2005 AMC 12B 第 21 题

Write $n = 7^k Q$ with $Q$ not divisible by $7$; let $d$ be the number of divisors of $Q$

Then $n$ has $(k+1)d$ divisors and $7n$ has $(k+2)d$ divisors

Write $n = 7^k Q$ where $Q$ is not divisible by $7,$ and let $d$ be the number of divisors of $Q.$ Then $n$ has $(k + 1)d = 60$ divisors and $7n = 7^{k+1}Q$ has $(k + 2)d = 80$ divisors.

Dividing, $\dfrac{k + 2}{k + 1} = \dfrac{80}{60} = \dfrac43,$ so $3(k + 2) = 4(k + 1),$ giving $k = 2.$

Thus, the correct answer is C.