# 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?

bibhash jha
15 Points
11 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

Now you can win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.

Bibhash

Ojash
11 Points
4 years ago
Attic dark in men men churn end JHVH he heaven did icy DVD ebbs JFK beg ebbs JFK men end mm BC JVC band he BBC I dud hen den end NFC be be be be be be Bibb CBBC BBC Venn end be BBC hfbdbbdndnf. F BBC BBC mm end henge Hugh end gush fund BBC
Nasheed Jafri
11 Points
4 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