Let the speeds of Mira and Amal be \( M \) and \( A \) rounds per minute respectively.
\[ (A - M) \times 45 = 3 \Rightarrow A - M = \frac{1}{15} \]
\[ (A + M) \times 3 = 1 \Rightarrow A + M = \frac{1}{3} \]
Now solving the two equations:
\[ A - M = \frac{1}{15} \quad \text{(1)} \\ \] \[ A + M = \frac{1}{3} \quad \text{(2)} \]
Add (1) and (2):
\[ 2A = \frac{1}{15} + \frac{1}{3} = \frac{1 + 5}{15} = \frac{6}{15} = \frac{2}{5} \Rightarrow A = \frac{1}{5} \]
Substitute back into equation (2):
\[ \frac{1}{5} + M = \frac{1}{3} \Rightarrow M = \frac{1}{3} - \frac{1}{5} = \frac{5 - 3}{15} = \frac{2}{15} \]
So, Mira walks \( \frac{2}{15} \) rounds per minute.
In one hour (i.e., 60 minutes), Mira walks: \[ 60 \times \frac{2}{15} = 8 \text{ rounds} \]
Answer: 8 rounds
For any natural number $k$, let $a_k = 3^k$. The smallest natural number $m$ for which \[ (a_1)^1 \times (a_2)^2 \times \dots \times (a_{20})^{20} \;<\; a_{21} \times a_{22} \times \dots \times a_{20+m} \] is: