80
Let A, B, C’s work rates be \( a, b, c \) (work/day).
Given: \( a + b = \frac{1}{12}, b + c = \frac{1}{15}, c + a = \frac{1}{20} \).
Add all:
\[ 2(a + b + c) = \frac{1}{12} + \frac{1}{15} + \frac{1}{20} = \frac{5 + 4 + 3}{60} = \frac{12}{60} = \frac{1}{5} \] \[ a + b + c = \frac{1}{10} \] \[ a = (a + b + c) - (b + c) = \frac{1}{10} - \frac{1}{15} = \frac{3 - 2}{30} = \frac{1}{30} \] Time for A alone = \( \frac{1}{\frac{1}{30}} = 30 \) days.
Thus, the answer is 30.
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: