nicetry007 wrote: Suppose $ n = a \cdot b$. We have $ n-20 \; \leq \; a \;\leq \; n-12$ $ n-20 \; \leq \; b \;\leq \; n-12$. multiplying the two inequalities, we get $ (n-20)^{2}\; \leq \; n \;\leq \; (n-12)^{2}$ The first inequality $ (n-20)^{2}\; \leq \; n$ gives us $ 16 \; \leq \; n \; \leq \;25$. The second inequality $ n \; \leq \;(n-12)^{2}$ gives us $ n\; \leq \;9$ or $ n \; \geq \;16$. Since we need to satisfy both the inequalities simultaneously, we get $ 16 \; \leq \; n \; \leq \;25$. Suppose $ n$ is even. Then we have $ n-20 \; \leq \; 2 \;\leq \; n-12$, which gives $ 14 \; \leq \; n \; \leq \;22$ and $ n-20 \; \leq \; \frac{n}{2}\;\leq \; n-12$, which gives $ 24 \; \leq \; n \; \leq \;40$. Since there is no $ n$ that satisfies both the inequalities, $ n$ cannot be even. This leaves us with $ n = 21$ and $ n=25$. It is easy to verify that both of them satisfy the required conditions.

n is one such composite number let $ \ d_{1} $ is a proper divisor of n. $\ d_{2}= \frac{n}{d_{1}} $ so $ \ n-20 \leq d_{1} \leq n-12 $ and $ \ n-20 \leq d_{2} \leq n-12 $ so $ \ (n-20)^{2} \leq d_{1}d_{2} \leq (n-12)^{2} $ [ if $ \ n \geq 20 $] $ \ (n-20)^{2} \leq n \leq (n-12)^{2} $ [ if $ \ n \geq 20 $] for $ \ n \geq 20 $ from $ \ (n-20)^{2} \leq n $ we get $ \ n \leq 25 $ for $ \ n < 20 $ from $ \ d_{1} \leq n-12 $ we see $ \ n \geq 14 $ so $ \ 14 \leq n \leq 25 $ so the only possibilities are 14,15,16,18,20,21,22,24,25. from here easy to see 21 and 25 are only solutions.