Divisor Count of 6n
If $ n $ is a positive integer such that $ 2n $ has 28 positive divisors and $ 3n $ has 30 positive divisors, then how many positive divisors does $ 6n $ have?
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
Let $ \, 2^{e_1} 3^{e_2} 5^{e_3} \cdots \, $ be the prime factorization of $ \, n $. Then the number of positive divisors of $ \, n \, $ is $ \, (e_1 + 1)(e_2 + 1)(e_3 + 1) \cdots \; $. In view of the given information, we have \[
28 = (e_1 + 2)(e_2 + 1)P
\]and \[
30 = (e_1 + 1)(e_2 + 2)P,
\]where $ \, P = (e_3 + 1)(e_4 + 1) \cdots \; $. Subtracting the first equation from the second, we obtain $ \, 2 = (e_1 - e_2)P,
\, $ so either $ \, e_1 - e_2 = 1 \, $ and $ \, P = 2, \, $ or $ \, e_1
- e_2 = 2 \, $ and $ \, P = 1 $. The first case yields $ \, 14 = (e_1
+ 2)e_1 \, $ and $ \, (e_1 + 1)^2 = 15 $; since $ \, e_1 \, $ is a nonnegative integer, this is impossible. In the second case, $ \,
e_2 = e_1 - 2 \, $ and $ \, 30 = (e_1 + 1)e_1, \, $ from which we find $ \, e_1 = 5 \, $ and $ \, e_2 = 3 $. Thus $ \, n = 2^5 3^3, \, $ so $ \, 6n = 2^6 3^4 \, $ has $ \, (6+1)(4+1) = \boxed{35} \, $ positive divisors.
Freeflows