Let n be a natural number with exactly 2 positive prime divisors. Of n^2 has 27 divisors, how many does n have?
+26367 Let n be a natural number with exactly 2 positive prime divisors.
If n^2 has 27 divisors, how many does n have?
\(\text{ Let $ n = p^n\times q^m $ and $p$ and $q$ are prime }\)
The number of devisors of \(n\) is \((n+1)(m+1)\)
\(\text{ Let $ n^2 = p^{2n} \times q^{2m} $ } \)
The number of devisors of \(n^2\) is \((2n+1)(2m+1)\)
\(\begin{array}{|rcll|} \hline (2n+1)(2m+1) &=& 27 \quad & | \quad 27=3^3 \\\\ \underbrace{(2n+1)}_{=3}\underbrace{(2m+1)}_{=3^2} &=& 3^3\\\\ 2n+1 &=& 3 \\ 2n &=& 2 \\ \mathbf{n} &\mathbf{=} & \mathbf{ 1 } \\\\ 2m+1 &=& 3^2 \\ 2m+1 &=& 9 \\ 2m &=& 8 \\ \mathbf{m} &\mathbf{=} & \mathbf{4} \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline && (n+1)(m+1) \quad & | \quad n=1 \text{ and } m = 4 \\ &=& (1+1)(4+1) \\ &=& 2\cdot 5 \\ &\mathbf{=} & \mathbf{10} \\ \hline \end{array}\)
n has 10 divisors.
Example:
\(\text{Let $n = 162 = 2\times 3^4$ } \\ \text{Divisors $162: 1,2,3,6,9,18,27,54,81,162\ $ ($10$ divisors)}\)
\(\text{Let $n^2 = 162^2 = 26244 = 2^2\times 3^8 $} \\ \text{Divisors 162^2:$\\1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, 81, 108, 162, 243, 324, 486,\\ 729, 972, 1458, 2187, 2916, 4374, 6561, 8748, 13122, 26244\ $ ($27$ divisors)}\)
