Waste less time on Facebook — follow Brilliant.
×

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 week, 5 days ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

@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 Anirudh Sreekumar · 1 week, 4 days ago

Log in to reply

I had an idea about the proof somehow using modular arithmetic, lets see if we can come up with fruitful outcomes. Chinmay Sangawadekar · 1 week, 5 days ago

Log in to reply

@Chinmay Sangawadekar How to use modular arithmetic here? Harsh Shrivastava · 1 week, 5 days ago

Log in to reply

@Harsh Shrivastava 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 Chinmay Sangawadekar · 1 week, 5 days ago

Log in to reply

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...