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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?

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?

Grade:11

3 Answers

bibhash jha
15 Points
10 years ago

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

Ojash
11 Points
2 years ago
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Nasheed Jafri
11 Points
2 years ago
 
33! contains of product of 1 through 33.
Consider 33, it has zero number of 2 as factor.
Consider 32, it has five 2’s as factor.
Consider 31, it has zero 2 as factor.
Consider 30, it has one 2 as factor.
(Dont consider odd numbers now)
Consider 28, it has four 2’s.
26 has one 2.
24 has three 2’s.
22 has one.
20 has two.
18 has one.
16 has four.
14 has one.
12 has two.
10 has one.
8 has three.
6 has one.
4 has two.
2 has one.
 
So total number of twos
= 5 + 1 + 2 + 1 + 3 + 1 + 2 + 1 + 4 + 1 + 2 + 1 + 3 + 1 + 2 + 1
= 31

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