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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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2 \times 3

2^{34}

a_{i-1}

\frac{2}{3}

\sqrt{2}

\sum_{i=1}^3

\sin \theta

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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

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Comment deleted 5 months ago

How to use modular arithmetic here?

@Akshat Sharda @Harsh Shrivastava

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italics`**bold**`

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boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

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Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

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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

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Comment deleted 5 months ago

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How to use modular arithmetic here?

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@Akshat Sharda @Harsh Shrivastava

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