Question 618060
Let


*[tex \LARGE n = 7^{k_1} \prod_{i=2}^{m} p_i^{k_i}]


where *[tex \LARGE p_i] are the prime divisors of n. Then,


*[tex \LARGE 7n = 7^{k_1 + 1} \prod_{i=2}^{m} p_i^{k_i}].


To count the number of factors of an integer, we take the k_i's, add 1 to each, and multiply. The number of factors of n is *[tex \LARGE (k_1 + 1) \prod_{i=2}^{m} (k_i + 1) = 60]


The number of factors of 7n is *[tex \LARGE (k_1 + 2)\prod_{i=2}^{m} (k_i + 1) = 80]


If we divide the first equation by the second equation, the products will cancel out, so we are left with


*[tex \LARGE \frac{k_1 + 1}{k_1 + 2} = \frac{60}{80} = \frac{3}{4}]


Therefore, *[tex \LARGE 4(k_1 + 1) = 3(k_1 + 2)], solving for k_1 we obtain *[tex \LARGE k_1 = 2]. The answer is 2.