2002 AMC 10A Problem 1

Every term is a power of $10$, so factor the lowest power out of the numerator and the denominator separately:
$$\frac{10^{2000}+10^{2002}}{10^{2001}+10^{2001}} = \frac{10^{2000}\,(1+10^2)}{10^{2001}\cdot 2} = \frac{101}{10\cdot 2} = \frac{101}{20}.$$

Now $\frac{101}{20} = 5.05$, which is closest to $5$.

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

Sums of powers with enormous exponents are meant to be factored, not computed: dividing top and bottom by the smallest power present reduces the whole expression to small numbers. The general lesson is that $10^{n} + 10^{n+2} = 10^{n}(1 + 100)$ no matter how big $n$ is, so the exponent $2000$ is pure decoration.

Estimate: the numerator is dominated by $10^{2002}$ and the denominator equals $2\cdot 10^{2001}$, so the ratio is roughly $\frac{10^{2002}}{2\cdot 10^{2001}} = 5$. Since $10^{2000}$ adds only $1\%$ to the numerator, the answer stays closest to $5$.

• Cancelling the 10-powers term by term (as if the fraction were 10^{-1} + 10^{1} over 2) and answering 1 or 10.
• Writing $10^{2001} + 10^{2001} = 10^{2002}$ instead of $2\cdot 10^{2001}$, which gives approximately $1$.