2020 AMC 10B Problem 12

Split the power: $20^{20} = 2^{20} \cdot 10^{20}$, so
$$\frac{1}{20^{20}} = \frac{1}{2^{20}} \cdot 10^{-20}.$$

Now $2^{20} = 1{,}048{,}576$, which is just over $10^6$. Hence
$$\frac{1}{2^{20}} = \frac{1}{1.048576 \times 10^6} \approx 0.9537 \times 10^{-6} = 9.537 \times 10^{-7}.$$
(The problem's mention of a leading $9$ confirms this.)

Multiplying by $10^{-20}$ gives $9.537\ldots \times 10^{-27}$. A number of the form $9.5\ldots \times 10^{-27}$ has its first nonzero digit in the $27$th decimal place, so there are $26$ zeros before it.

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

Powers of $10$ only shift the decimal point, so peel them off and study the leftover factor. $2^{10} \approx 10^3$ is the key estimate every contestant should know; $2^{20} \approx 1.05 \times 10^6$ slightly exceeds a million, which is what makes the leading digit $9$ rather than $1$. Then $a \times 10^{-k}$ with $1 \le a < 10$ has exactly $k - 1$ leading zeros.

Bound directly: $10^{26} < 20^{20} < 10^{27}$ because $20^{20} = 2^{20} \cdot 10^{20}$ and $10^6 < 2^{20} < 10^7$. Therefore $10^{-27} < \frac{1}{20^{20}} < 10^{-26}$, which places the first nonzero digit in position $27$, after $26$ zeros.

• Counting the zeros in 1/10^20 only (19 or 20) and ignoring the extra factor 1/2^20, or miscounting the shift by one.
• Confusing "first nonzero digit in place 27" with "27 zeros"; the number of leading zeros is one less than the position of the first nonzero digit.