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Abhishek Arora Grade: 11
        Prove that 33! is divisible by 2 raise to power 15.What is the largest integor n such that 33! is divisible by 2 raise to power n?
7 years ago

Answers : (1)

bibhash jha
15 Points

Dear Abhishek,


33! contains product of 1 through 33 , therefore 16 even numbers . Therefore , it must be divisible by 2^15 .

Now , 33! contains 2x1 , 2x2 , ....2x16 as even terms  . Out of these there are 2,2^2,2^3 ,2^4

therefore total 2's in 33! = 16+4=20

therefore 33! is divisible by 2^n for a maximum value of n=20


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7 years ago
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