# Need a proof

Prove that for all $$n \geq 1$$ , $\tau(n) \leq 2\sqrt n$

Where $$\tau (n)$$ denotes total number of positive divisors of $$n$$

Note by Chinmay Sangawadekar
1 year, 7 months ago

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@Chinmay Sangawadekar- you see divisors come in pairs

if $$d$$ divides $$n$$ then $$\dfrac{n}{d}$$ also divides n

thus you can generate pairs like $$(1,n),(2,\dfrac{n}{2}),(3,\dfrac{n}{3})..............(\sqrt{n},\dfrac{n}{\sqrt{n}})$$

thus n can have atmost $$\sqrt n$$ pairs or

$$2\sqrt n$$ divisors

- 1 year, 7 months ago

Comment deleted 5 months ago

How to use modular arithmetic here?

- 1 year, 7 months ago