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Let n be a natural number with exactly 2 positive prime divisors. Of n^2 has 27 divisors, how many does n have?

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Let n be a natural number with exactly 2 positive prime divisors. Of n^2 has 27 divisors, how many does n have?

Guest Jun 24, 2018
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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:

$$\text{Let n = 162 = 2\times 3^4 } \\ \text{Divisors 162: 1,2,3,6,9,18,27,54,81,162\  (10 divisors)}$$

$$\text{Let n^2 = 162^2 = 26244 = 2^2\times 3^8 } \\ \text{Divisors 162^2:\\1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, 81, 108, 162, 243, 324, 486,\\ 729, 972, 1458, 2187, 2916, 4374, 6561, 8748, 13122, 26244\  (27 divisors)}$$

heureka  Jun 25, 2018