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`        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

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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Bibhash
```
7 years ago
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