2020 AMC 10B Problem 12

1. A small example first. Take $\frac{1}{400}$. The number $400$ has $3$ digits, so it lies between $100$ and $1000$. Then $\frac{1}{400}$ lies between $\frac{1}{1000} = 0.001$ and $\frac{1}{100} = 0.01$. Indeed $\frac{1}{400} = 0.0025$: there are $2$ zeros after the decimal point. A $3$-digit number gave $2$ zeros.
2. So count the digits of $20^{20}$. Split it up:
$$20^{20} = 2^{20} \times 10^{20}.$$
3. $2^{10} = 1024$, so $2^{20} = 1024 \times 1024 = 1{,}048{,}576$. That has $7$ digits.
4. Multiplying by $10^{20}$ just writes $20$ zeros after it. So $20^{20}$ has $7 + 20 = 27$ digits. That means
$$10^{26} \le 20^{20} < 10^{27}.$$
(It cannot equal $10^{26}$, since $1{,}048{,}576$ is not a power of $10$.)
5. Flip everything over:
$$10^{-27} < \frac{1}{20^{20}} < 10^{-26}.$$
6. $10^{-27}$ has its first nonzero digit in the $27$th place after the decimal point, and $10^{-26}$ has it in the $26$th place. Our number is strictly between them, so its first nonzero digit is in the $27$th place. That leaves $26$ zeros in front of it.

The answer is $\boxed{\textbf{(D)}\ 26}$.

Multiplying or dividing by $10$ only moves the decimal point, so pull out the powers of $10$ and look at what is left. The fact $2^{10} = 1024$, just over $1000$, is worth remembering. It tells you $2^{20}$ is just over a million, and that is why the first nonzero digit of $\frac{1}{20^{20}}$ is a $9$ (since $\frac{1}{1.05}$ is about $0.95$).

Estimate the leading digit directly. $\frac{1}{2^{20}} = \frac{1}{1{,}048{,}576}$ is a little less than one millionth, about $0.00000095$, which is $9.5 \times 10^{-7}$. Dividing by another $10^{20}$ moves the decimal point $20$ more places, giving about $9.5 \times 10^{-27}$. Written out, the $9$ sits in the $27$th decimal place, so $26$ zeros come before it.