# The maximum value of n such that 2^n divides 21!

2
by ANEERUDH
how can 2^n divides 21!

2015-04-25T14:15:59+05:30

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In order to find the maximum value of such that divides , it is enough to find the exponent of in the prime factorization of :

That means is the maximum value of such that divides .
2015-04-25T15:27:06+05:30

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21 !  =  2 * 3 * 4 * 5 * ... 18 * 19 * 20 * 21

The factors of 21 !  which are multiples of 2 are :  (ignore odd integers)
= 2 * 4 * 6 * 8 ...  * 18 * 20
=  2¹°  *  ( 1 * 2 * 3 * .....  * 8 * 9 * 10 ),    now we ignore the odd integers.
= 2¹° * 2⁵ * ( 1 * 2 * 3 * 4 * 5 )      ,  ignore the odd integers
= 2¹° * 2⁵ * 2²  * ( 1 * 2 )
= 2¹°⁺⁵⁺²⁺¹
=2¹⁸
maximum value of n such that 2^n divides 21!  is  18

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Another way is:

The power of 2 in the value of  N! is given by the sum:

[ N / 2 ]  + [ N / 4 ]  + [ N / 8 ]  + ...  + 2  + 1
where  [ X ]   is the greatest integer function with value <= X.

= [21/2 ]  +  [ 21/4 ] + [ 21 / 8 ] + [21/16 ]
= 10 + 5 + 2 + 1
= 18

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