Find the smallest positive integer that has exactly 16 positive divisors.
I kinda know how to approach this problem but I'm not sure how to answer it in a formal way.
Please help!!! Thank you in advance.
Let $\displaystyle n = \prod_{i=1}^k p_i^{e_i}$ be the prime factorization of $\displaystyle n$.
If you recall the properties of $\displaystyle \tau (n)$, we want: $\displaystyle \tau (n) = \prod_{i=1}^k \left(e_i+1\right) = 16$
Now, $\displaystyle 16 = 2^4$ can be written as a product of integers in only 5 ways:
- $\displaystyle 2^4 = 16$
- $\displaystyle 2^3 \cdot 2 = 8 \cdot 2$
- $\displaystyle 2^2 \cdot 2^2 = 4 \cdot 4$
- $\displaystyle 2^2 \cdot 2 \cdot 2 = 4 \cdot 2 \cdot 2$
- $\displaystyle 2 \cdot 2 \cdot 2 \cdot 2 $
So our possibilities:
- $\displaystyle \tau (n) = e_1 + 1 = 16 \ \Rightarrow \ e_1 = 15$
- $\displaystyle \tau (n) = (e_1 + 1)(e_2 + 1) = (8)(2) \ \Rightarrow \ e_1 = 7, \ e_2 = 2$
- $\displaystyle \tau (n) = (e_1 + 1)(e_2 + 1) = (4)(4) \ \Rightarrow \ e_1 = 3, \ e_2 = 3$
- $\displaystyle \tau (n) = (e_1 + 1)(e_2 + 1)(e_3 + 1) = (4)(2)(2) \ \Rightarrow \ e_1 = 3, \ e_2 = 1, \ e_3 = 1$
- $\displaystyle \tau (n) = (e_1 + 1)(e_2 + 1)(e_3 + 1)(e_4 + 1) = (2)(2)(2)(2) \ \Rightarrow \ e_1 = 1, \ e_2 = 1, \ e_3 = 1, \ e_4 = 1$
Since we want the smallest integer, we consider the smallest primes in the factorization of $\displaystyle n$. So this narrows our list down to:
$\displaystyle n = 2^{15} \quad 2^7 \cdot 3^1\quad 2^3 \cdot 3^3\quad 2^3 \cdot 3^1 \cdot 5^1\quad 2 \cdot 3 \cdot 5 \cdot 7$