2005 AMC 12B Problem 21

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.