The following are the prime factorizations of 1134 and 168:
\(168 = 2^3 × 3 × 7\)
1134 is equal to \(2 × 3^4 × 7. \)
Evidently, three is the least positive integral number of n that allows 168 to be a factor of 1134n.
\(1134^3 = 2^3 × 3^{12} × 7^3 = 1134^n \)
It is evident that 12 is the least positive integral value of m that allows \(1134^3\) to be a factor of 168m.
It follows that \(m + n = 12 + 3 = 15\)
The correct answer is 15.
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: