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Find the number of divisors of \(20!\) of the form \(4k+1\) (where \(k\) belongs to whole numbers ).

Note by Deepansh Jindal 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

\boxed{123}

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Hey Deep i think that i am very close . Check if it is right

The only numbers satisfying this form are

[5^{1,2,3,4}, 3^{2,4,6,8} , 7^{2}, 13,17}. Therefore u can figure from these numbers the possible combinations.

I have used the fact that (4m+1)(4n+1)= 4x+1 or the same form and that 4k+1 is odd.

plz do comment.

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Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

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)`

`> This is a quote`

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

Sort by:

TopNewestHey Deep i think that i am very close . Check if it is right

The only numbers satisfying this form are

[5^{1,2,3,4}, 3^{2,4,6,8} , 7^{2}, 13,17}. Therefore u can figure from these numbers the possible combinations.

I have used the fact that (4m+1)(4n+1)= 4x+1 or the same form and that 4k+1 is odd.

plz do comment.

Log in to reply

????

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