just find the exponent of 2 for this q as ketan has said. ..(since 2 is a prime).
Prove that 33! is divisible by 215. What is the largest integer n such that 33! is divisible by 2n?
Plz give detailed ans.
-
UP 0 DOWN 0 0 3
3 Answers
souradipta Sen
·2012-04-24 02:41:57
there are 33 numbers out of them every second number are divisible by 2
1,(2),3,(4),5,(6)........(32)
taking 2 common from all of them we get
33!=216K where k is an integer
33! is divisible by 215
now number of 2s the even numbers are divisible by starting from 2
1+2+1+3+1+2+1+4+1+2+1+3+1+2+1+5= 31
the maximum value of n is 31
Ketan Chandak
·2012-04-24 05:37:47
remember dis...
power of y in x! is [xy]+[xy2]+[xy3] ..........
where [] denotes G.I.F...
Sigma
·2012-04-24 06:47:10