^1 + 10^1} = \frac{1 + 100}{10 + 10} = \frac{101}{20}.$$
We will see that the real fraction has exactly the same value.

2. The top. $10^{2002}$ is $10^{2000}$ times $100$. So
$$10^{2000} + 10^{2002} = 10^{2000} \times (1 + 100) = 10^{2000} \times 101.$$

3. The bottom. It is two copies of $10^{2001}$, and $10^{2001}$ is $10^{2000}$ times $10$. So
$$10^{2001} + 10^{2001} = 2 \times 10 \times 10^{2000} = 10^{2000} \times 20.$$

4. Cancel. Now $10^{2000}$ multiplies the whole top and the whole bottom, so it cancels:
$$\frac{10^{2000} \times 101}{10^{2000} \times 20} = \frac{101}{20} = 5.05.$$

1. Of the choices, $5.05$ is closest to $5$.

The answer is $\boxed{\textbf{(D)}\ 5}$.

When the exponents are huge, the exponents are not the point: every term is a power of the same number, so the smallest one can be taken out of each sum. After that only small numbers are left. Next time you see sums of powers with big exponents, take out the smallest power first.

Estimate. On top, $10^{2002}$ is $100$ times bigger than $10^{2000}$, so the top is almost exactly $10^{2002}$. The bottom is $2 \times 10^{2001}$. So the fraction is about
$$\frac{10^{2002}}{2 \times 10^{2001}} = \frac{10}{2} = 5.$$
The small term $10^{2000}$ only adds about $1\%$, so the answer stays closest to $5$.