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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, 1 month ago

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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. Achal Jain · 1 year ago

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???? Deepansh Jindal · 1 year ago

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