Solution

See that $2^{3}=8$ is less than 10, so

$$
2^{100}2^{98}=\left(2^{7}\right)^{14}>\left(10^{2}\right)^{14}=10^{28}
$$

Since $10^{28}$ is the smallest number with 29 digits, $2^{100}$ has at least 29 digits. It is thus proven that $k=29$ satisfies the condition given that $29 \leq N \leq 34$.

#
How many digits does the number $3^{100}$ have? Well, we can give an approximate answer to this question without using a calculator, simply by comparing it with powers of 10. Notice that $3^{2}<10$ allows us to conclude that $\left(3^{2}\right)^{50}=3^{100}<10^{50}$. Therefore, $3^{100}$ has at most 50 digits since $10^{50}$ is the first number with 51 digits. The number $3^{100}$ actually has 48 digits! Now it's your turn. Let $N$ be the number of digits in $2^{100}$, determine an integer $k$ such that $k \leq N \leq k+5$.