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# 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 $$n$$

11 months, 1 week 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

- 11 months, 1 week ago

I had an idea about the proof somehow using modular arithmetic, lets see if we can come up with fruitful outcomes.

- 11 months, 1 week ago

How to use modular arithmetic here?

- 11 months, 1 week ago

We can write n in its prime factor form , similarly we can right $$\tau(n)$$ as (1+k1)... (1+kr) where k1 ...kr are powers of the prime factors,and then i just tried by establishing simple congruences and relate prime factor and its power but somehow i could not manage that

- 11 months, 1 week ago