Question

Let $n$ be the least positive integer such that $168$ is a factor of $1134^n$. If $m$ is the least positive integer such that $1134^n$ is a factor of $168^m$, then $m+n$ equals

Accepted Answer

C. Prime factorizations: $$168 = 2^3 	imes 3 	imes 7$$ $$1134 = 2 	imes 3^4 	imes 7$$ For $168$ to divide $1134^n = 2^n 	imes 3^{4n} 	imes 7^n$: - Power of $2$: $n \ge 3$ - Power of $3$: $4n \ge 1 \implies n \ge 1$ - Power of $7$: $n \ge 1$ Thus, the least positive integer $n = 3$. So $1134^n = 1134^3 = 2^3 	imes 3^{12} 	imes 7^3$. For $1134^3$ to divide $168^m = 2^{3m} 	imes 3^m 	imes 7^m$: - Power of $2$: $3m \ge 3 \implies m \ge 1$ - Power of $3$: $m \ge 12$ - Power of $7$: $m \ge 3$ Thus, the least positive integer $m = 12$. Therefore, $m + n = 12 + 3 = 15$.

Divisibility and Prime Factorization

Explanation

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