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

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

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

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

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TopNewest@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 · 6 months, 1 week ago

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I had an idea about the proof somehow using modular arithmetic, lets see if we can come up with fruitful outcomes. – Chinmay Sangawadekar · 6 months, 1 week ago

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– Harsh Shrivastava · 6 months, 1 week ago

How to use modular arithmetic here?Log in to reply

– Chinmay Sangawadekar · 6 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 thatLog in to reply

@Akshat Sharda @Harsh Shrivastava – Chinmay Sangawadekar · 6 months, 1 week ago

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