Problem: For any positive integer $n$, prove that $\tau(n)\leq 2\sqrt{n}.$

My Attempt: For every divisor $d_1<d_2...<d_k$ less than $\sqrt{n}$ we have divisors $n/d_1,n/d_2......n/d_k$. Thus $\tau(n)\leq 2k.$ How do we show that $k\leq \sqrt{n}$?

Note: Please do not mark this question as a duplicate. I have read other answers to this question and most of them provide hints or unjustified assertions such as the number of divisors less that $\sqrt{n}$ are less than $\sqrt{n}$ itself. I specifically want to know why is $k\leq \sqrt{n}$ true in order to complete the proof.

  • $\begingroup$ What is $\tau(n)$? The number of divisor of $n$? $\endgroup$ – Arthur Nov 10 '16 at 9:03
  • $\begingroup$ @Arthur Yes. $\tau(n)$ is the number of positive divisors of $n$. $\endgroup$ – nls Nov 10 '16 at 9:23

Well, your proof is almost complete. Notice that each of the $d_i$ is less than $\sqrt{n}$ and they are different. There could be no more than $\sqrt{n}$ positive integers less than $\sqrt{n}$. So $k\leq\sqrt{n}$.


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