×

# 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 $$n$$

3 months, 2 weeks ago

Sort by:

@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 · 3 months, 2 weeks ago

I had an idea about the proof somehow using modular arithmetic, lets see if we can come up with fruitful outcomes. · 3 months, 2 weeks ago

How to use modular arithmetic here? · 3 months, 2 weeks 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 · 3 months, 2 weeks ago