211. How many digits does the number $2^{100}$ have?
211. It is easy to check that $2^{10}=1024$; thus, $2^{100}=$ $=1024^{10}$. Since $1000^{10}=10^{30}$ represents a number consisting of a one followed by 30 zeros, and $1024^{10}>1000^{10}$, the number $2^{100}=$ $=1024^{10}$ cannot have fewer than 31 digits. On the other hand,

$$
\begin{aligned}
\frac{1024^{10}}{1000^{10}}<\left(\frac{1025}{1000}\right)^{10} & =\left(\frac{41}{40}\right)^{10}= \\
& =\frac{41}{40} \cdot \frac{41}{40} \cdot \frac{41}{40} \cdot \frac{41}{40} \cdot \frac{41}{40} \cdot \frac{41}{40} \cdot \frac{41}{40} \cdot \frac{41}{40} \cdot \frac{41}{40} \cdot \frac{41}{40}< \\
& <\frac{41}{40} \cdot \frac{40}{39} \cdot \frac{39}{38} \cdot \frac{38}{37} \cdot \frac{37}{36} \cdot \frac{36}{35} \cdot \frac{35}{34} \cdot \frac{34}{33} \cdot \frac{33}{32} \cdot \frac{32}{31}=\frac{41}{31}<10
\end{aligned}
$$

since

$$
\begin{gathered}
\frac{41}{40}<\frac{40}{39}<\frac{39}{38} \ldots, \text { and so on. } \\
\left(\text { because } \frac{41}{40}=1+\frac{1}{40} ; \frac{40}{39}=1+\frac{1}{39}, \text { and so on. }\right) .
\end{gathered}
$$

Thus,

$$
2^{100}=1024^{10}<10 \cdot 1000^{10}
$$

from which it follows that $2^{100}$ contains fewer than 32 digits. Therefore, the number $2^{100}$ consists of 31 digits.

Note. This problem is very easy to solve using logarithms: since $\log 2=0.30103$, then $\log 2^{100}=$ $=100 \log 2=30.103$ and, consequently, the number $2^{100}$ has 31 digits. The point of the problem is to obtain this result without using logarithms.