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**demode** If d(n) denotes the number of divisors (+ve) of integer n. Prove that $\displaystyle d(n) \leq 2 \sqrt{n}$.

So, we are not told if n is prime or composite, so I don't know how to approach this proble. But if we suppose the prime factorization of n is $\displaystyle n=p_1^{\alpha_1} p_2^{\alpha_2}...p_r^{\alpha_r}$,

then number of divisors of n are: $\displaystyle d(n)=(\alpha +1)...(\alpha_r+1)$.

So, how do I show that $\displaystyle d(n)=(\alpha +1)...(\alpha_r+1) \leq 2 \sqrt{n}$?

I also know that if n is composite then the number of its divisors are less than or equal to $\displaystyle \left\lfloor \sqrt{n} \right\rfloor$ (the notation means the integer part of $\displaystyle \sqrt{n}$),

This is false: 36 has 9 positive divisors, and $\displaystyle 9>6=\sqrt{36}$ . What did you __really__ mean here?

Tonio

but I don't know if that helps at all.

Any helps is really appreciated.